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Earth/Space LWP---01

Live/Work/Play in Earth Orbit 01

Bob & George • Bob Albrecht & George Firedrake • [email protected]

This eBook is published under a Creative Commons Attribution-NonCommercial-ShareAlike license. You may give all or part of this eBook free to anyone. You may use it in other ways described at http://creativecommons.org/licenses/by-nc-sa/4.0/.

Reality expands to fill the available fantasies. To expand your personal reality, fantasize, fantasize, and keep on fantasizing. – Bob & George

This eBook is for teachers, tutors, librarians, parents, siblings, grandparents, and others who help learners learn math and science. It is A-OK to copy stuff from this eBook, modify it, enhance it, and paste it into your documents that you give free to students – or to anyone.

IMAGINE/FANTASIZE Now: Live, work and play in Earth orbit in the International Space Station (ISS). Soon: Live, work and play in Earth orbit in resort hotels, research centers, factories, sports arena, junkyards and other space habitats. Far out: Millions of people live, work, and play in communities in Earth orbit. IMAGINE/FANTASIZE   Today and tomorrow and beyond: You and your students imagine living, working and playing in a resort hotel, research center, factory, sports arena, junkyard or other community in Earth orbit. We will search the Internet for information and inspiration, and contrive mathemagical alakazams related to living, working and playing in Earth orbit. We are not experts, so much of this eBook will consist of links to Internet sites where we and you and your students can learn about living, working and playing in Earth orbit.

Bob & George? Bob is a 90-year-old human (as of February 2020). George is a dragon. Read about Bob & George at Information Age Education (IAE): http://iae-pedia.org/Robert_Albrecht. Bob & George write math & science eBooks and post them on the Internet for free download and use as PDF files and Word files at http://i-a-e.org/downloads/cat_view/86-free-ebooks-by-bob-albrecht.html

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Table of Contents (TOC)

Today in Low-Earth Orbit (LEO): The International Space Station (ISS) Live, Work and Play in Earth Orbit: On the Moon Things to Come in (LEO) The High Frontier: 1976 and 2019 Appendix 01 Earth and Moon Data Appendix 02 Orbital Equations

DragonFun image by Marcie Hawthorne http://marciehawthorne.com/

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Today in Low Earth Orbit (LEO): International Space Station (ISS) | TOC

Today astronauts and scientists (the ISS crew) live, work and play in the International Space Station (ISS) in Low Earth Orbit (LEO).

• International Space Station https://www.nasa.gov/mission_pages/station/main/index.html • Current ISS Crew http://www.ariss.org/current-iss-crew.html • International Space Station https://en.wikipedia.org/wiki/International_Space_Station • Low Earth orbit (LEO) https://en.wikipedia.org/wiki/Low_Earth_orbit

Paraphrased from International Space Station https://en.wikipedia.org/wiki/International_Space_Station:

The International Space Station (ISS) is a habitable artificial in low Earth orbit (LEO). Its first component launched into orbit in 1998, and the ISS is now the largest human-made body in low Earth orbit and can often be seen with the naked eye from Earth.

Your students can spot the ISS and watch it as it traverses the sky above their  neighborhoods. Imagine living/working/playing aboard the ISS.  • Spot the ISS. Where? When? https://spotthestation.nasa.gov/

Table ISS-01 ISS size, mass, and pressure data from Wikipedia International Space Station https://en.wikipedia.org/wiki/International_Space_Station Data retrieved 2019-05-11. This data might be different today. Length: 72.8 m (meters) Width: 108.5 m (meters) Height: approximately 20 m (meters)

Mass: 419,725 kg (kilograms)

Pressurized volume (May 2016): 931.57 m3 (meters cubed – cubic meters) Atmospheric pressure: 1 atmosphere (1 Earth atmosphere at sea level)

ISS picture purloined from Wikipedia https://en.wikipedia.org/wiki/International_Space_Station Thank you, Wikipedia. You are always there when we need you.

Reality expands to fill the available fantasies. Your students can imagine living,  working and playing as astronauts/scientists aboard the ISS.  Spot the ISS https://spotthestation.nasa.gov/

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Table ISS-01 up yonder  presents data retrieved from Wikipedia, our favorite source of information. Table ISS-02 below  displays data we found at NASA’s official ISS site.

Table ISS-02 ISS facts and figures from NASA International Space Station facts and figures https://www.nasa.gov/feature/facts-and-figures Data retrieved 2019-05-11. It may change as ISS goes ‘round and ‘round Earth.

Data from NASA ISS Facts and Figures 2019-05-11 Data in metric units ☺ Pressurized module length: 240 feet 73.2 meters Truss length: 357. 5 feet 109 meters Solar array length: 239.4 feet 73.0 meters Mass: 925,335 pounds (See Note 1) 419,725 kilograms (See Note 1) Habitable volume: 13,696 cubic feet 388.83 cubic meters Pressurized volume: 32,333 cubic feet 915.57 cubic meters Power generation by solar arrays 75 to 90 kilowatts Note 1: Oops, pound is a unit of force, not mass, in archaic US customary units.  https://en.wikipedia.org/wiki/United_States_customary_units Mass is a unit of amount of matter in the International System of Units (SI). ☺ https://www.nist.gov/pml/weights-and-measures/si-units The ISS’s weight depends on its distance from Earth’s center: 925,335 pounds at Earth sea level. The ISS’s mass (419,725 kilograms) is the same at any distance from Earth’s center. Yeah! The unit of mass in US customary units is the slug https://en.wikipedia.org/wiki/Slug_(unit). The mass of the ISS in US customary units is approximately 28,760 slugs.

The Banana Slug is the mascot of the University of California, Santa Cruz. https://www.ucsc.edu/about/mascot.html Go, Banana Slug, Go!

The mass of the ISS is 419,725 kilograms (2019-05-11). The mass of the water in a 1-liter bottle of water is 1 kilogram. The mass of the ISS is equal to the mass of the water in 419,725 1-liter bottles of water.

• If you drink 3 liters of water a day (3 kilograms a day), how long will it take you to drink 419,725 kilograms (419,725 liters) of water in days? In weeks? In months? In years?

The pressurized volume of the ISS is approximately 916 cubic meters (2019-05-11). The volume of one liter of water is 1000 cubic centimeters, or 1/1000 cubic meter. 1000 liters of water has a volume of 1 cubic meter. The ISS’s 916 cubic meters of pressurized volume can hold 916 cubic meters of water, or 916,000 liters of water. The mass of water that would fill the ISS’s pressurized volume is 916,000 kilograms. Are all those numbers correct? Your students can check and say yea or nay.

Fantasize: Would a satellite full or partially full of water be a great aquatic environment for dolphins and bioengineered humans with gills? Imagine aquahumans and dolphins living, working and playing in an Earth-orbit habitat full or partially full of water. Supply water and life support from Earth? From the Moon? From the asteroid belt? Your students can investigate and fantasize.

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ISS, Snoopy and the Red Baron

The ISS’s solar array wingspan is 73 meters (2019-05-11). The wingspan of Snoopy’s Sopwith Camel is 8.53 meters.

How many Sopwith Camel wingspans placed end to end equals the wingspan of the ISS solar array?

Sopwith Camel picture filched from Wikipedia https://en.wikipedia.org/wiki/Sopwith_Camel Snoopy’s Sopwith Camel The wingspan of the Red Baron’s Fokker DR 1 triplane is 7.19 meters.

How many Fokker DR 1 wingspans placed end to end equals the wingspan of the ISS solar array?

Fokker DR 1 picture pirated from https://en.wikipedia.org/wiki/Fokker_Dr.I Red Baron’s Fokker Dr 1

The ISS’s solar arrays generate 75 to 90 kilowatts of electrical power (2019-05-11).

• How many 20-watt LED light bulbs will 75 kilowatts (kW) light up? • How many 20-watt LED light bulbs will 90 kW light up? • How many laptop computers will 75 kW power? How many will 90 kW power? • How many smart phones will 75 kW power? How many will 90 kW power?

The above comparisons are based on data posted by NASA 2019-05-11. As Earth and the ISS go ‘round and ‘round in their , the ISS may change in size, mass, electrical power and other ways. To view the latest data, visit https://www.nasa.gov/feature/facts-and-figures.

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ISS Orbital Alakazams

The International Space Station’s orbit is almost circular, slightly elliptical. The orbital plane is inclined 51.64 degrees to Earth’s equatorial plane.

Table ISS-03 ISS : 51.64 degrees from Earth’s Orbital inclination https://en.wikipedia.org/wiki/Orbital_inclination Earth observatory https://earthobservatory.nasa.gov/Features/OrbitsCatalog/ About the picture over yonder →

The horizontal line labeled ‘Earth’s rotation’ is Earth’s equator. The rotation is from west to east as shown by the arrow (→).

The line labeled ‘satellite orbit’ is the path of a satellite whose orbit is inclined to Earth’s equator. The satellite’s direction of motion is shown by the arrow ().

Picture heisted from Earth observatory Catalog of Earth Satellite Orbits https://earthobservatory.nasa.gov/Features/OrbitsCatalog/ Great place to learn about low Earth orbit (LEO), mid Earth orbit (MEO), high Earth orbit (HEO), Global Positioning System (GPS) orbits and geostationary orbits.

As the ISS wends its way around Earth at an orbital inclination of 51.64 degrees from Earth’s equator, Earth rotates from west to east, counterclockwise (↺) as seen from above the . At various points in its orbit, the ISS can be seen by keen-eyed observers almost anywhere on Earth.

YOU and YOUR STUDENTS can spot the ISS as a bright dot crossing the sky in your neighborhoods.

• Spot the ISS. When? Where? Go to https://spotthestation.nasa.gov

Orbital info from Earth Observatory https://earthobservatory.nasa.gov/Features/OrbitsCatalog/ :

• Low Earth orbit (LEO): 180 to 2000 km (kilometers) above Earth’s surface. • Mid Earth orbit (MEO): 2000 to 35,780 km above Earth’s surface. Global Positioning Satellite (GPS) orbits are in MEO at 20,180 km above Earth’s surface. Global positioning system https://en.wikipedia.org/wiki/Global_Positioning_System • High Earth (HEO) and : 35,780 km above Earth’s surface. A geostationary orbit is also known as a . Geostationary orbit https://en.wikipedia.org/wiki/Geostationary_orbit

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ISS Images and Data from Heavens Above

While orbiting the Internet looking for information about the International Space Station, we found ISS Orbit at Heavens Above (https://www.heavens-above.com/orbit.aspx?satid=25544 ). Wow, great site!

We recommend: You and your students visit Heavens Above. Images and data are updated frequently, so what you see may be different than what we saw 2019-05-11, shown below.

ISS orbit at Heavens Above https://www.heavens-above.com/orbit.aspx?satid=25544 We added comments below the images. Any misteaks are ours.

View from above the orbital plane 2019-05-11 View from above the ISS 2019-05-11 The orbit is almost circular with perigee 408 km This is an ‘edgewise’ view of the ISS’s orbit from a above Earth’s surface and apogee 410 km above point in the plane of the orbit above the orbit. Earth’s surface. The ISS makes 15.5 orbits per The orbital plane is inclined 51.64 degrees to day. Earth’s equator.

Geometry teacher: This view might be on the Geometry teacher: The plane of the orbit is axis of revolution of the ISS, a ray with endpoint inclined 51.64 degrees to the plane of Earth’s at Earth’s center perpendicular to the plane of equator. See Heavens Above to view the path of revolution. the ISS as seen on Earth.

Orbital data from Heavens Above 2019-05-11 Internet information about orbital data eccentricity: 0.0001262 (See Note 1 ) https://en.wikipedia.org/wiki/Orbital_eccentricity inclination: 51.6404 degrees https://en.wikipedia.org/wiki/Orbital_inclination perigee height above Earth’s surface: 408 km https://en.wikipedia.org/wiki/Apsis apogee height above Earth’s surface: 410 km apsis https://en.wikipedia.org/wiki/Apsis revolutions per day: 15.52655989 Approximately 15.5 revolutions per day

Note 1. The orbit is nearly circular, eccentricity = 0.0001262. The eccentricity of a is 0. Explore at https://en.wikipedia.org/wiki/Orbital_eccentricity.

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A satellite in LEO (Low Earth Orbit) experiences orbital decay caused by atmospheric drag. The satellite’s height above Earth’s surface decreases. This is not good!

• Orbital decay https://en.wikipedia.org/wiki/Orbital_decay

The atmosphere at the ISS’s orbit is very thin, yet causes slow orbital decay, reducing the height of the ISS above Earth’s surface. If orbital decay continues, the ISS’s height will become less and less (lower and lower) until the ISS enters denser atmosphere and – alas – burns up.

Reboost to the rescue! The ISS is periodically reboosted to keep it at a desired height.

• How does the International Space Station maintain its orbit? https://www.quora.com/How-does-the-International-Space-Station-maintain-its-orbit-and-what- propellant-does-it-use

A most excellent graph showing recent ISS orbital decays and reboosts resides at

• Height of the ISS – Heavens-Above https://heavens-above.com/IssHeight.aspx

On 2019-05-11, this graph displayed the ISS’s mean height above Earth’s surface from May 2018 to May 2019. The graph showed orbital decay and reboosts.

• ISS mean height above Earth’s surface on 2019-05-11 = 409 km (kilometers)

Earth is almost, but not quite, a sphere. She is bit fatter at her equator than at her poles. At NASA’s Earth Fact Sheet http://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html), we found three measures of Earth’s radius.

• Equatorial radius = 6378.137 km. [For some calculations, we will round to 6378 km.] • Polar radius = 6356.752 km [For some calculations, we will round to 6357 km] • Volumetric mean radius = 6371.000 km [For some calculations, we will round to 6371 km]. Volumetric mean radius = radius of a sphere that has a volume equal to Earth’s volume

Which radius – equatorial, polar or volumetric mean – is the right stuff for calculating the distance of the ISS from Earth’s center? We choose the volumetric mean radius: 6371 km.

• 2019-05-11: mean distance of ISS from Earth’s center = 6371 km + 409 km = 6780 km • 2019-05-11: distance of ISS perigee from Earth’s center = 6371 km + 408 km = 6779 km • 2019-05-11: distance of ISS apogee from Earth’s center = 6371 km + 410 km = 6781 km

Height of the ISS – Heavens-Above https://heavens-above.com/IssHeight.aspx  is updated frequently. If your students visit Height of the ISS – Heavens-Above,  they will see an updated graph of orbital decay and reboosts

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We used data from Heavens Above May 11, 2019 to draw a picture of Earth and the almost circular orbit of the ISS. The picture is almost, but not exactly, to scale.

In the picture over yonder →

C is Earth center. P is ISS perigee, 408 km above Earth’s surface [2019-05-11]. A is ISS apogee, 410 km above Earth’s surface [2019-05-11]. P A Perigee (P) is 6779 km above Earth center (C) [2019-05-11]. C Apogee (A) is 6781 km above Earth center (C) [2019-05-11].

The orbit is an ellipse. The distance from perigee (P) to apogee (A) is the major axis of the ellipse. The semimajor axis of the ellipse is half the major axis. Ellipse https://en.wikipedia.org/wiki/Ellipse C = Earth center P = ISS perigee Major axis = 6779 km + 6781 km = 13,560 km (kilometers) A = ISS apogee Major axis = 13,560,000 m (meters)

Semimajor axis = (13,560 km) / 2 = 6780 km (kilometers) Semimajor axis = 6,780,000 m (meters)

Let’s visit Heavens Above again. https://www.heavens-above.com/orbit.aspx?satid=25544

View from above the orbital plane 2019-05-11 View from above the ISS 2019-05-11 Perigee: 6779 km above Earth’s center The orbital plane is inclined 51.64 degrees to Apogee: 6781 km above Earth’s center. Earth’s equator. Major axis = 6779 km + 6781 km = 13,560 km Semimajor axis = 13,560 km / 2 = 6780 km Semimajor axis = 6,780,000 m

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The T of an Earth satellite is the time it takes the satellite to make one 3 a orbit. Use the equation over yonder → to calculate the ISS’s orbital period in seconds. T = 2 Orbital period https://en.wikipedia.org/wiki/Orbital_period GM

T = orbital period in seconds (s) a = semimajor axis in meters 2019-05-11 = 6,780,000 m G = gravitational constant = 6.6743  10−11 m3/kg·s2 (See Note ) M = mass of Earth = 5.9723  1024 kg

 This equation is A-OK for calculating the period of a satellite whose mass is very small compared to the mass of the object about which it orbits. The mass of the ISS (419,725 kg) is very much smaller than the mass of Earth (5.9723  1024 kg).

Note. We show G in units m3/kg·s2. G may be shown elsewhere in units m3·kg−1·s−2 or Nm2/kg2.

Calculate T: Substitute the values of the semimajor axis a, gravitational constant G, and Earth’s mass M into the orbital period equation and crunch the numbers. Round the answer to four significant digits.

(6,780,000 m)3 T ==2 5556 s (6.6743 10−11 m 3 /kg  s 2 )(5.9723  10 24 kg)

We used a TI-84 graphing calculator to crunch the numbers.

TI-84: 2π√(6780000^3 / (6.6743E-11 * 5.9723E24)) Enter Answer: 5555.852916 [Round to 5556]

Rounded to 4 significant digits, the ISS’s orbital period on 2019-05-11 was 5556 s. Divide 5556 s by 60 s/min to get the orbital period in minutes.

T = (5556 s) / (60 s/min) = 92.60 min

The semimajor axis of the ISS’s orbit (6,780,000 m) has 4 significant digits. Gravitational constant G = 6.6743  10−11 m3/kg·s2 (5 significant digits) is from National Institute of Science and Technology Fundamental Physical Constants. https://physics.nist.gov/cgi-bin/cuu/Value?bg|search_for=universal_in Earth’s mass M = 5.9723  1024 kg (5 significant digits)is from NASA’s Earth Fact  Sheet https://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html  We rounded the ISS’s orbital period to 4 significant digits: 5556 s. • Rounding and Significant Digits https://www.purplemath.com/modules/rounding2.htm • Significant Figures – Purdue University http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch1/sigfigs.html

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The orbital velocity v of an Earth satellite in a circular orbit is the speed the satellite GM travels everywhere in its orbit. Use the equation over yonder → to calculate the orbital v = velocity in meters per second (m/s) of a satellite in a circular orbit at the ISS’s mean r height on 2019-05-11. https://en.wikipedia.org/wiki/Orbital_speed.

v = orbital velocity in m/s (meters per second) G = gravitational constant = 6.6743  10−11 m3/kg·s2 (See Note ) M = mass of Earth = 5.9723  1024 kg r = distance of the satellite from Earth’s center in kilometers (km) and meters (m) ISS perigee on 2019-05-11: r = 6779 km = 6,779,000 m ISS apogee on 2019-05-11: r = 6781 km = 6,781,000 m ISS mean height on 2019-05-11: r = 6780 km = 6,780,000 m

 This equation is A-OK for calculating the velocity of a satellite whose mass is small compared to the mass of the object about which it orbits. The mass of the ISS (419,725 kg) is much smaller than the mass of the Earth (5.972  1024 kg).

Note. We show G in units m3/kg·s2. G may be shown elsewhere in units m3·kg−1·s−2 or Nm2/kg2.

Calculate: Substitute values of the gravitational constant G, Earth’s mass M, and distance of a satellite in a circular orbit at ISS’s mean height from Earth’s center into the equation and crunch the numbers.

(6.6743 10−11 m 3 /kg  s 2 )(5.9723  10 24 kg) v ==7668 m/s 6,780,000 m

TI-84: √(6.6743E-11 * 5.9723E24 / 6780000) Enter Answer: 7667.588943 [Round to 7668]

Wow! A satellite in a circular orbit 6780 km (6,780,000 m) above Earth’s center is traveling 7668 m/s (meters per second) everywhere in its orbit. Convert that velocity to km/h (kilometers per hour).

Convert orbital velocity in meters per second to kilometers per hour:

7668 m 3600 s 1 km 27,600 km   = 1 s 1 h 1000 m 1 h

TI-84: 7668 * 3600 / 1000 Enter Answer: 27604.8 [Round to 27600]

Zowie! A satellite in a circular orbit 6780 kilometers (6,780,000 meters) above Earth’s center is traveling 27,600 km/h (kilometers per hour) everywhere in its orbit.

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A satellite in a circular orbit 409 km above Earth’s surface (6780 km = 6,780,000 m above Earth’s center) is zooming along at constant orbital velocity 7668 m/s (27,600 km/h).

• Circular orbit 6780 km (6,780,000 m) above Earth’s center: orbital velocity = 7668 m/s (27,600 km/h)

The orbit of the ISS is an ellipse that is almost – but not quite – a circle. The ISS’s orbital velocity is greatest at perigee and least at apogee.

Alakazam! You can use the equation over yonder → to calculate the orbital velocity v of a satellite in an elliptical orbit at any point in its orbit. 21 v=− GM  ra v = orbital velocity in meters per second (m/s)

G = gravitational constant = 6.6743  10−11 m3/kg·s2 (See Note ) Orbital velocity at M = mass of Earth = 5.9723  1024 kg an orbital point r = distance of the satellite from Earth’s center in meters (m) distance r from a = semimajor axis of the elliptical orbit in meters (m) Earth’s center.

 This equation is A-OK for calculating the velocity of a satellite whose mass is small compared to the mass of the object about which it orbits. The mass of the ISS (419,725 kg) is much smaller than mass of the Earth (5.9723  1024 kg).

Note. We show G in units m3/kg·s2. G may be shown elsewhere in units m3·kg−1·s−2 or Nm2/kg2.

Calculate the orbital velocity v of the ISS at perigee on 2019-05-11. r = 6779 km = 6,779,000 m.

−11 3 2 24 21 v =(6.6743  10 m /kg  s )(5.9723  10 kg) − = 7669 m/s 6,779,000m 6,780,000m

TI-84: √(6.6743E-11 * 5.9723E24 (2/6779000 − 1/6780000)) Enter Ans: 7668.719939 [Round to 7669]

Calculate the orbital velocity v of the ISS at apogee on 2019-05-11. r = 6781 km = 6,781,000 m.

−11 3 2 24 21 v =(6.6743  10 m /kg  s )(5.9723  10 kg) − = 7666 m/s 6,781,000m 6,780,000m

TI-84: √(6.6743E−11 * 5.9723E24 (2/6781000 − 1/6780000)) Enter Ans: 7666.458113 [Round to 7666]

• ISS perigee 6779 km = 6,779,000 m above Earth’s center; orbital velocity = 7669 m/s • ISS apogee 6781 km = 6,781,000 m above Earth’s center; orbital velocity = 7666 m/s

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A month has passed since we collected ISS data at Heavens Above. Since then, the ISS has experienced a bit of orbital decay and a reboost. We return to Heavens Above 2019-06-11.

Table ISS-04 Heavens Above https://www.heavens-above.com/orbit.aspx?satid=25544. . ISS orbital data on 2019-05-11 ISS orbital data on 2019-06-11 Eccentricity: 0.0001262 Eccentricity: 0.0007752 Inclination: 51.6404 degrees Inclination: 51.6435 degrees Perigee height above Earth’s surface: 408 km Perigee height above Earth’s surface: 408 km Apogee height above Earth’s surface: 410 km Apogee height above Earth’s surface: 418 km Revolutions per day: 15.52655989 Revolutions per day: 15.51152714

Calculations by Bob & George. Errrors are ours. Calculations by Bob & George. Errrors are ours. Perigee height above Earth’s center: 6779 km Perigee height above Earth’s center: 6779 km Perigee height above Earth’s center: 6,779,000 m Perigee height above Earth’s center: 6,779,000 m Apogee height above Earth’s center: 6781 km Apogee height above Earth’s center: 6789 km Apogee height above Earth’s center: 6,781,000 m Apogee height above Earth’s center: 6,789,000 m Major axis: 6779 km + 6781 km = 13,560 km Major axis: 6779 km + 6789 km = 13,568 km Major axis: 13,560,000 m Major axis: 13,568,000 m Semimajor axis: 13,560 km / 2 = 6780 km Semimajor axis: 13,568 km / 2 = 6784 km Semimajor axis: 13,560,000 m / 2 = 6,780,000 m Semimajor axis = 13,568,000 m / 2 = 6,784,000 m Orbital period: 5556 s Orbital period: 5561 s Orbital period: 92.60 minutes Orbital Period: 92.68 minutes Orbital velocity at perigee = 7669 m/s Orbital velocity at perigee = 7671 m/s Orbital velocity at apogee = 7666 m/s Orbital velocity at apogee = 7660 m/s

Orbital period equation → T = orbital period in seconds (s) a = semimajor axis of orbit in meters (m) a 3 −11 3 2 T = 2 G = gravitational constant = 6.6743  10 m /kg·s GM M = mass of Earth = 5.9723  1024 kg

Orbital velocity equation → v = orbital velocity in meters per second (m/s)  −11 3 2 G = gravitational constant = 6.6743 10 m /kg·s 21 M = mass of Earth = 5.9723  1024 kg v=− GM  r = distance of satellite from Earth’s center in meters (m) ra a = semimajor axis of elliptical orbit in meters (m)

Down yonder  are our calculations of the ISS’s orbital period, orbital velocity at perigee, and orbital velocity at apogee on 2019-05-11 and 2019-06-11.

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ISS orbital period and orbital velocity at perigee and apogee 2019-05-11 and 2019-06-11

(6,780,000 m)3 Orbital period 2019-05-11: T ==2 5556 s (6.6743 10−11 m 3 /kg  s 2 )(5.9723  10 24 kg)

TI-84: 2π√(6780000^3 / (6.6743E-11 * 5.9723E24)) Enter Answer: 5555.852916 [Round to 5556]

(6,784,000 m)3 Orbital period 2019-06-11: T ==2 5561 s (6.6743 10−11 m 3 /kg  s 2 )(5.9723  10 24 kg)

TI-84: 2π√(6784000^3 / ( 6.6743E-11 * 5.9723E24)) Enter Answer: 5560.770325 [Round to 5561]

Orbital velocity at perigee 2019-05-11:

−11 3 2 24 21 v =(6.6743  10 m /kg  s )(5.9723  10 kg) − = 7669 m/s 6,779,000 m 6,780,000 m

TI-84: √(6.6743E-11 * 5.9723E24 (2/6779000 − 1/6780000)) Enter Ans: 7668.719939 [Round to 7669]

Orbital velocity at apogee 2019-05-11:

−11 3 2 24 21 v =(6.6743  10 m /kg  s )(5.9723  10 kg) − = 7666 m/s 6,781,000 m 6,780,000 m

TI-84: √(6.6743E-11 * 5.9723E24 (2/6781000 − 1/6780000)) Enter Ans: 7666.458113 [Round to 7666]

Orbital velocity at perigee 2019-06-11:

−11 3 2 24 21 v =(6.6743  10 m /kg  s )(5.9723  10 kg) − = 7671 m/s 6,779,000 m 6,784,000 m

TI-84: √(6.6743E-11 * 5.9723E24 (2/6779000 − 1/6784000)) Enter Ans: 7670.979764 [Round to 7671]

Orbital velocity at apogee 2019-06-11:

−11 3 2 24 21 v =(6.6743  10 m /kg  s )(5.9723  10 kg) − = 7660 m/s 6,789,000 m 6,784,000 m

TI-84: √(6.6743E-11 * 5.9723E24 (2/6789000 − 1/6784000)) Enter Ans: 7659.680634 [Round to 7660]

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We just remembered that it is good to repeat information and to present previous information in a different way. Table ISS-05 repeats ISS orbital data and calculations 2019-05-11 and 2019-06-11. Orbits were nearly circular – eccentricity 0.0001262 on 2019-05-11 and 0.0007752 on 2019-06-11

• The eccentricity of an elliptical orbit is a number greater than or equal to 0 and less than 1. 0  eccentricity < 1. The eccentricity of a circular orbit is 0. • Orbital eccentricity https://en.wikipedia.org/wiki/Orbital_eccentricity

Table ISS-05 ISS orbital data and calculations 2019-05-11 & 2019-06-11 Data from Heavens Above https://www.heavens-above.com/orbit.aspx?satid=25544

Orbital alakazam 2019-05-11 2019-06-11 Notes Eccentricity 0.0001262 0.0007752 Nearly circular Perigee Closest to Earth’s center Ht. above Earth’s surface 408 km 408 km 408,000 m 408,000 m Ht. above Earth’s center 6779 km 6779 km 6,779,000 m 6,779,000 m Apogee Farthest from Earth’s center Ht. above Earth’s surface 410 km 418 km 410,000 m 418,000 m Ht. above Earth’s center 6781 km 6789 km 6,781,000 m 6,789,000 m Major axis 13,560 km 13,568 km Distance between perigee and apogee 13,560,000 m 13,568,000 m Semimajor axis 6780 km 6784 km Major axis / 2 6,780,000 m 6,784,000 m Orbital period 5556 s 5561 s Time required for one orbit. 92.60 min 92.68 min Orbital velocity at perigee 7669 m/s 7671 m/s Greatest orbital velocity 27,610 km/h 27,620 km/h Orbital velocity at apogee 7666 m/s 7660 m/s Least orbital velocity 27,600 km/h 27,580 km/h

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International Space Station Tourism

Tourists visited the International Space Station in 2001, 2002, 2005, 2006 and 2007. They were transported to the ISS by the Russian Soyuz spacecraft.

• How Space Tourism Works https://science.howstuffworks.com/space-tourism.htm • Soyuz spacecraft https://en.wikipedia.org/wiki/Soyuz_(spacecraft)

More tourists beginning 2020? [Data retrieved 2019-08-15. May be different today.]

NASA to open International Space Station to tourists from 2020 https://phys.org/news/2019-06-nasa-international-space-station-tourists.html

• Price tag? Tens of millions of dollars for a round trip ticket and $35,000 a night. • Travelers would be ferried to the orbiter exclusively by the two US companies currently developing transport vehicles for NASA: SpaceX, with its Crew Dragon capsule, and Boeing, which is building a spacecraft called Starliner.

Space-X’s Dragon is the first private spacecraft to the ISS.

• Space X | Dragon https://www.spacex.com/dragon

Under construction (2019-10-26): the Boeing Starliner.

• Boeing CST-100 Starliner https://en.wikipedia.org/wiki/Boeing_CST-100_Starliner

Your students can find more information about visiting the ISS by using search keys such as

• NASA International Space Station tourism • Visit the International Space Station

Imagine! Fantasize! The International Space Station Lottery. Ticket price: $1 Who may win the lottery? A high school student.   When enough money has been collected (several 10s of millions of dollars), a name is drawn. If the winner is a physically fit high school student and parents allow, she or he visits the ISS. Reality expands to fill the available fantasies.

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Live, Work and Play in Earth Orbit: On the Moon | TOC

You and your students can cruise the Internet and learn about Moon bases, Moon settlements, Moon habitats and Moon villages. When you and they have done so, they and you will know as much as we know about living, working and playing in Earth orbit on the Moon.

Image of the full moon is from http://www.clker.com/. They say:

• Free clip art you can use for anything you like

This full moon image is at http://www.clker.com/clipart-6955.html.

For many more moon images from www.clker.com, go to

• http://www.clker.com/search/moon/1

National Geographic Earth’s Moon Wall Map is available from www.amazon.com. It will look great on your classroom wall while you and your students fantasize living, working and playing on the Moon.

• Shows near side and far side of the Moon • Size: 108 cm  72 cm (42.5  28.5 inches)

At www.amazon.com, search for ‘moon map’.

Replogle Lunar Wonder Globe is a small Moon globe (11-cm diameter) available from www.amazon.com. Search for ‘moon globe’.

• Shows craters, ‘seas’, rilles, valleys and mountain ranges. • Diameter: 11 cm (4.3 in). Small: names of lunar features are in small type – Bob uses a 3X magnifying glass to read them. • You can find larger, more expensive Moon globes at Amazon and elsewhere online.

Teams of students can locate sites for their Moon bases on this globe. What might be a good name for a team researching living, working and playing on the Moon? Moonauts? Lunanauts? Loonies? Lunatics?

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About living on the Moon: Internet sites and a science fiction novel

• What Would It Be Like to Live on the Moon? (2015) https://www.space.com/28494-how-to-live-on-the-moon.html

• Living on the Moon: What Would It Be Like? (2015) https://www.space.com/27203-living-on-the-moon-explained-infographic.html

• Home on the Moon: How to Build a Lunar Colony (2013) https://www.space.com/21588-how-moon-base-lunar-colony-works-infographic.html

• Novel: Red Moon by Kim Stanley Robinson. ISBN-10: 031626237. ISBN-13: 978-0316262378. 2018. What it might be like on the Moon in 2047. We read the Kindle edition and found a cornucopia of ideas that we can use in writing eBooks about living, working and playing on the Moon. Read a review of Red Moon at The Space Review http://www.thespacereview.com/article/3627/1.

When? Where on the Moon? China, the European Space Agency and others are thinking about a Moon base in the next 10 to 20 years (2029 – 2039).

• Colonization of the Moon https://en.wikipedia.org/wiki/Colonization_of_the_Moon Updated April 2019.

• China Plans to Build a Moon Base Near the Lunar Pole (2019) https://www.space.com/china-moon-base-10-years.html

• Lunar Leap: is Reaching for a Moon Base by the 2030s (2015) https://www.space.com/31488-european-moon-base-2030s.html

• Moon Village Association (2019) https://moonvillageassociation.org/

Internet searches using search keys such as ‘moon base’, ‘China moon base’, ‘Europe moon base’ and ‘ moon base’ will yield more sites. What about a US moon base? Browse Project Artemis:

• What is Artemis https://www.nasa.gov/what-is-artemis • Artemis Program https://www.nasa.gov/artemis • NASA Moon and Mars https://www.nasa.gov/specials/moon2mars/

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Building a Moon Base based on research by Haym Benaroya and Leonhard Bernold

Book: Building Habitats on the Moon: Engineering Approaches to Lunar Settlements by Haym Benaroya. ISBN-10: 3319682423, ISBN-13: 978-3319682426. 2018. Read a review of this book at The Space Review http://www.thespacereview.com/article/3437/1 .

Internet:

Building a Moon base: • Building a Moon Base: Part 1 – Challenges and Hazards (2019) https://www.universetoday.com/12726/building-a-base-on-the-moon-challenges-and-hazards/ • Building a Moon Base: Part 2 – Habitat Concepts [Contains gun-related ads] https://www.universetoday.com/12758/building-a-base-on-the-moon-part-2-habitat-concepts/ • Building a Moon Base: Part 3 – Structural Design https://www.universetoday.com/12864/building-a-base-on-the-moon-part-3-structural-design/ • Building a Moon Base: Part 4 – Infrastructure and Transportation https://www.universetoday.com/13216/building-a-base-on-the-moon-part-4-infrastructure-and- transportation/

National Space Society: Lunar Bases and Settlements https://space.nss.org/lunar-bases-and-settlements/

Moon Village Association https://moonvillageassociation.org/

Lunar Outpost (NASA) https://en.wikipedia.org/wiki/Lunar_outpost_(NASA)

China plans to build a Moon base https://www.livescience.com/65312-china-moon-base-10-years.html

Europe Aiming for International ‘Moon Village’ https://www.space.com/32695-moon-colony-european- space-agency.html

China and Europe May Build a ‘Moon Village’ in the 2020s https://futurism.com/china-and-europe-may-build-a-moon-village-in-the-2020s/

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Moon data and orbital alakazams

Tables Moon-01 and Moon-02 display data from NASA’s Moon Fact Sheet and Earth Fact Sheet.

Table Moon-01 Moon and Earth data from NASA’s Fact Sheets (2020-01-01) Moon Fact Sheet – NASA https://nssdc.gsfc.nasa.gov/planetary/factsheet/moonfact.html Earth Fact Sheet – NASA https://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html Ratio Data Moon Earth Notes Moon/Earth Mass 0.07346  1024 kg 5.9723  1024 kg 0.0123 See Notes 1 & 2. Volume 2.1968  1010 km3 108.321  1010 km3 0.0203 Equatorial radius 1738.1 km 6378.137 km 0.2725 Polar radius 1736.0 km 6356.752 km 0.2731 Volumetric mean radius 1737.4 km 6371.000 km 0.2727 See Note 3. Surface gravity 1.62 m/s2 9.80 m/s2 0.165 2.38 km/s 11.2 km/s 0.213 See Note 4.

Note 1. Moon’s mass 0.07346  1024 kg has 4 significant digits. We will frequently round calculations using Moon’s mass to 4 significant digits. Rounding and Significant Digits https://www.purplemath.com/modules/rounding2.htm

Note 2. Earth’s mass 5.9723  1024 kg has 5 significant digits. We will frequently round calculations using Earth’s mass to 5 significant digits. Significant Figures – Purdue University http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch1/sigfigs.html

Note 3. Calculate the volume of spheres with volumetric mean radii 1737.4 km (Moon) and 6371.000 km (Earth) and compare with the volumes of the Moon (2.1968  1010 km3) and Earth (108.321  1010 km3). • Moon: V = (4/3) π r3 = (4/3) π (1737.4 km)3 = 2.1968  1010 km3 [Rounded to 5 digits] • Earth: V = (4/3) π r3 = (4/3) π (6371.000 km)3 = 108.321  1010 km3 [Rounded to 6 digits]

Note 4. Escape velocity vesc is the velocity an orbital object must attain to escape from Moon or Earth.

2GM 2(6.6743 10−11 m 3 /kg  s 2 )(0.07346  10 24 kg) Moon: v = = = 2376 m/s [ 2.38 km/s] esc r 1,737,400 m

TI-84: √(2 * 6.6743E-11 * 0.07346E24 / 1737400) Enter Answer: 2375.709931 [Round to 2376]

2GM 2(6.6743 10−11 m 3 /kg  s 2 )(5.9723  10 24 kg) Earth: v = = = 11,186 m/s [ 11.2 km/s] esc r 6,371,000 m

TI-84: √(2 * 6.6743E-11 * 5.9723E24 / 6371000) Enter Answer: 11186.25885 [Round to 11186]

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Table Moon-02 displays orbital data retrieved from NASA’s Moon Fact sheet in 1) units shown in the fact sheet and 2) units we will use to calculate orbital alakazams such as orbital period and orbital velocity.

Table Moon-02 Moon orbital data from NASA’s Moon Fact Sheet Moon Fact Sheet https://nssdc.gsfc.nasa.gov/planetary/factsheet/moonfact.html Moon Fact Sheet Units to calculate Orbital data (Moon) Notes units orbital alakazams Mean distance from Earth’s 0.3844106 km 0.3844109 m Semimajor axis center to Moon’s center 384,400 km 384,400,000 m 4 significant digits 0.3633106 km 0.3633109 m Point in orbit closest to Earth Perigee 363,300 km 363,300,000 m 4 significant digits 0.4055106 km 0.4055109 m Point in orbit farthest from Earth Apogee 405,500 km 405,500,000 m 4 significant digits See Note 1, Note 2 & Note 4. Revolution period 27.3217 days 2,360,590 s 6 significant digits See Note 3 and Note 5. Synodic period 29.53 days 2,551,000 s 4 significant digits See Note 6. Sidereal rotation period 655.728 h 2,360,620 s 6 significant digits From Notes on the Fact Sheets https://nssdc.gsfc.nasa.gov/planetary/factsheet/fact_notes.html

Note 1. Day = mean time from noon at a point on Earth’s equator to noon at that point the next day. 24 hours. This is a solar day. Solar time https://en.wikipedia.org/wiki/Solar_time

Note 2. Time for the Moon to make one orbit. Sidereal orbit period. https://en.wikipedia.org/wiki/Orbit_of_the_Moon Sidereal time https://en.wikipedia.org/wiki/Sidereal_time

Note 3. Time from a Moon phase to a recurrence of that Moon phase. Example: From new moon to the next new moon. Example: From full moon to the next full moon. Lunar Sidereal and Synodic Periods https://community.dur.ac.uk/john.lucey/users/lunar_sid_syn.html

Note 4. 27.3217 days  86,400 s/day = 2,360,594.88 s Round to 2,360,590 s (6 significant digits) 27.3217 has 6 significant digits. 86,400 seconds per solar day (24 hours) is exact.

Note 5. 29.53 days  2, 86,400 s/day = 2,551,392 s Round to 2,551,000 s (4 significant digits) 29.53 has 4 significant digits. 86,400 seconds per solar day (24 hours) is exact.

Note 6. Time for the Moon to rotate 360 degrees about its axis of rotation (line through its poles). 655.728 h  3600 s/h = 2,360,620 s [6 significant digits] 655.728 has 6 significant digits. 3600 seconds per hour is exact.

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In Table Moon-02, Moon’s sidereal orbit period and sidereal rotation period are approximately equal.

• Sidereal orbit period: 2,360,590 seconds [6 significant digits] 2.360,590 s  2,360,620 s • Sidereal rotation period: 2,360,620 seconds [6 significant digits]

In fact, Moon’s sidereal orbit period and sidereal rotation period are equal. As Moon orbits Earth, it rotates so that the same side of the Moon always faces Earth. This side is the near side. The other side, the far side, always faces away from Earth. Read about near side and far side:

• Near side of Moon https://moon.nasa.gov/resources/77/the-near-side-of-the-moon/ Great picture! • Near side of Moon https://en.wikipedia.org/wiki/Near_side_of_the_Moon • Far side of Moon https://moon.nasa.gov/resources/59/far-side-of-the-moon/ Great picture! • Far side of Moon https://en.wikipedia.org/wiki/Far_side_of_the_Moon

National Geographic Earth’s Moon Wall Map is available from www.amazon.com. It will look great on your classroom wall while you and your students investigate living, working and playing on the Moon.

• Shows near side and far side of the Moon • Size: 108 cm  72 cm (42.5  28.5 inches)

At www.amazon.com, search for ‘moon map’.

Far side faces away Near side faces Earth from Earth

Replogle Lunar Wonder Globe is a small Moon globe (11-cm diameter) available from www.amazon.com. Search for ‘moon globe’.

• Shows craters, ‘seas’, rilles, valleys and mountain ranges. • Diameter: 11 cm (4.3 in). Small: names of lunar features are in small type – Bob uses a 3X magnifying glass to read them. • You can find larger, more expensive Moon globes at Amazon and elsewhere online.

Teams of students can locate sites for their Moon bases on this globe. What might be a good name for a team researching living, working and playing on the Moon? Moonauts? Lunanauts? Loonies? Lunatics?

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Moon’s synodic orbit period is 29.53 solar days. It is the length of time between two successive occurrences of a Moon phase at a point on Earth. For example, the length of time between two successive new moons or between two successive full moons. Also called synodic month.

• Synodic orbit period = 29.53 d  86,400 s/d = 2,551,000 s [rounded to 4 significant digits] • Lunar month, lunation; synodic month https://www.timeanddate.com/astronomy/moon/lunar-month.html

Imagine. Fantasize. You are aboard an Enterprise-class starship in orbit around Polaris, the North Star. Polaris is 433 light years (4.10  1015 kilometers) above Earth’s North Pole.

• Polaris https://en.wikipedia.org/wiki/Polaris

Earth’s axis of rotation is a line that passes through Earth’s north and south poles. To Polaris This line passes close to Polaris. Your orbit around Polaris is a splendid place for observing Earth’s orbit around the Sun and Moon’s orbit around Earth. N

Using amazingly farseeing technology (fantasize!), you observe Earth’s orbit around the Sun and Moon’s orbit around Earth. As Moon orbits Earth, Earthlings see it go through Moon phases from New Moon to New Moon or from Full Moon to Full Moon or from any Moon phase until that phase occurs again. The length of time from a Moon phase until that phase occurs again is Moon’s synodic orbit S period, approximately 29.53 solar days (solar day = 24 hours = 86,400 seconds).

At Phases of the Moon https://simple.wikipedia.org/wiki/Phases_of_the_Moon, there is a most excellent diagram and description of the phases of the Moon as seen from Earth.

Wikipedia Phases of the Moon https://simple.wikipedia.org/wiki/Phases_of_the_Moon

Diagram of the Moon's phases: The Earth is at the center of the diagram and the Moon is shown orbiting (dashed circle). The Sun lights half of the Moon and half of the Earth from the right-side. The phase of the Moon is shown next to the corresponding position of the Moon in its orbit around Earth. The phase is as seen in the of Earth.

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Table Moon-03 down yonder  displays diagrams of Earth and Moon as Moon revolves around Earth in its sidereal orbit (sidereal month) and synodic orbit (synodic month).

Lunar month https://en.wikipedia.org/wiki/Lunar_month#Synodic_month

• Day 0. New Moon. Moon is exactly between Earth and Sun. • Day 13.6609. Moon is halfway through its sidereal orbit. Sidereal orbit period = 27.3217 days. • Day 14.77. Full Moon. Moon is halfway through its synodic orbit. Synodic orbit period = 29.53 days. • Day 27.3217. Moon has completed a sidereal orbit. • Day 29.53. New Moon. Moon has completed a synodic orbit.

Table Moon-03 Moon’s sidereal orbit (27.32 days) and synodic orbit (29.53 days) Orbit of the Moon https://en.wikipedia.org/wiki/Orbit_of_the_Moon  DIAGRAMS ARE NOT TO SCALE. 

New Moon Day 0: New Moon. Moon is between Earth and Sun. Earth’s center, Moon’s center and Sun’s center lie on Sun the same straight line. Moon’s near side faces toward Earth and away from Sun. Near side is dark.

Sun Day 13.6609. Moon is halfway through its 27.3217-day sidereal orbit. Full Moon will occur in approximately 1.1 days.

Sun Day 14.77 Full Moon. Moon is halfway through its Full Moon synodic orbit. Earth is between Moon and Sun. Moon’s center, Earth’s center and Sun’s center lie on the same straight line. Moon’s near side faces Earth and Sun and is illuminated by the Sun. It is very bright.

Day 27.32 End of sidereal orbit. Moon has revolved 360 Sun degrees around Earth. New Moon will occur in approximately 29.53 – 27.32 = 2.21 days.

Day 29.53 New Moon. End of synodic orbit. Moon is New Moon between Earth and Sun. Earth’s center, Moon’s center and Sun’s center lie on the same straight line. Moon’s Sun near side faces toward Earth and away from Sun. Near side is dark.

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Earth and a satellite orbiting Earth both revolve around a point called the barycenter located at the center of mass of the system on the line that connects Earth’s center and the satellite’s center. If the mass of the satellite is very much smaller than the mass of Earth, the barycenter is close enough to

a 3 Earth’s center so that we and you can use the equation T = 2 to calculate the orbital period. This GM equation works for the International Space Station because the ISS’s mass is very much smaller than Earth’s mass. ISS mass = 419,725 kilograms. Earth mass = 5.9723  1024 kilograms.

Mass of Earth: 5,972,300,000,000,000,000,000,000 kilograms ISS’s mass is very, very, very much Mass of ISS: 419,725 kilograms smaller than Earth’s mass!

Alas, this equation doesn’t work for Moon’s obit. Moon’s mass is 0.07346  1024 kilograms.

Mass of Earth: 5,972,300,000,000,000,000,000,000 kilograms Moon’s mass is approximately Mass of Moon: 73,460,000,000,000,000,000,000 kilograms 1.23% of Earth’s mass.

The Earth-Moon barycenter is located inside Earth 4,671 kilometers (4,671,000 meters) from Earth’s center on the line connecting Earth’s center and Moon’s center.

Table Moon-04 Earth and Moon revolve around a barycenter located inside Earth The barycenter is the center of mass of the Earth-Moon system. Barycenter https://en.wikipedia.org/wiki/Barycenter  Earth-Moon diagram is not to scale. 

Earth, M1

Moon, M2

• • • C1 r1 B C2

a = semimajor axis

C1 = Earth’s center B = Earth-Moon barycenter C2 = Moon’s center

24 24 M1 = Earth’s mass = 5.9723  10 kilograms M2 = Moon’s mass = 0.07346  10 kilograms a = semimajor axis of Moon’s orbit: 384,400 kilometers = 384,400,000 meters r1 = distance from Earth’s center C1 to barycenter B = 4671 kilometers = 4,671,000 meters

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Fantasize. You are observing Earth and Moon from an Enterprise-class starship orbiting the star Polaris many light years above Earth’s North Pole. You see Moon and Earth orbiting the barycenter (center of mass) of Earth and Moon. It is located inside Earth on the line connecting Earth’s center and Moon’s center. The diagram in Table Moon-04 up yonder  appears again in Table Moon-05 below .

Table Moon-05 Calculate location of the Earth-Moon barycenter and Moon’s orbital period The Earth-Moon barycenter is the center of mass of the Earth-Moon system. Barycenter https://en.wikipedia.org/wiki/Barycenter  The Earth-Moon diagram is not to scale. 

Earth, M1

Moon, M2

• • •

C1 r1 B C2

a = semimajor axis

C1 = Earth’s center B = barycenter C2 = Moon’s center 24 24 M1 = Earth mass = 5.9723  10 kg M2 = Moon mass = 0.07346  10 kg a = semimajor axis of Moon’s orbit = distance from Earth center C1 to Moon center C2 a = 384,400 km = 384,400,000 m

M 2 r1 = distance from Earth center C1 to barycenter B ra1 = MM12+ 0.07346 1024 kg r =384,400 km  = 4671 km [4,671,000 m] 1 5.9723 1024 kg + 0.07346  10 24 kg TI-84: 384400 * 0.07346E24 / (5.9723E24 + 0.07346E24) Enter Ans: 4670.715344 [Round to 4671]

Use the equation over yonder → to calculate the sidereal orbital 3 a period of Earth and Moon around the barycenter. Equation is T = 2 GMM()+ from Orbital period https://en.wikipedia.org/wiki/Orbital_period . 12 (384,400,000 m)3 T ==2 2,357,000 s (6.6743 10−11 m 3 / kg  s 2 )(5.9723  10 24 kg + 0.07346  10 24 kg) TI-84: 2π√(384400000^3/(6.6743E-11 (5.9723E24 + 0.07346E24))) Enter Ans: 2357363.873 [Round to 2357000]

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Recap. Calculate Moon’s orbital period.

NASA’s Moon Fact Sheet 3 a (https://nssdc.gsfc.nasa.gov/planetary/factsheet/moonfact.html) displays the T = 2 6 GMM()+ semimajor axis of Moon’s orbit as 0.384410 kilometers, which we converted 12 to 384,400,000 meters for use in the orbital period equation. Orbital period T of TI-84: 2π√(384400000 / (6.6743E-11 (5.9723E24 + 0.07346E24))) Enter two bodies with Answer: 2357363.873 masses M1 and M2 orbiting a barycenter Alas, 384,400,0000 has only four significant digits, so we rounded the TI-84’s (center of mass) with 10-digit answer to four significant digits, inserted commas to enhance semimajor axis a. readability and added s for seconds: 2357363.873 → 2,357,000 s

NASA’s Moon Fact Sheet: Moon’s revolution period = 27.3217 days. This is Moon’s sidereal orbit period in solar days (24-hour days). Convert the four-significant-digit TI-84 result (2,357,000 s) to solar days:

• (2,357,000 s) / (86,400 s/day) = 27.28 days [4 significant digits]

Our calculated orbital period (27.28 days) is close to NASA’s value (27.3217 days). We rounded NASA’s value to 27.32 days and calculated our percent error.

NASA's value −− our value 27.32 days 27.28 days % error=  100% =  100% = 0.1464% NASA's value 27.32 days

TI-84: ((27.32 − 27.28) / 27.32) * 100 Enter Ans: 0.1464128843 [Round to 0.1464]

Our orbital period calculation differs from the value at NASA’s Moon Fact Sheet by only 0.1464%.

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Things to Come in Low Earth Orbit (LEO) | TOC

Soon: resort hotels, research centers, factories, sports arenas and junkyards in LEO. • LEO: Low Earth Orbit https://en.wikipedia.org/wiki/Low_Earth_orbit

Fantasize. Imagine visiting a resort hotel in Earth orbit. Describe your stay. Search the Internet for space tourism information and ideas. On 2019-12-16, we found: • Space tourism https://en.wikipedia.org/wiki/Space_tourism • Search the Internet for ‘space tourism’.

Fantasize: Imagine working in a research center or factory in Earth orbit and living in an attached or nearby orbital habitat. What type of research? What does the factory manufacture? What is your job? Search the Internet for space manufacturing information and ideas. On 2019-12-16, we found: • Space manufacturing https://en.wikipedia.org/wiki/Space_manufacturing • Search the Internet using search keys such as ‘space factory’, ‘space manufacturing’ and ‘science center in earth orbit’.

Fantasize: Imagine living in an orbital sports arena and playing on a sports team whose games are televised to Earth. What sports? We suggest Quidditch (https://en.wikipedia.org/wiki/Quidditch).

In orbit, you seem to be weightless. You can float or fly in the orbiting sports arena. Design a broom that you can use to fly in the arena and play Quidditch. Idea: start with the Harry Potter Nimbus 2000 broom shown over yonder → and add compressed air tanks for propulsion and control devices for maneuvering. The Nimbus 2000 image is from www.Amazon.com .

At Hogwarts, quidditch is played in an oval pitch (playing field). An orbital quidditch pitch might be a cylinder with hemispherical ends or an ellipsoid or an oblate spheroid. It would be good to pad the walls because players may bounce off the walls during a quidditch match.

Fantasize: LEO is littered with space junk – millions of items of orbital debris. Imagine orbiting junkyards that send out robotic junk collectors to collect orbital debris. • Space Junk: Tracking & Removing Orbital Debris https://www.space.com/16518-space-junk.html • https://en.wikipedia.org/wiki/Space_debris • Search the Internet for ‘space junk’ and ‘space debris’.

Fantasize and then enter a space-settlement contest.

• National Space Society Space Settlement Contest https://space.nss.org/settlement/nasa/Contest/ • National Space Society Space Settlement Contest 2019 Results https://space.nss.org/settlement/nasa/Contest/Results/2019/ • Join a Contest/Competition | National Space Society https://space.nss.org/contests/ • Lesson Plans & Activities – Space Settlement https://settlement.arc.nasa.gov/teacher/lessons/

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Fantasize. Design space settlements. Where might your students locate their Earth-orbiting settlements? We suggest circular equatorial orbits high enough above the equator so that orbital decay is not a big problem and low enough so that Earth’s magnetic field provides protection from solar radiation and galactic cosmic radiation (GCR).

• Circular orbit: The path of the orbit is a circle with orbital radius r. • Near-equatorial orbit https://en.wikipedia.org/wiki/Near-equatorial_orbit • Orbital decay https://en.wikipedia.org/wiki/Orbital_decay • Van Allen radiation belt https://en.wikipedia.org/wiki/Van_Allen_radiation_belt • Health threat from Cosmic rays https://en.wikipedia.org/wiki/Health_threat_from_cosmic_rays

As a small contribution to your students’ fantasies, we will calculate orbital distances and related orbital alakazams for circular equatorial orbits at various heights above Earth’s equator and corresponding heights above Earth’s center. We searched the Internet for an accurate value (many significant digits) of Earth equatorial radius in meters.

• π WGS-84 Earth equatorial radius (meters) https://www.vcalc.com/equation/?uuid=e6cfcccb-da27-11e2-8e97-bc764e04d25f ‘The Equatorial radius of the Earth is 6,378,137.0 meters.’ • NASA’s Earth Fact Sheet (https://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html) displays Earth’s equatorial radius in kilometers: 6378.137 kilometers.

For calculations, we will us:

  • Earth’s equatorial radius = 6,378,137 meters [7 significant digits] • Earth’s equatorial radius = 6378.137 kilometers [7 significant digits]

The height above Earth’s center of a satellite in a circular equatorial obit is the height of the satellite above the equator plus Earth’s equatorial radius.

• Height above Earth’s center = height in kilometers above Earth’s equator + 6378.137 kilometers • Height above Earth’s center = height in meters above Earth’s equator + 6,378,137 meters

Earth rotates eastward. Observed from above the North Pole (NP), Earth’s rotation is counterclockwise ↺. An orbiting resort hotel, research center, factory, sports arena or junkyard also rotates eastward. Its orbit is also counterclockwise ↺. Earth makes one 360-degree rotation in one sidereal • day: 23 hours 56 minutes 4 seconds (86,164 seconds). NP

• Sidereal day = 23 h  3600 s/h + 56 min  60 s/min + 4 s = 86,164 s

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Earth’s equator is a circle. Radius: R = 6378.137 kilometers = 6,378,137 meters. Calculate the circumference C of Earth’s equator.

• Circumference of equator = 2π  radius of equator. C = 2πR • C in kilometers: C = 2πR = 2π  6378.137 km = 40,075.01669 km [Round to 40,075.02 km] • C in meters: C = 2πR = 2π  6,378,137 m = 40,075,016.69 m [Round to 40,075,020 m]

Imagine you are standing on Earth’s equator at rest, not moving – you think. Oho! The place where you are standing moves 40,075.02 kilometers (40,075,020 meters) per sidereal day (86,164 seconds).

How fast are you moving? What is your velocity? Here are calculations in meters per second (m/s), kilometers per second (km/s) and kilometers per hour (km/h). 86,164 seconds has 5 significant digits, so we rounded each answer to 5 digits.

• Circumference of equator = 40,075.02 kilometers (km) = 40,075,020 meters (m) • Sidereal day = 86,164 seconds (s) [5 significant digits] • m/s: (40,075,020 m) / (86,164 s) = 465.1016666 m/s Round to 465.10 m/s • km/s: (40,075.02 km) / (86,164 s) = 0.4651016666 km/s Round to 0.46510 km/s • km/h: (40,075.02 km) / (86,164 s)  (3600 s/h) = 1674.365874 km/h Round to 1674.4 km/h

Wow! Standing on the equator at rest, you are travelling 1674.4 kilometers per hour as Earth rotates.

Earth’s equator is a circle. Radius = 6378.137 kilometers = 6,378,137 meters. Circumference of the equator = 40,075.02 kilometers = 40,075,020 meters.  Earth rotates 360 degrees in one sidereal day (86,164 seconds).  A point on the equator rotates 360 around Earth’s center in one sidereal day. The velocity of a point on the equator is 1674.4 kilometers per hour.

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Fantasize! Imagine an orbiter (resort hotel, research center, factory, sports arena, junkyard) in a circular equatorial orbit exactly 600 kilometers (600,000 meters) above Earth’s equator.

• Earth’s equatorial radius = 6,378,137 meters • Height of orbiter above the equator = 600,000 meters • Height of orbiter above Earth’s center = 6,378,137 meters + 600,000 meters = 6,978,137 meters • Circular orbit, orbital radius in meters: r = 6,978,137 m

Calculate the orbital period T and orbital velocity v of the orbiter.

r 3 T = orbital period, seconds (s), r = orbital radius, meters (m) Orbital period equation: T = 2 −11 3 2 24 GM G = 6.6743  10 m /kg·s ; M = 5.9723  10 kg

(6,978,137 m)3 Orbital period, r = 6,968,137 m: T ==2 5801.2 s (6.6743 10−11 m 3 /kg  s 2 )(5.9723  10 24 kg)

TI-84: 2π√(6978137^3 / (6.6743E-11 * 5.9723E24)) Enter Ans: 5801.167916 [Round to 5801.2]

v = orbital velocity in meters/second (m/s) GM Orbital velocity equation: v = G = 6.6743  10−11 m3/kg·s2 ; M = 5.9723  1024 kg r r = orbital radius in meters (m)

(6.6743 10−11 m 3 /kg  s 2 )(5.9723  10 24 kg) Orbital velocity: r = 6,978,137: v ==7557.9 m/s 6,978,137 m

TI-84: √(6.6743E-11 * 5.9723E24 / 6978137) Enter Answer: 7557.948417 [Round to 7557.9]

Orbital period in seconds, minutes, and hours: Orbital velocity in kilometers/hour: • Seconds: 5801.2 s 7557.9 m 3600 s 1 km 27,208 km   = • Minutes: (5801.2 s) / (60 s/min) = 96.687 min 1 s 1 h 1000 m 1 h • Hours: (6307.0 s) / (3600 s/h) = 1.6114 h

An orbiter in a circular equatorial obit 600 kilometers (600,000 meters) above the equator makes one orbit in 5801.2 seconds. How many orbits does it make each sidereal day (86,164 seconds)?

• Orbits per sidereal day = (86,164 s) / (5801.2 s) = 14.053

In one sidereal day (86,164 s), a point on the equator makes one 360-degree rotation around Earth’s center while the orbiter makes 14.053 360-degree revolutions around Earth’s center. The orbiter will pass over that point on the equator 14.053 – 1 = 13.053 times.

The orbiter is zooming eastward in its orbit at 27,208 kilometers per hour. A point on the equator is travelling eastward at 1674.4 kilometers per hour. The orbiter is traveling much faster than the point on the equator!

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Table Orbits-01 displays data for a bunch of circular equatorial orbits. Read about Earth satellite orbits at Catalog of Earth Satellite Orbits https://earthobservatory.nasa.gov/features/OrbitsCatalog.

Table Orbits-01 Low Earth Orbits, Medium Earth Orbits, and Geostationary Orbit Circular orbits in the plane of Earth’s equator (circular equatorial orbits) Catalog of Earth Satellite Orbits https://earthobservatory.nasa.gov/features/OrbitsCatalog

r 3 T = orbital period in seconds (s) Orbital period: T = 2 r = orbital radius in meters (m) GM v = orbital velocity in meters/second (m/s) GM G = 6.6743  10−11 m3/kg·s2 Orbital velocity: v = r M = 5.9723  1024 kg

Orbital radius = ht. above equator (km) + 6378.137 km = ht. above equator (m) + 6,378,137 m 1 sidereal day = 23 h 56 min 4 s = 86,164 s = 23.934 h Height above Height above Orbital Orbital Orbits per Orbital equator Earth center radius r period T sidereal day velocity v 400 km   6778.137 km 7668.6 m/s 6,778,137 m 5553.6 s 15.515 400,000 m 6,778,137 m 27,607 km/h 500 km  6878.137 km 7612.7 m/s 6,878,137 m 5676.9 s 15.178 500,000 m 6,878,137 m 27,406 km/h 600 km  6978.137 km 7557.9 m/s 6,978,137 m 5801.2 s 14.853 600,000 m 6,978,137 m 27,209 km/h 650 km  7028.137 km 7531.0 m/s 7,028,137 m 5863.6 s 14.695 650,000 m 7,028,137 m 27,112 km/h 20,200 km  26,578.137 km 3872.7 m/s 26,578,137 m 43,121 s 1.9982 20,200,000 m 26,578,137 m 13,942 km/h 35,786 km  42,164.137 km 3074.7 m/s 42,164,000 m 86,163 s 1.0000 35,786,000 m 42,164,137 m 11,069 km/h 35,786.313 km  42,164.450 km 3074.7 m/s 42,164,450 m 86,164 s 1.0000 35,786,313 m 42,164,450 m 11,069 km/h

 Approximate height of the International Space Station (ISS) above Earth’s surface.  Orbit enjoys protection from radiation by Earth’s magnetic field. Great orbits for resort hotels, research centers, factories, sports arenas, junkyards and other space habitats.  Orbit is in the Van Allen radiation belt https://en.wikipedia.org/wiki/Van_Allen_radiation_belt and may experience an undesirable amount of cosmic radiation.  Orbital height of Global Positioning System (GPS) . Calculated number of orbits per sidereal day = 1.9982. Conjecture: a GPS satellite makes exactly two orbits per sidereal day.  Geostationary orbit https://en.wikipedia.org/wiki/Geostationary_orbit. Calculated orbital period (86,163 s) is approximately equal to Earth’s sidereal rotation period (86,164 s). Orbiter appears to be almost stationary above a point on the equator.  Geostationary orbit. We adjusted the orbital radius so that the orbital period equals Earth’s sidereal rotation period (86,164 s). Orbiter appears to be stationary above a point on the equator.

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What can you see from a circular equatorial orbit?

Earth rotates 360 in one sidereal day: 23 hours 56 minutes 4 seconds. Sidereal day = 23 h  3600 s/h + 56 min  60 s/min + 4 s = 86,164 s.

As seen from above the North Pole, Earth’s rotation is counterclockwise .

A point on the equator travels eastward. •

NP

As seen from above the North Pole, an orbiter (space habitat) revolves 360 counterclockwise ↺ around Earth’s center in one orbital period. It travels eastward. An observer on the equator sees the orbiter travel from west to east across the sky.

The equator is 0. If your space habitat is in orbit 600 km above the equator, how far north of the equator can you see? How far south of the equator can you see?

In the diagram over yonder →, an orbiter (●) is 600 kilometers above the equator. Line-of-sight arrows ( and ) are tangent to Earth’s surface. The angle φ is the north or south latitude of the points where the line-of-sight arrows are tangent to Earth’s surface. φ  24. An observer on the orbiter φ • can view Earth between 24 north and 24 south, from the Tropic φ of Cancer (23.4 north) to the (23.4 south). • https://en.wikipedia.org/wiki/Tropic_of_Cancer • Tropic of Capricorn https://en.wikipedia.org/wiki/Tropic_of_Capricorn

 Idea: Put camera satellites (C) in the orbiter’s (O) orbit, one leading the ● C orbiter 60 and one trailing the orbiter 60. Transmit the camera images to the orbiter. Observers on the orbiter can view more than half the Earth. The orbiter and cameras are in a circular equatorial orbit 600 km (600,000 m) ● O above the equator. [Oops? The diagram is not to scale. Is the line of sight from a camera satellite to the orbiter clear, or does Earth get in the way? Might it be better to locate the cameras a bit closer to the orbiter? ● C • Orbital radius: r = 6,378.137 km + 600.000 km = 6,978,137.0 km

Math interlude. φ is the latitude angle between 1) the line from Earth’s center to the orbiter and 2) Earth’s radius at the point where the line-of-sight arrow ( or ) is tangent to Earth’s surface. We estimated the measure of φ on the diagram as approximately 30 and used Earth radius by latitude (https://rechneronline.de/earth-radius/) to calculate Earth’s radius at that latitude. And then we used trigonometry to calculate an approximate value of φ:

Earth radius at latitude  6373 km  =cos−−11 = cos  24 Height of orbiter above Earth center6978 km

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Imagine. You are aboard an orbiter (space habitat) 600 kilometers above the equator. Camera satellites occupy the same orbit, one leading the orbiter and one trailing the orbiter. With the aid of the camera satellites, you can see Earth from the North Pole to the .

What might you see? Behold Table Orbit-02!

Table Orbit-02 Earth views along parallels of latitude thanks to Wikipedia

Latitude Equator, north & south parallels of latitude, North Pole & South Pole Equator https://en.wikipedia.org/wiki/Equator Countries on the Equator 0 https://www.thoughtco.com/countries-that-lie-on-the-equator-1435319 Cities on the Equator https://www.touristmaker.com/blog/the-largest-equatorial-cities/ 15th parallel north https://en.wikipedia.org/wiki/15th_parallel_north 15 15th parallel south https://en.wikipedia.org/wiki/15th_parallel_south

30th parallel north https://en.wikipedia.org/wiki/30th_parallel_north 30 30th parallel south https://en.wikipedia.org/wiki/30th_parallel_south

45th parallel north https://en.wikipedia.org/wiki/45th_parallel_north 45 https://en.wikipedia.org/wiki/45th_parallel_south

60th parallel north https://en.wikipedia.org/wiki/60th_parallel_north 60 60th parallel south https://en.wikipedia.org/wiki/60th_parallel_south

75th parallel north https://en.wikipedia.org/wiki/75th_parallel_north 75 75th parallel south https://en.wikipedia.org/wiki/75th_parallel_south

North Pole https://en.wikipedia.org/wiki/North_Pole 90 South Pole https://en.wikipedia.org/wiki/South_Pole

Imagine: You are aboard a space habitat (orbiter) in a circular equatorial orbit 600 km above the equator. You are watching the Earth go by way down below. • Do you see Earth go by east to west or west to east? • Do you see Earth go by slowly? Or rapidly? Or somewhere in between? • Assume: your space habitat travels faster than a point on the equator. In one orbit, how many times will you pass that point?

Imagine: You are aboard an orbiter (space habitat) in a geostationary orbit. You are stationary above a point on the equator. Watching the Earth is a bit boring – the same vista all day every day. Oops, in a geostationary orbit, you may be adversely affected by cosmic rays!

Imagine: Several space habitats orbit between 450 kilometers and 600 kilometers above the equator. In your personal spacecraft, you visit these habitats and spend time aboard each. Describe your adventure. Aha! A best-selling science fiction novel?

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The High Frontier: 1976 and 2019 | TOC

We will not explore The High Frontier in this eBook – another story for another time. You and your students can explore The High Frontier by means of books and the Internet.

1976: O’Neill, Gerard K (Author), Donald Davis (Illustrator) & Kathy Sullivan (Preface). The High Frontier: Human Colonies in Space. ISBN-10: 1686872720 ISBN-13: 978-1686872723 [This edition published by Space Institute Studies Press in 2019.]

The High Frontier: Human Colonies in Space was first published in 1976. Here is a review of the 3rd edition published by Apogee Books in 2000.

Book Review: The High Frontier: Human Colonies in Space, 3rd Edition https://space.nss.org/book-review-the-high-frontier-human-colonies-in-space- 3rd-edition/

www.Amazon.com Internet: The High Frontier: Human Colonies in Space https://en.wikipedia.org/wiki/The_High_Frontier:_Human_Colonies_in_Space

2018: Marotta, Tom & Al Globus. The High Frontier: An Easier Way. Hardcover ISBN-10: 0464706300 ISBN-13: 978-0464706304 Paperback ISBN-10: 1719231745 ISBN-13: 978-1719231749 Kindle ASIN: B07FYWM41W

Book review: The High Frontier: An Easier Way https://space.nss.org/book-review-the-high-frontier-an-easier-way/

Book review: The High Frontier: An Easier Way https://thespacereview.com/article/3568/1

www.Amazon.com

FANTASIZE. IMAGINE. You and your students boldly exploring – in your creative imaginations – a future in which people live, work, and play in communities in Earth orbit, on the Moon,  in Earth’s orbit around the Sun, on Mars, in Mars orbit, on Jupiter’s moon Europa,  on Saturn’s moon Titan, and elsewhere in the Solar System. You and your students imagine creating these communities. Reality expands to fill the available fantasies.

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Appendix 01 Earth and Moon Data | TOC

Earth and Moon data are retrieved from NASA’s Earth Fact Sheet and Moon Fact Sheet.

Earth data from NASA’s Earth Fact Sheet 2019-12-20.

Table 01 Earth data selected from NASA’s Earth Fact Sheet (minor additions by B & G) Earth Fact Sheet https://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html Notes on the fact sheets https://nssdc.gsfc.nasa.gov/planetary/factsheet/fact_notes.html

Bulk parameters Mass 5.9723  1024 kg Volume 108.321  1010 km3 Equatorial radius 6378.137 km 6,378,137 m Polar radius 6356.752 km 6,356,752 m Volumetric mean radius 6371.000 km 6,371,000 m Mean density 5514 kg/m3 5.514 g/cm3 Surface gravity 9.798 m/s2 Surface acceleration 9.780 m/s2 Escape velocity 11.186 km/s 11,186 m/s; 40,270 km/h Solar irradiance 1361.0 W/m2 1.3610 kW/m2 Orbital parameters Semimajor axis  149.60  106 km 149.60  109 m Sidereal orbit period 365.256 days Solar day: 24 hours = 86,400 s Tropical orbit period 365.242 days Solar day: 24 hours = 86,400 s Perihelion  147.09  106 km 147.09  109 m Aphelion  152.10  106 km 152.10  109 m Mean orbital velocity 29.78 km/s 29,780 m/s Max. orbital velocity 30.29 km/s 30,290 m/s Min. orbital velocity 29.29 km/s 29,290 m/s Orbit eccentricity 0.0167 Sidereal rotation period 23.9345 h 86,164.2 s Length of day 24.0000 h Solar day: 86,400 s Obliquity to orbit 23.44 deg  (Perihelion + aphelion)/2 = semimajor axis. (147.09106 km + 152.10106 km)/2 = 149.60106 km

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Moon data from NASA’s Moon Fact Sheet 2019-12-20.

Table 02 Moon data selected from NASA’s Moon Fact Sheet (minor additions by B & G) Moon Fact Sheet https://nssdc.gsfc.nasa.gov/planetary/factsheet/moonfact.html Notes on the fact sheets https://nssdc.gsfc.nasa.gov/planetary/factsheet/fact_notes.html

Bulk parameters Mass 0.07346  1024 kg Volume 2.1968  1010 km3 Equatorial radius 1738.1 km 1,738,100 m Polar radius 1736.0 km 1,736,000 m Volumetric mean radius 1737.4 km 1,737,400 m Mean density 3344 kg/m3 3.344 g/cm3 Surface gravity 1.62 m/s2 Surface acceleration 1.62 m/s2 Escape velocity 2.38 km/s 2,380 m/s; 8570 km/h Solar irradiance 1361.0 W/m2 1.3610 kW/m2 Orbital parameters Semimajor axis  0.3844  10 6 km 384,400 km = 384,400,000 m Perigee  0.3633  106 km 363,300 km = 363,300,000 m Apogee  0.4055  106 km 405,500 km = 405,500,000 m Revolution period 27.3217 days 27.3217 d  86,400 s/d = 2,360,590 s Sidereal rotation period 655.728 h 655.728 h  3600 s/h = 2,360,620 s Synodic period 29.53 days 29.53 d  86,400 s.d = 2,551,000 s Mean orbital velocity 1.022 km/s 1022 m/s = 3679 km/h Max. orbital velocity 1.082 km/s 1083 m/s = 3899 km/h Min. orbital velocity 0.970 km/s 970 m/s = 3492 km/h Orbit eccentricity 0.0549  (Perihelion + aphelion) / 2 = semimajor axis. (363,300 km + 405,500 km) / 2 = 384,400 km

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Table 03 displays Moon data, Earth data and Moon/Earth ratio for bulk parameters.

Table 03 Moon data, Earth data and Moon/Earth ratio Moon Fact Sheet https://nssdc.gsfc.nasa.gov/planetary/factsheet/moonfact.html Earth Fact Sheet https://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html

Bulk parameters Moon Earth Moon / Earth Mass 0.07346  1024 kg 5.9723  1024 kg 0.0123 Volume 2.1968  1010 km3 108.321  1010 km3 0.0203 1738.1 km 6378.137 km Equatorial radius 0.2725 1,738,100 m 6,378,137 m 1736.0 km 6356.752 km Polar radius 0.2731 1,736,000 m 6,356,752 m 1737.4 km 6371.000 km Volumetric mean radius 0.2727 1,737,400 m 6,371,000 m 3344 kg/m3 5514 kg/m3 Mean density 0.606 3.344 g/cm3 5.514 g/cm3 Surface gravity 1.62 m/s2 9.798 m/s2 0.165 Surface acceleration 1.62 m/s2 9.780 m/s2 0.166 2.38 km/s 11.186 m/s Escape velocity 0.213 2380 m/s 11,186 m/s Solar irradiance 1361.0 W/m2 1361.0 W/m2 1.000 Orbital parameters Moon Notes (Moon data) Semimajor axis  0.3844  106 km 384,400 km = 384,400,000 m Perigee  0.3633  106 km 363,300 km = 363,300,000 m Apogee  0.4055  106 km 405,500 km = 405,500,000 m Revolution period 27.3217 days 27.3217 d  86,400 s/d = 2,360,590 s Sidereal rotation period 655.728 h 655.728 h  3600 s/h = 2,360,620 s Synodic period 29.53 days 29.53 d  86,400 s/d = 2,551,000 s Mean orbital velocity 1.022 km/s 1022 m/s Max. orbital velocity 1.082 km/s 1082 m/s Min. orbital velocity 0.970 km/s 970 m/s Orbit eccentricity 0.0549  (Perihelion + aphelion) / 2 = semimajor axis Moon: (0.3633  106 km + 0.4055  106 km) / 2 = 0.3844  106 km Moon: (363,300 km + 405,500 km) / 2 = 384,400 km Earth: (147.09  106 km + 152.10  106 km) / 2 = 149.60  106 km

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Appendix 02 Orbital Equations | TOC

Calculate orbital period T

Use the equations over yonder → to calculate the orbital period T of an Earth satellite a 3 in an elliptical orbit with semimajor axis a or a circular orbit with orbital radius r. T = 2 Orbital period https://en.wikipedia.org/wiki/Orbital_period. GM Elliptical orbit T = orbital period in seconds (s) a = semimajor axis of the orbit in meters (elliptical orbit) r 3 T = 2 r = radius of the orbit in meters (circular orbit) GM −11 3 2 G = gravitational constant = 6.6743  10 m /kg·s [5 significant digits] Circular orbit 24 M = mass of Earth = 5.9723  10 kg [5 significant digits]

 These equations are A-OK for calculating the period of a satellite whose mass is very small compared to the mass of the object about which it orbits.

Example. Elliptical orbit, perigee 550,000 meters (550 kilometers) from Earth’s equator, apogee 650,000 meters (650 kilometers) from Earth’s equator.

• Perigee: Equatorial radius + 550,000 m = 6,378,137 m + 550,000 m = 6,928,137 m • Apogee: Equatorial radius + 650,000 m = 6,378,137 m + 650,000 m = 7,028,137 m • Semimajor axis: a = (perigee + apogee) / 2 = (6,928,137 m + 7,028,137 m) / 2 = 6,978,137 m

(6,978,137 m)3 Orbital period: T ==2 5801.2 s (6.6743 10−11 m 3 /kg  s 2 )(5.9723  10 24 kg)

TI-84: 2π √(6978137^3 / (6.6743E-11 * 5.9723E24)) Enter Answer: 5801.167916 [Round to 5801.2]

Example: Circular equatorial orbit 600,000 meters (600 kilometers) above Earth’s equator.

• Orbital radius = equatorial radius + ht. above equator = 6,378.137 m + 600,000 m = 6.978,137 m

3 r (6,978,137 m)3 Orbital period: T = 2 T ==2 5801.2 s GM (6.6743 10−11 m 3 /kg  s 2 )(5.9723  10 24 kg)

TI-84: 2π √(6978137^3 / (6.6743E-11 * 5.9723E24)) Enter Answer: 5801.167916 [Round to 5801.2]

If semimajor axis a of an elliptical orbit = orbital radius r of a circular orbit then orbital period of the elliptical orbit = orbital period of the circular orbit.

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Calculate semimajor axis (elliptical orbit) or orbital radius (circular orbit)

Use the equations over yonder → to calculate the length of the semimajor axis a of GMT 2 an elliptical orbit or the orbital radius r of a circular orbit with orbital period T. a = 3 4 2 a = semimajor axis of the orbit in meters (elliptical orbit) Elliptical orbit r = radius of the orbit in meters (circular orbit) GMT 2 T = orbital period in seconds (s) 3 − r = G = gravitational constant = 6.6743  10 11 m3/kg·s2 [5 significant digits] 4 2 M = mass of Earth = 5.9723  1024 kg [5 significant digits] Circular orbit

 These equations are the right stuff for calculating the period of a satellite whose mass is very small compared to the mass of the object about which it orbits.

Example. Elliptical orbit, orbital period T = 5801.2 seconds. Up yonder  we calculated the orbital period for an elliptical orbit with semimajor axis = 6,978,137 meters. Result: 5801.2 seconds. Using T = 5801.2 seconds, it would be good if the semimajor axis equation produces a value close to 6,978,137 meters.

(6.6743 10−11m3 /kg  s 2 )(5.9723  10 24 kg )(5801.2 s ) 2 Semimajor axis: a ==3 6,978,200 m 4 2 TI-84: (6.6743E-11 * 5.9723E24 * 5801.22 / (4π2)) ^ (1/3) Enter Answer: 6978162.729 [Round to 6978200] Close. 6,978,200 m – 6,978,137 m = 63 m.

Example: Circular equatorial orbit with orbital period equal to 1/2 Earth’s sidereal rotation period. The orbiter makes two orbits per sidereal day while Earth makes one 360 rotation.

• Orbital period: T = length of Earth sidereal day / 2 = 86164 s / 2 = 43,082 s

GMT 2 (6.6743 10−11m3 /kg  s 2 )(5.9723  10 24 kg )(43,082 s ) 2 Orbital radius: r = 3 r ==3 26,562,000 m 4 2 4 2 TI-84: (6.6743E-11 * 5.9723E24 * 430822 / (4π2)) ^ (1/3) Enter Ans: 26561938.79 [Round to 26562000] Check: We rounded the orbital radius to 5 significant figures. Calculate the orbital period for a circular equatorial orbit 26,562,000 meters above Earth’s center and hope the result is close to 43,082 seconds.

3 r (26,562,000 m)3 Orbital period: T = 2 T ==2 43,082 s GM (6.6743 10−11 m 3 /kg  s 2 )(5.9723  10 24 kg)

TI-84: 2π √(26562000^3 / (6.6743E-11 * 5.9723E24)) Enter Answer: 43082.14892 [Round to 43082]

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Calculate location of barycenter and orbital period around barycenter

Two bodies that orbit each other revolve around a barycenter located at the center of mass of the system on the line that connects the centers of the two orbiting bodies.

• Barycenter https://en.wikipedia.org/wiki/Barycenter

The Earth-Moon barycenter is located inside Earth 4671 kilometers (4,671,000 meters) from Earth’s center on the line connecting Earth’s center and Moon’s center.

Use the equation over yonder → to calculate the distance r1 of the barycenter M ra= 2 from the center of orbiting body M1. 1 MM+ 12 r1 = distance of barycenter from M1 (Earth) a = semimajor axis of Moon’s orbit = 384,400 km [4 significant digits] r1 is the distance 24 of the barycenter M1 = mass of Earth = 5.9723  10 kg [5 significant digits] 24 from the center M2 = mass of Moon = 0.07346  10 kg [4 significant digits] of M1

0.07346 1024 kg r =384,400 km  = 4671 km [4,671,000 m] 1 5.9723 1024 kg + 0.07346  10 24 kg

TI-84: 384400 * 0.07346E24 / (5.9723E24 + 0.07346E24) Enter Ans: 4670.715344 [Round to 4671]

→ 3 Use the equation over yonder to calculate the sidereal orbital period of a Earth and Moon around the barycenter. T = 2 Orbital period https://en.wikipedia.org/wiki/Orbital_period . GMM()12+

T = orbital period in seconds a = semimajor axis in meters = 384,400,000 m [4 significant digits] G = gravitational constant = 6.6743  10−11 m3/kg·s2 [5 significant digits] 24 M1 = mass of Earth = 5.9723  10 kg [5 significant digits] 24 M2 = mass of Moon = 0.07346  10 kg [4 significant digits]

(384,400,000 m)3 T ==2 2,357,000 s (6.6743 10−11m 3 /kg  s 2 )(5.9723  10 24 kg + 0.07346  10 24 kg)

TI-84: 2π√(384400000^3 / (6.6743E-11 (5.9723E24 + 0.07346E24))) Enter Ans: 2357363.873 [Round to 2357000]

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Calculate orbital velocity (circular orbit)

The orbital velocity v of an Earth satellite in a circular orbit is the constant speed the GM satellite travels in its orbit. Use the equation over yonder → to calculate the orbital v = velocity in meters per second (m/s) of a satellite in a circular orbit. r Circular orbit v = orbital velocity in meters per second (m/s) G = gravitational constant = 6.6743  10−11 m3/kg·s2 [5 significant digits]

M = mass of Earth = 5.9723  1024 kg (kilogram) [5 significant digits] r = orbital radius = distance of the satellite from Earth’s center in meters (m)

 This equation is A-OK for calculating the velocity of a satellite whose mass is small compared to the mass of the object about which it orbits.

Example: Circular equatorial orbit 600,000 meters (600 kilometers) above Earth’s equator.

• Orbital radius = equatorial radius + height above equator = 6,378,137 m + 600,000 m = 6,978,137 m

(6.6743 10−11m 3 /kg  s 2 )(5.9723  10 24 kg ) Orbital velocity: v ==7557.9 m/s 6,978,137 m

TI-84: √(6.6743E-11 * 5.9723E24 / 6978137) Enter Answer: 7557.948417 [Round to 7557.9]

Check: The orbital velocity of a circular orbit with orbital radius r is equal to the circumference of the circular orbit divided by the orbital period.

• Orbital velocity v of circular orbit = orbital circumference / orbital period. v = 2πr / T

Circular equatorial orbit with orbital radius r = 6,978,137 meters and orbital period T = 5801.2 seconds. Calculate orbital velocity v.

• v = 2πr / T = 2π  6,978,137 m / 5801.2 s = 7557.9 m/s

TI-84: 2π * 6978317 / 5801.2 Enter Answer: 7557.906618 [Round to 7557.9]

Convert orbital velocity in meters per second to kilometers per hour:

7557.9 m 3600 s 1 km 27,208 km   = 1 s 1 h 1000 m 1 h

TI-84: 7557.9 * 3600 / 1000 Enter Answer: 27208.44 [Round to 27208]

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Calculate orbital velocity at perigee and apogee (elliptical orbit)

Use the equation over yonder → to calculate the orbital velocity v of an Earth 21 satellite in an elliptical orbit at any point in its orbit. v=− GM  ra v = orbital velocity in meters per second (m/s) G = gravitational constant = 6.6743  10−11 m3/kg·s2 [5 significant digits] Orbital velocity at M = mass of Earth = 5.9723  1024 kg [5 significant digits] orbital point r = distance of the satellite from Earth’s center in meters (m) distance r from a = semimajor axis of the elliptical orbit in meters (m) Earth’s center.

 This equation is the right stuff for calculating the velocity of a satellite whose mass is small compared to the mass of the object about which it orbits.

Example. Elliptical orbit, perigee 550,000 meters (550 kilometers) from Earth’s equator, apogee 650,000 meters (650 kilometers) from Earth’s equator.

• Perigee: Equatorial radius + 550,000 m = 6,378,137 m + 550,000 m = 6,928,137 m • Apogee: Equatorial radius + 650,000 m = 6,378,137 m + 650,000 m = 7,028,137 m • Semimajor axis: a = (perigee + apogee) / 2 = (6,928,137 m + 7,028,137 m) / 2 = 6,978,137 m

Orbital velocity at perigee: r = 6,928,137 m

−11 3 2 24 21 v =(6.6743  10 m /kg  s )(5.9723  10 kg) − = 7612.3 m/s 6,928,137 m 6,978,137 m

TI-84: √(6.6743E-11 * 5.9723E24 (2/6928137 − 1/6978137)) Enter Ans: 7612.298316 [Round to 7612.3]

Orbital velocity at apogee: r = 7,028,137 m

−11 3 2 24 21 v =(6.6743  10 m /kg  s )(5.9723  10 kg) − = 7504.0 m/s 7,028,137 m 6,978,137 m

TI-84: √(6.6743E-11 * 5.9723E24 (2/7028137 − 1/6978137)) Enter Ans: 7503.986564 [Round to 7504.0]

Convert orbital velocity at perigee to km/h Convert orbital velocity at apogee to km/h

7612.3 m 3600 s 1 km 27,404 km 7504.0 m 3600 s 1 km 27,014 km   =   = 1 s 1 h 1000 m 1 h 1 s 1 h 1000 m 1 h

TI-84: 7612.3 * 3600 / 1000 Enter Ans: 27404.28 TI-84: 7504.0 * 3600 / 1000 Enter Ans: 27014.4 Round to 27404 Round to 27014

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Moon’s orbital velocity at perigee and apogee

Use the equation over yonder → to calculate the Moon’s orbital 21 velocity v at any point in its orbit. v= G() M + M − 12 ra v = orbital velocity in meters per second (m/s) −11 3 2 G = gravitational constant = 6.6743  10 m /kg·s [5 significant digits] Moon’s (M2) orbital velocity 24 M1 = mass of Earth = 5.9723  10 kg [5 significant digits] at distance r from Earth’s 24 M2 = mass of Moon = 0.07346  10 kg [4 significant digits] (M1) center r = distance from Earth’s center to Moon’s center in meters (m) a = semimajor axis of Moon’s orbit in meters = 384,400,000 m [4 significant digits]

Moon: orbital velocity at perigee. r = 363,300 km (NASA: Moon Fact Sheet). r = 363,300,000 m

−11 3 2 24 24 21 v =(6.6743  10m /kg  s )(5.9723  10 kg + 0.07346  10 kg) − = 1082 m/s 363,300,000 m 384, 400,000 m

TI-84: √(6.6743E-11 (5.9723E24 + 0.07346E24) (2/363300000 − 1/384400000) ) Enter Ans: 1082.428838 Round to 1082

Moon: orbital velocity at apogee. r = 405,500 km (NASA: Moon Fact Sheet). r = 405,500,000 m

−11 3 2 24 24 21 v =(6.6743  10m /kg  s }(5.9723  10 kg + 0.07346  10 kg) − = 969.8 m/s 405,500,000 m 384, 400,000 m

TI-84: √(6.6743E-11 (5.9723E24 + 0.07346E24) (2/405500000 − 1/384400000) ) Enter Ans: 969.7814964 Round to 969.8

Our calculated orbital velocity NASA Moon Fact Sheet % difference

Perigee: 1082 m/s = 1.082 km/s Perigee: 1.082 km/s 0% Apogee: 969.8 m/s = 0.9698 km/s Apogee: 0.970 km/s 0.02%

Convert orbital velocity at perigee to km/h Convert orbital velocity at apogee to km/h

1082 m3600 s 1 km 3895 km 969.8 m3600 s 1 km 3491 km   =   = 1 s 1 h 1000 m 1 h 1 s 1 h 1000 m 1 h

TI-84: 1082 * 3600 / 1000 Enter Ans: 3895.2 TI-84: 969.8 * 3600 / 1000 Enter Ans: 3491.28 Round to 3895 Round to 3491

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NASA’s Moon Fact Sheet says Moon’s mean orbital velocity is 1.022 km/s (kilometers/second). We don’t know how NASA calculated this velocity. We thought about calculating orbital velocities at 100 points in Moon’s orbit, adding them, and dividing the sum by 100. The thought quickly passed.

Moon’s orbit is almost circular – eccentricity = 0.0549 (a circular orbit has zero eccentricity). The orbital velocity in a circular orbit is constant – the same everywhere in the orbit. Might the orbital velocity of a circular orbit with radius equal to Moon’s semimajor axis (384,400 km = 384,400,000 m) be close to Moon’s mean orbital velocity? Let’s investigate.

We will use the equation over yonder → to calculate the orbital velocity of GMM()+ Moon in a circular orbit with orbital radius r. v = 12 r v = orbital velocity in meters/second (m/s) G = gravitational constant = 6.6743  10−11 m3/kg·s2 [5 significant digits] Orbital velocity v in 24 a circular orbit with M1 = mass of Earth = 5.9723  10 kg [5 significant digits] 24 orbital radius r M2 = mass of Moon = 0.07346  10 kg [4 significant digits] r = radius of circular orbit = 384,400,000 m [4 significant digits]

(6.6743 10−11m 3 /kg  s 2 )(5.9723  10 24 kg + 0.07346  10 24 kg ) v ==1025 m/s = 1.025 km/s 384,400,000 m

TI-84: √(6.6743E-11 (5.9723E24 + 0.07346E24) / 384400000) Enter Ans: 1024.558177 [Round to 1025]

Convert m/s to km/s: 1025 m/s = 1.025 km/s

Our calculated orbital velocity NASA Moon Fact Sheet % difference Notes

Mean: 1025 m/s = 1.025 km/s  Mean: 1.022 km/s 0.3% Close Perigee: 1082 m/s = 1.082 m/s Perigee: 1.082 km/s 0% ☺ Apogee: 969.8 m/s = 0.9698 km/s Apogee: 0.970 km/s 0.02% Very close

 Circular orbit, radius = 384,400,000 m

Convert orbital velocity to km/h:

1025 m3600 s 1 km 3690 km   = 1 s 1 h 1000 m 1 h

TI-84: 1025 * 3600 / 1000 Enter Answer: 3690

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Another way to calculate the orbital velocity of a circular orbit: calculate the circumference of the circular orbit and then divide the circumference by the orbital period.

• Orbital velocity, circular orbit = circumference of orbit / orbital period

Imagine that Moon’s orbit is circular with orbital radius equal to Moon’s semimajor axis a. r = 384,400 kilometers (384,400,000 meters). Calculate the circumference of the circular orbit.

• Circumference C of circular orbit = 2π  orbital radius. C = 2πr • r = 384,400,000 m: C = 2π  384,400,000 m = 2,415,256,432 m [Round to 2,415,000,000 m]

NASA’s Moon Fact Sheet: Moon’s orbital period is 27.3217 days. Convert to seconds:

• (27.3217 days / 86,400 s/day) = 2,360,594.88 s [Round to 2,360,590 s]

Calculate the orbital velocity:

• Orbital velocity v, circular orbit = circumference of orbit / orbital period. v = C / T • v = (2,415,000,000 m) / (2,360,590 s) = 1023.049322 m/s [Round to 1023 m/s] • Convert m/s to km / s: 1023 m/s / 1000 m/km = 1.023 km/s

Huzzah! This value of Moon’s mean orbital velocity is close to NASA’s value of 1.022 meters/second.

Here there be a summary of our thrashing about calculating Moon’s mean orbital velocity, orbital velocity at perigee, and orbital velocity at apogee.

Our calculated orbital velocity NASA: Moon Fact Sheet % difference Notes

Mean: 1025 m/s = 1.025 km/s  Mean: 1.022 km/s 0.3% Close Mean: 1023 m/s = 1.023 km/s  Mean: 1.022 km/s 0.1% Close Perigee: 1082 m/s = 1.082 m/s  Perigee: 1.082 km/s 0% ☺ Apogee: 969.8 m/s = 0.9698 km/s Apogee: 0.970 km/s 0.02% Very close

GMM()+  Circular orbit, radius = 384,400,000 m v = 12 r  Circular orbit, radius = 384,400,000 m v = C / T 21 v= G() M + M −  Elliptical orbit at perigee: a = 384,400,000 m, r = 363,300,000 m 12 ra

 Elliptical orbit at apogee: a = 384,400,000 m, r = 405,500,000 m

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Perhaps you know that an accurate way to calculate the mean orbital velocity of an elliptical orbit is to calculate the perimeter of the orbit and then divide by the orbital period. [We will use ‘circumference’ for circles and ‘perimeter’ for ellipses.]

• Orbital velocity = perimeter of elliptical orbit / orbital period

We have avoided this method because calculating the perimeter of an ellipse requires calculus. If you are calculus savvy, cruise the Internet using the search key ‘perimeter of an ellipse using calculus’. Enjoy.

If you are not calculus savvy, take heart. You can use algebraic approximations to calculate an accurate approximate value of the perimeter of an ellipse. All you b need are the ellipse’s semimajor axis a and semiminor axis b, both shown in the diagram over yonder →. a

NASA’s Moon Fact Sheet provides Moon’s semimajor axis a, but not Moon’s semiminor axis b. Fortunately, it also provides Moon’s orbital eccentricity e. Use a = semimajor axis b = semiminor axis b=− a1 e 2 this equation to calculate Moon’s semiminor axis b: NASA: Moon Fact Sheet: • Semimajor axis: a = 0.3844  106 km [a = 384,400 km] • Eccentricity: e = 0.05490 [Wikipedia: "The orbit of the Moon is distinctly elliptical with an average eccentricity of 0.054900489.” The eccentricity varies. We will assume 4 significant digits (0.0540) in calculations.]

Calculate • Semiminor axis: b =(384,400 km) 1 − 0.054902 = 383,800 km TI-84: 384400 √(1 – 0.054902) Enter Answer: 383820.2701 [Round to 383800]

Armed with Moon’s semimajor axis (a = 384,400 km) and semiminor axis (b = 383,800 km), we embarked on a voyage of discovery into cyberspace using search key ‘perimeter of an ellipse’ and found several sites. Hallelujah! We found a most excellent site:

• Perimeter of Ellipse – mathisfun.com https://www.mathsisfun.com/geometry/ellipse-perimeter.html

Perimeter of Ellipse – mathisfun.com presents five ways to calculate the perimeter of an ellipse. Serendipity! It also provides an interactive Ellipse Perimeter Calculations Tool you can use to calculate the perimeter of an ellipse – no math required!

Scroll down to Ellipse Perimeter Calculations Tool. After entering data

• In box a: enter Moon’s semimajor axis 384400 (no comma, no km). a: 384400 • In box b: enter Moon’s semiminor axis 383800 (no comma, no km). b: 383800

Oops! We are running out of room on this page. We continue next page.

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After we entered Moon’s semimajor axis in box a: and semiminor axis in box b: Ellipse Perimeter Calculations Tool presented five approximations to Moon’s perimeter.

Our entries, 384400 and 383800, are in kilometers (km), so the Approx. 1: 2413372.213 perimeter is in kilometers (km). 384400 and 383800 each have Approx. 2: 2413371.845 4 significant digits. Round the approximations to 4 significant Approx. 3: 2413371.845 digits, add commas for readability, and km for kilometers: Series 1 (100 terms): 2413371.845 • Perimeter of Moon’s orbit = 2,413,000 km Series 2 (7 terms): 2413371.845

Orbital velocity calculation:

• Perimeter of orbit = 2,413,000 km [4 significant digits] • Sidereal orbit period = 27.3217 days  86,400 s/day = 2,360,590 s [6 significant digits] [27.3217 has 6 significant digits. 86,400 is exact. Therefore 2,360,590 has 6 significant digits. • Mean orbital velocity = perimeter of orbit / orbital period • Mean orbital velocity = 2,413,000 km / 2,360,590 s = 1.022202077 km/s [Round to 1.022 km/s]

About rounding and significant digits, also called significant figures: • Rounding and Significant Digits  https://www.purplemath.com/modules/rounding2.htm  • Significant Figures – Purdue University http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch1/sigfigs.html

Summary: Moon’s mean orbital velocity, orbital velocity at perigee, and orbital velocity at apogee.

Our calculated orbital velocity NASA: Moon Fact Sheet % difference Notes

Mean: 1025 m/s = 1.025 km/s  Mean: 1.022 km/s 0.3% Close Mean: 1023 m/s = 1.023 km/s  Mean: 1.022 km/s 0.1% Close Mean: 1.022 km/s  Mean: 1.022 km/s 0% ☺ Perigee: 1082 m/s = 1.082 m/s  Perigee: 1.082 km/s 0% ☺ Apogee: 969.8 m/s = 0.9698 km/s Apogee: 0.970 km/s 0.02% Very close

GMM()+  Circular orbit, radius = 384,400,000 m v = 12 r  Circular orbit, radius = 384,400,000 m v = C / T  Elliptical orbit: v = perimeter of orbit / orbital period 21 v= G() M + M −  Elliptical orbit at perigee: a = 384,400,000 m, r = 363,300,000 m 12 ra

 Elliptical orbit at apogee: a = 384,400,000 m, r = 405,500,000 m

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Postscript: Perimeter of Ellipse – mathisfun.com presents five ways to calculate the perimeter of an ellipse: Approximation 1, Approximation 2, Approximation 3, Infinite Series 1 and Infinite Series 2. We will explore Approximation 2 and Approximation 3, These approximations were created by mathematician Srinivasa Ramanujan https://en.wikipedia.org/wiki/Srinivasa_Ramanujan.

Perimeter of Ellipse – mathisfun.com Approximation 2

a = semimajor axis. Moon: 384,400 km perimeter: p 3( a + b ) − (3 a + b )( a + 3 b ) b= semiminor axis. Moon: 383,800 km

Calculate perimeter p in kilometers:  p  3(384, 400km + 383,800 km ) − (3  384, 400 km + 383,800 km) (384, 400 km + 3  383,800 km ) = 2, 413,000 km TI-84: π(3*(384400+383800)−√((3*384400+383800)(384400+3*383800))) Enter Ans: 2413371.845 [2413000]

Perimeter of Ellipse – mathisfun.com Approximation 3

()ab− 2 3h a = semimajor axis. Moon: 384,400 km h = Perimeter: p( a + b ) 1 + ()ab+ 2 10+− 4 3h b= semiminor axis. Moon: 383,800 km

(384,400 km− 383,800 km )2 h = =6.100  10−7 (384,400 km+ 383,800 km )2

TI-84: (384400 − 383800)2 / (384400 + 383800)2 Enter Answer: 6.100337952E-7 [Round to 6.100E-7]

3 6.100 10−7 p (384,400 km + 383,800 km ) 1 + = 2,413,000 km 10+ 4 − 3  6.100  10−7

TI-84: π(384400+383800)(1+3*6.100E-7/(10+√(4–3*6.100E-7))) Enter Ans 2413371.845 [Round to 2413000]

We are happy to report: Approximation 2 and Approximation 3 values of the perimeter of Moon’s orbit calculated by Perimeter of Ellipse’s Ellipse Perimeter Calculations Tool and the values calculated by our TI-84 – are equal.

Perimeter of Ellipse Perimeter TI-84 Round to 4 significant digits and Moon’s orbit Calculations Tool insert comas Approximation 2 2413371.845 km 2413371.845 km 2,413,000 km [4 significant digits] Approximation 3 2413371.845 km 2413371.845 km 2,413,000 km [4 significant digits]

 Perimeter of Moon’s orbit, rounded to 4 significant digits = 2,413,000 km = 2,413,000,000 m 

The end of this eBook. Reality expands to fill the available fantasies.

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