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Home , Circular orbit, Escape velocity, Orbital period, Orbital speed

Chapter 13 Newton’s Universal Law of

Today: • Force • Energy • Circular

Weds: • Kepler’s Laws mm • Types of Orbits FrG 12ˆ • Escape 12r2 12 • Tidal Forces •

Sun at Center Orbits are Circular Tycho Brahe 1546-1601

Tycho was the greatest observational astronomer of his time. Tycho did not believe in the Copernican model because of the lack of observational parallax. He didn’t believe that the Moved. The Rejection of the Copernican Heliocentric Model Tycho Brahe 1546-1601

Kepler worked for Tycho as his mathematician. Kepler derived his laws of planetary motion from Tycho’s observational data. Kepler’s Laws are thus empirical - based on observation and not Theory. Kepler’s 3 Laws of Planetary Motion

1: The of each about the is an with the sun at one .

2. Each planet moves so that it sweeps out equal areas in equal times.

3. The square of the of any planet is proportional to the 2 cube of the semimajor axis of the T1 elliptical orbit. 3  constant r1 Planet Orbits are Elliptical "The next question was - what makes go around the sun? At the time of Kepler some people answered this problem by saying that there were angels behind them beating their wings and pushing the planets around an orbit. As you will see, the answer is not very far from the truth. The only difference is that the angels sit in a different direction and their wings push inward." -Richard Feynman Man of the Millennium Sir Issac Newton

(1642 -1727) Galaxy

Orbital of : 220 km/s Orbital Period: 225 Million : 48 km/s : 35 km/s Earth: 30 km/s : 24 km/s : 13 km/s : 5 km/s Although the is always  lit from the Sun, we see different amounts of the lit portion from Earth depending on where the Moon is located in its -long of Moon: ~ 1 km/s orbit. Gravitational Force is Universal

The same force that makes the apple fall to Earth, causes the moon to fall around the Earth. Universal Law of Gravity 1687 Every particle in the Universe attracts every other particle with a force along a line joining them. The force is directly proportional to the product to their and inversely proportional to the square of the between them. M d m mM F ~ d 2 Measuring G: Cavendish 1798 mM GmM F ~ 2 F  d d 2 Nm2 Gx 6.67 1011 kg 2

G is the same everywhere in the Universe. G’s small size is a measure of relative strength of gravity. By comparison, the proportionality constant for the electric force is k~109!! Universal Law of Gravity mm FrG 12ˆ 12r2 12

Minus because of the direction of the unit vector. Attractive Central Forces are negative! Ignore when finding magnitude! The sign is critical for solvng Energy Problems. Nm2 Gx 6.67 1011 kg 2 Gravity: Inverse Square Law

GmM F  d 2 Gravitational Force INSIDE the Earth How would the force of gravity and the due to gravity change as you fell through a hole in the Earth? What would your motion be? Assume you jump from rest. Ignore Air Resistance. Gravitational Force INSIDE the Earth Inside the Earth the Gravitational Force is Linear. Acceleration decreases as you fall to the center (where your speed is the greatest) and then the acceleration increases but in the opposite direction, slowing you down to a stop at the other end…but then you would fall back in again, bouncing back and forth forever! Earth-Moon Gravity Calculate the magnitude of the force of gravity between the Earth and the Moon. The distance between the Earth and Moon centers is 3.84x108m GmM F  EM d 2

6.673x 1011 Nm 2 / kg 2 5.98 x 10 24 kg 7.35 x 10 22 kg FEM  2 3.84xm 108 

20 FEM  2.01 x 10 N Earth-Moon Gravity Calculate the acceleration of the Earth due to the Earth-Moon gravitational interaction.

FEM aE  mE

2.01xN 1020  5.98x 1024 kg

52 aE  3.33 x 10 m / s Earth-Moon Gravity Calculate the acceleration of the Moon due to the Earth-Moon gravitational interaction.

FEM aM  mM

2.01xN 1020  7.35x 1022 kg

32 aM  2.73 x 10 m / s Earth-Moon Gravity The acceleration of gravity at the Moon due to the Earth is:

32 aM  2.73 x 10 m / s

The acceleration of gravity at the Earth due to the moon is:

52 aE  3.33 x 10 m / s

Why the difference?

FORCE is the same. Acceleration is NOT!!! BECAUSE MASSES ARE DIFFERENT! Force is not Acceleration!

FFEarth on Moon Moon on Earth

The forces are equal but the are not! Superposition of Gravitational Force Vectors

mm12 What is the net force Fr12G 2 ˆ 12 on the at the r origin? Problem Solving Strategy 1. Draw FBD 2. Calculate Force Magnitudes 3. Resolve into components 4. Add components 5. Express as ijk vector 6. Find magnitude and direction of resultant force. Problem Finding little g Calculate the acceleration of gravity acting on you at the surface of the Earth. What is g? Gm M you E F m a F  2 you RE

Source of the Force Response to the Force Gmyou M E This is your 2  mayou WEIGHT! RE

Independent of your mass! GM E This is why a rock and a  2 feather fall with the RE same acceleration! Finding little g Calculate the acceleration of gravity acting on you at the surface of the Earth. What is g?

Gmyou M E F myou a F  2 RE GM Source of the Force E Reaction to the Force a  2 RE

6.673x 1011 Nm 2 / kg 2 5.98 x 10 24 kg a  2 6.38xm 106 

a 9.81 m / = g! In general, g for any Planet:

GM g  planet planet R2 planet g field is an acceleration field The gravitational field describes the “effect” that any object has on the empty space around itself in terms of the force that would be present if a object were somewhere in that space F GM grg   ˆ mr2 In Sum, Gravitational Force & Field

Gm12 m FGG rˆ F m g r 2 2 F GM grg   ˆ mr2 Gravity field is the force that +1kg would feel if it were placed at this location. It is an acceleration field. The gravitational Field of the Earth at the surface of the Earth is simply, g: GM gE 9.81 N / kg  9.81 m / s2 r2 42 Electric Force and Field

qq Fr k 12ˆ 12 r2 F  qE Two flavors of charge (+/-)

kq Electric Field E due to q : E  1 1 r2

Curvature of Earth Curvature of the Earth: Every 8000 m, the Earth curves by 5 meters!

If you threw the ball at 8000 m/s off the surface of the Earth (and there were no buildings or mountains in the way) how far would it travel in the vertical direction in 1 second? 1 y  gt2~ 5 m / s 2  (1 s 2 )  5 m 2

The ball will achieve orbit. Orbital Velocity If you can throw a ball at 8000m/s, the Earth curves away from it so that the ball continually falls in around the Earth – it is in orbit around the Earth!

Ignoring air resistance.

Above the Projectile Motion/Orbital Motion

Projectile Motion is Orbital motion that hits the Earth! Orbital Motion| &

8km/s: Between 8 & 11.2 km/s: Elliptical orbit 11.2 km/s: Escape Earth 42.5 km/s: Escape Solar System! Orbits

Circular Orbit Elliptical Orbit Circular Orbits

As the ball falls around the Earth in a circular orbit, does the acceleration due to gravity change its orbital speed? It only changes its direction!

Ignoring air resistance.

Above the atmosphere Circular Orbital Velocity The force of gravity is perpendicular to the velocity of the ball so it doesn’t speed it up – it changes only the direction of the ball. Gravity provides a centripetal acceleration the keeps it in a circle! The PE and KE are the same throughout the orbit. Since F is parallel to r, is also conserved.

v Elliptical Orbits As the falls around the Earth in an elliptical orbit, does the acceleration due to gravity change its orbital speed?

There is a component of force (and A acceleration) in the direction of motion! Gravity changes the satellite’s speed when in elliptical orbits. Mechanical Energy is conserved but KE and PE change throughout the orbit. Is angular momentum conserved? Why? Where is the speed greatest- A or B? B Satellite Orbits Circular Orbit Speed With Increasing Altitude g and v Above the Earth’s Surface GMm If an object is some F  distance h above the r2 Earth’s surface, r GM g  E becomes RE + h 2 RhE  

v2 The tangential speed of an  g r object is its orbital speed v2 GM and is given by the  E Rh 2 centripetal acceleration, g: E RhE   GM Orbital speed decreases with v  E increasing altitude! ()RhE  Orbit Question Find the orbital speed of a satellite 200 km above the Earth. 24 6 Assume a circular orbit. MEE5.97 x 10 kg , R 6.38 x 10 m v2 msE M G F  F ms a a  r 2 r 2 msE M G v MGE Fm2 s v  rr RhE  What is this?

(5.97x 1024 kg )(6.67 x 10 11 Nm 2 / kg 2 ) v  6.58xm 106 Notice that this 3 v 7.78 x 10 m / s is less 8km/s! Orbit Question What is the period of a satellite orbiting 200 km above the Earth? 24 6 Assume a circular orbit. MEE5.97 x 10 kg , R 6.38 x 10 m

2 r If you don’t know the velocity: v  GMm v2  Fm r2 r 2 r (2 r / )2   m v r 2 (6.58xm 106 )  4 2 3 23r 7.78x 10 m / s GM rd  5314s  88min Kepler’s 3 ! Period increases with r! Orbital Sum…. with increasing altitude:

GM • g, acceleration decreases g  E r 2

MG • v, orbital speed decreases v  E r

4 2 23r •T, orbital period increases GM Grav. – Work Since the Force is Conservative, the Work is independent of path.

• The work done by F along any radial segment is dWFr  d  F() r dr • The work done by a force that is perpendicular to the displacement is 0

rf • The total work is W F( r ) dr r i Recall that the work done by a conservative force on an object is: W  PE   KE (As a rock falls, it loses PE but gains KE!!!) Energy • As a particle moves from A to B, its gravitational potential energy changes by

rf U  Ufi  U   W   F( r ) dr r i

• Choose the zero for the gravitational potential energy

where the force is zero: Ui = 0 where ri = GM m Ur() E r

This is valid only for r ≥ RE and not valid for r < RE U is negative because of the choice of Ui Work Done by a Varying Force: Gravity

The interplanetary probe is attracted to the sun by a force given by: 1.3x 1022 F  12 r 2 The negative sign indicates that the force is attractive. This is because of Probe the way that the unit vectors are defined. With the origin located at the sun and the radial vector pointing towards the probe, the force of gravity acting on the probe is in Sun the negative direction. Work Done by a Varying Force: Gravity

The probe is moving away from the sun so the work done ON the probe BY the sun is slowing it down. Thus, the work should be negative.

22 2.3x 1011 1.3x 10 W dx 1.5x 1011 2 x

2.3x 1011 1.3x 1022 ()x1 1.5x 1011 1.3x 1022 F  10 12 r 2 3xJ 10 Attractive force versus distance for interplanetary probe. The area under the curve is negative since curve is below x-axis. Systems with Three or More Particles Particles STATIC

• The total gravitational potential energy of the system is the sum over all pairs of particles

• Gravitational potential energy Mm obeys the superposition UG principle r

UUUUtotal 12  13  23 • The absolute value of U total m m m m m m represents the work needed to  G 1 2  1 3  2 3 separate the particles by an r12 r 13 r 23 infinite distance Problem Find the potential energy of this ensemble. Assume the masses are equal and have mass of 1.00 kg and the separation are 1m.

UUUUtotal 12  13  23 m m m m m m  G 1 2  1 3  2 3 r12 r 13 r 23 Gravitational Potential Energy

. Suppose two masses a

distance r1 apart are released from rest. . How will the small mass move

as r decreases from r1 to r2? . At r1 U is negative. . At r2 |U| is larger and U is still negative, meaning that U has decreased. . As the system loses potential energy, it gains

while conserving Emech. . The smaller mass up as it falls.

Slide 13-45 Problem At the Earth's surface a projectile is launched straight up at a speed of 10.0 km/s. To what height will it rise? Ignore air resistance and the of the Earth.

The height attained is not small compared to the radius of the Earth, so U = mgh does not apply. Use: U = -GmM/r

: KUKUi i  f  f

1 2 GME M p GM E M p Mvpi 0  2 REE R h

h 2.54 107 m Energy and Satellite Motion Circular Orbit Total energy E = K +U: 1 Mm E mv2 G 2 r Rewrite the Kinetic Energy: GMm v2 1 GMm Fm mv2  KE = ½ U r2 r 22r Rewrite the Energy:

GMm GMm GMm In a bound E  E  system, E <0 2rr 2r Orbital Energetics Energy and Satellite Motion Circular Orbit . The figure shows the kinetic, potential, and total energy of a satellite in a circular orbit. . Notice how, for a circular

orbit, Emech = Ug/2. . It requires positive energy in order to lift a satellite into a higher orbit.

GMm E  2r Slide 13-70 Orbit Question What minimum energy does it take to put a 200 kg satellite in orbit 200 km above the Earth? Assume a circular orbit.

24 6 MEE5.98 x 10 kg , R 6.38 x 10 m EEif (from last time) v7.78 x 103 m / s ,   88min 1122GmM GmM mvi  E input  mv f  22rrif The minimum initial speed is 1 1 1 E m()() v22  v  GMm  just the rotational speed of the input f i Earth: 2REE ( R h ) 2 R v E 4.63  102 m s i 86 400 s Substituting in the values: 9 Einput 6.43 10 J Orbital Motion| & Escape Velocity

8km/s: Circular orbit Between 8 & 11.2 km/s: Elliptical orbit 11.2 km/s: Escape Earth 42.5 km/s: Escape Solar System! Escape Speed from Earth An object of mass m is projected upward from the Earth’s surface with an initial speed, vi .Use energy considerations to find the minimum value of the initial speed needed to allow the object to move infinitely far away from the Earth: 0 1122GMEE m GM m mvif  mv  22RrEf

2GM E r  vesc  r RE

2(6.67x 1011 Nm 2 / kg 2 )(5.98 x 10 24 kg ) v esc 11.2 km s 6.37xm 106 Solar System Escape Speeds In General: 2GM v  esc R

Complete escape from an object is not really possible. The gravitational field is infinite and so some gravitational force will always be felt no matter how far away you can get. Cosmic : Conservation of Angular Momentum Why are rotating systems FLAT?

https://www.youtube.com/watch?v=tmNXKqeUtJM Planetary System Formation

https://www.youtube.com/watch?v=3YmeajE- TT8&feature=emb_logo https://www.youtube.com/watch?v =mQAdYWcA7ig The gravitational force between Earth and Moon results in tides

Large spring tides occur when the Sun and Moon are aligned such that they BOTH pull on the Earth’s water.

Smaller neap tides occur when the Sun and Moon are misaligned such that they pull on the Earth’s water in different directions. How many High and Low Tides per day? Earth’s Ocean Tides

Neap Tides Spring Tides Spring

Spring

Neap

Workbook 43 Questions! Which exerts a greater gravitational force on the Earth, the moon or the Sun? Which has a greater effect on the Earth’s tides? Why?

The Sun exerts a greater force on the Earth. It also causes tides, though it has about half the effect as the moon. Tides are due to differences in force on opposite sides of the Earth. The sun is so far away relative to the size of the Earth that the difference in force is not as significant as the difference due to the Moon since it is closer! TOTAL Angular Momentum Orbital and Spin Angular Momentum The TOTAL angular momentum includes both the angular momentum due to revolutions (orbits) and rotations (spin). It is the TOTAL angular momentum that is conserved. Earth- Moon System: Total Angular Momentum is Conserved!

•Earth Rotation Slowing due to friction of ocean on bottom •.0023 s per century: 900 Million yrs ago, Earth day was 18 hrs! •Decrease of Earth’s spin angular momentum, increases the orbital angular momentum of the Moon by increasing the distance, r, in order to keep L conserved! •Earth is slowing down and Moon is moving further away! Bouncing laser beams off the Moon demonstrates that it slowly moving away from the Earth ~0.25 cm/month On Earth we always see the same side of the moon.

Does the Moon Spin? Earth- Moon System: Tidal Lock Moon pulls on Earth and causes tidal bulges in Oceans. Earth pulls on moon causing tidal bulge in Moon such that the CM is off from CG.

Earth’s pull on the moon at its center of gravity produces a torque which rotates it. The moon is ‘tidally locked’ so that it rotates at the same rate at which it revolves, showing Earth only one face. Eventually the Earth's will be identical to the Moon's orbital period. This situation is called synchronous (1:1) rotation. In the distant future (many billions of years from now), the Earth will have a day which is 47 current days long, and the Moon will only be visible from one side of the Earth!! https://www.youtube.com/watch?v=gftT3wHJGtg ‘The 3-Body Problem’

Newton’s Law of Gravity can solve orbits for two-body systems such as the Earth and Sun, resulting in elliptical orbits orbiting the CM of the system. The three-body problem (where more than one body is moving) is much more complicated and, in general, cannot be solved analytically. The orbits that result are chaotic. In fact, chaos theory evolved from attempts to solve the 3-Body Problem. Interplanetary Super Highway Gravitational Potential Contour Map Lagrange Points Lagrange points are locations in space where gravitational forces and the orbital motion of a body balance each other. They were discovered by French mathematician Louis Lagrange in 1772 in his gravitational studies of the 3-body problem: how a third, small body would orbit around two orbiting large ones. The L1 point of the Earth-Sun system affords an uninterrupted view of the sun and is currently home to the Solar and Heliospheric Observatory Satellite (SOHO). The L2 point of the Earth-Sun system is home to the Microwave Anisotropy Probe (MAP). The L1 and L2 points are unstable on a time scale of approximately 23 days, which requires parked at these positions to undergo regular course and attitude corrections. The L4 and L5 points are stable and would be ideal locations for space habitats or solar power stations. L-5 Society Albert Einstein 1916

The Field Equation:

Mass WARPS Space-time Mass grips space by telling it how to curve and space grips mass by telling it how to move! - John Wheeler Proof: Precession of the Perihelion of Mercury Gravitational Frame Dragging Space-Time Twist Black Holes & Worm Holes (RS) . For any mass M, if that mass were compressed to the extent that its radius becomes less than the Schwarzschild radius, then the mass will collapse to a singularity, and anything that passes inside that radius cannot escape. Once inside RS , the arrow of time takes all things to the singularity. (In a broad mathematical sense, a singularity is where the value of a function goes to infinity. In this case, it is a point in space of zero volume with a finite mass. Hence, the mass density and gravitational energy become infinite.) The Schwarzschild radius is given by:

• The space distortion becomes more noticeable around increasingly larger masses. Once the mass density reaches a critical level, a forms and the fabric of space-time is torn. The curvature of space is greatest at the surface of each of the first three objects shown and is finite. The curvature then decreases (not shown) to zero as you move to the center of the object. But the black hole is different. The curvature becomes infinite: The surface has collapsed to a singularity, and the cone extends to infinity. (Note: These diagrams are not to any scale.) Gravity Waves

LISA https://www.youtube.com/watch?v =-zqN1kwKjkQ KEPLER’S 3 LAWS OF PLANETARY MOTION

1: The orbit of each planet about the sun is an ellipse with the sun at one focus.

2. Each planet moves so that it sweeps out equal areas in equal times.

3. The square of the orbital period of any planet is proportional to the 2 cube of the semimajor axis of the T1 elliptical orbit. 3  constant r1 How Does Newton’s Universal Law of Gravity (ULG) Explain Kepler’s Laws of Planetary Motion? • Kepler’s First Law (Orbits are ) - Express F = ma as a second order differential equation in polar coordinates, substitute in F as an inverse square law and the radial solutions are ellipses! (See Wikipedia.com for a simple and elegant solution!) • Kepler’s Second Law (Equal Areas in Equal Time) - Conservation of Angular Momentum leads to it : dA L constant dt2 M p

2 3 • Kepler’s Third Law (T ~ r ) 2 234 – Direct substitution into ULG of the Tr  centripetal acceleration (see book): GMSun FIGURE 13.20

The shaded regions shown have equal areas and represent the same time interval. Mercury: 48 km/s Venus: 35 km/s Earth: 30 km/s Mars: 24 km/s Jupiter: 13 km/s Neptune: 5 km/s https://www.youtube.com/watch?v=2aNC9kv0Ukc Types of Orbits SATELLITE ORBITS GLOBAL GEOSTATIONARY SATELLITE COVERAGE

USA USA

Euro

Japan

USSR China SUN-SYNCHRONOUS NEAR POLAR ORBITS With an orbital period of about 100 minutes, these satellites will complete slightly more than 14 orbits in a single day. Orbital Energetics Energy and Satellite Motion Circular Orbit . The figure shows the steps involved to lift a satellite to a higher circular orbit. . The first kick increases K

without increasing Ug, so K is not Ug/2, and the orbit is elliptical. . The satellite then slows down as r increases. . The second kick increases

K again so that K = Ug/2, and the orbit is circular. Slide 13-72 FIGURE 13.19

The transfer ellipse has its perihelion at Earth’s orbit and aphelion at Mars’ orbit. Discussion Session: Higher Orbit Problem

A 50-kg satellite circles the Earth in an orbit with a period of 120 min. What minimum energy is required to change the orbit to another circular orbit with a period of 180 min? 1071 operational satellites in orbit around the Earth. 50 percent of which were launched by the United States.Half of that 1071 are in Low-Earth Orbit, just a few hundred kilometers above the surface.

NASA EARTH OBSERVATION https://www.youtube.com/watch?v=O64KM4GuRPk

Orbiting Space Trash Man-made debris orbits at a speed of roughly 17,500 miles/ (28,000 km/h)!

More than 4,000 satellites have been launched into space since 1957. All that activity has led to large amounts of space trash. More than 13,000 objects that are at least three to four inches (seven to ten centimeters) wide. Of those objects, only 600 to 700 are still in use. 95 percent of everything up there that the United States is tracking is trash. There are millions of smaller parts that are too small to track.

2007 China ASAT Test

2,000 pieces of debris > 5 cm Increased LEO debris by 22% Increased the risk of debris collision by 15% No international laws restricting the testing of ground based anti-satellite systems! https://www.youtube.com/watch?v=eYVsVRgiS0w

Orbiting Space Trash Fast Trash Go Boom

Australia, in 1979. Orbiting Space Trash What Goes Up Must Come Down

This is the main propellant tank of the crashed onto second stage of a Delta 2 Australia in 1979. which landed near Georgetown, TX, on 22 January 1997. This approximately 250 kg tank is primarily a stainless steel structure and survived reentry relatively intact. Nuclear Power in Space

Our Spaceship Earth

One island in one ocean...from space “...we’re all aboard a little spaceship called Earth” - Bucky Fuller Orbital Energetics Energy and Satellite Motion Circular Orbit . The figure shows the steps involved to lift a satellite to a higher circular orbit. . The first kick increases K

without increasing Ug, so K is not Ug/2, and the orbit is elliptical. . The satellite then slows down as r increases. . The second kick increases

K again so that K = Ug/2, and the orbit is circular. Slide 13-72 Discussion Session: Higher Orbit Problem

A 50-kg satellite circles the Earth in an orbit with a period of 120 min. What minimum energy is required to change the orbit to another circular orbit with a period of 180 min? 5.5 Satellites in Circular Orbits What altitude for a ? 2 r 3 2 T  24 T  GME

Global Positioning System