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Algebra Based Physics

Newton's Law of Universal Gravitation

2015-11-30

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Newton's Law of Universal Gravitation

Click on the topic to go to that section · Gravitational Force

·

· Surface · Gravitational Field in Space Gravitational Force · Orbital Motion

· Kepler's Third Law of Motion

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https://www.njctl.org/video/?v=IP_u0xQvP04 Contents Slide 5 / 57 Slide 6 / 57

Newton’s Law of Universal Gravitation Newton’s Law of Universal Gravitation Newton connected the idea that objects, like apples, fall towards the center of with the idea that the It has been well around Earth...it's known since ancient also falling towards the times that Earth is a center of Earth. and objects that are near the The moon just stays in surface tend fall circular motion since it has down. a velocity perpendicular to its .

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Newton’s Law of Universal Gravitation

Newton concluded that all objects attract one another with a G = 6.67 x 10-11 N-m2 /kg2 "gravitational force". The magnitude of the gravitational force decreases as the centers of the increases in distance. In 1798, MORE Gravitational attraction measured G using a torsion beam M1 M2 balance. He did not initially set out to measure G, he was instead trying r to measure the of the Earth.

M1 LESS Gravitational attraction M2 r

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Newton’s Law of Universal Gravitation Newton’s Law of Universal Gravitation

Mathematically, the magnitude of the gravitational force decreases with the inverse of the square of the distance The direction of the force is along the line connecting the between the centers of the masses and in proportion to the centers of the two masses. Each feels a force of product of the masses. attraction towards the other mass...along that line.

r

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Newton’s Law of Universal Gravitation 1 What is the magnitude of the gravitational force between two 1 kg objects which are located 1.0 m apart?

Newton's third law tells us that the force on each mass is A 3.3 x 10-11 N equal. B 1.7 x 10-11 N That means that if I drop a pen, the force of Earth pulling the C 2.7 x 10-10 N pen down is equal to the force of the pen pulling Earth up. D 6.7 x 10-11 N However, since the mass of Earth is so much larger, that force causes the pen to accelerate down, while the movement of Earth up is completely unmeasurable.

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1 What is the magnitude of the gravitational 2 What is the magnitude of the gravitational force between two 1 kg objects which are force acting on a 4.0 kg object which is 1.0 m located 1.0 m apart? from a 1.0 kg object?

-11 A 3.3 x 10-11 N A 3.3 x 10 N B 1.7 x 10-11 N B 1.7 x 10-11 N C 2.7 x 10-10 N C 2.7 x 10-10 N D 6.7 x 10-11 N D 6.7 x 10-11 N

Answer D

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2 What is the magnitude of the gravitational 3 What is the magnitude of the gravitational force acting on a 4.0 kg object which is 1.0 m force acting on a 1.0 kg object which is 1.0 m from a 1.0 kg object? from a 4.0 kg object?

A 3.3 x 10-11 N A 3.3 x 10-11 N B 1.7 x 10-11 N B 1.7 x 10-11 N -10 C 2.7 x 10-10 N C 2.7 x 10 N D 6.7 x 10-11 N D 6.7 x 10-11 N

Answer C

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3 What is the magnitude of the gravitational 4 What is the magnitude of the gravitational force acting on a 1.0 kg object which is 1.0 m force acting on a 1.0 kg object which is 2.0 m from a 4.0 kg object? from a 4.0 kg object?

A 3.3 x 10-11 N A 3.3 x 10-11 N B 1.7 x 10-11 N B 1.7 x 10-11 N C 2.7 x 10-10 N C 2.7 x 10-10 N D 6.7 x 10-11 N D 6.7 x 10-11 N

Answer C

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4 What is the magnitude of the gravitational 5 What is the magnitude of the gravitational force force acting on a 1.0 kg object which is 2.0 m between Earth and its moon? from a 4.0 kg object? r = 3.8 x 108m 24 mEarth = 6.0 x 10 kg -11 22 A 3.3 x 10 N mmoon = 7.3 x 10 kg B 1.7 x 10-11 N A 2.0 x 10 18 N C 2.7 x 10-10 N 19 D 6.7 x 10-11 N B 2.0 x 10 N C 2.0 x 1020 N

Answer D D 2.0 x 1021 N

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5 What is the magnitude of the gravitational force 6 What is the magnitude of the gravitational between Earth and its moon? force between Earth and its ? 11 r = 3.8 x 108m r = 1.5 x 10 m 24 24 mEarth = 6.0 x 10 kg mEarth = 6.0 x 10 kg 30 22 msun = 2.0 x 10 kg mmoon = 7.3 x 10 kg -18 A 2.0 x 10 18 N A 3.6 x 10 N B 3.6 x 1019 N B 2.0 x 1019 N C 3.6 x 1021 N

20 Answer C C 2.0 x 10 N D 3.6 x 1022 N D 2.0 x 1021 N [This object is a pull tab]

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6 What is the magnitude of the gravitational force between Earth and its sun? r = 1.5 x 1011 m 24 mEarth = 6.0 x 10 kg 30 msun = 2.0 x 10 kg A 3.6 x 10-18 N B 3.6 x 1019 N C 3.6 x 1021 N * 22

D 3.6 x 10 N Answer D Gravitational Field

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Return to Table of https://www.njctl.org/video/?v=mAC5GoXjJGE https://www.njctl.org/video/?v=p_OteaRhSsk Contents Slide 19 / 57 Slide 20 / 57 *Gravitational Field *Gravitational Field While the force between two objects can always be computed The magnitude of the gravitational field created by an object varies from location to location in space; it depends on the distance from by using the formula for FG ; it's sometimes convenient to consider one mass as creating a gravitational field and the the object and the object's mass. other mass responding to that field. Gravitational field, g, is a vector. It's direction is always towards the object creating the field.

That's the direction of the force that a test mass would experience if placed at that location. In fact, g is the acceleration that a mass would experience if placed at that location in space.

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*Gravitational Field *Gravitational Field 7 Where is the gravitational field the strongest? 7 Where is the gravitational field the strongest? E E B B

D D

A A A Answer

C C [This object is a pull tab]

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8* What happens to the gravitational field if the 8* What happens to the gravitational field if the distance from the center of an object doubles? distance from the center of an object doubles?

A It doubles A It doubles B It quadruples B It quadruples

C It is cut to one half C It is cut to one half Answer D D It is cut to one fourth D It is cut to one fourth

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9* What happens to the gravitational field if the 9* What happens to the gravitational field if the mass of an object doubles? mass of an object doubles?

A It doubles A It doubles B It quadruples B It quadruples C It is cut to one half C It is cut to one half

D It is cut to one fourth D It is cut to one fourth Answer A

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Slide 24 / 57 Slide 25 / 57 * , , , all have a gravitational field...since they all have mass.

That field is largest at the object's surface, where the distance from the center of the object is the smallest...when "r" is the radius of the object.

By the way, only the mass of * Surface Gravity the that's closer to the center of the planet than you are contributes to its R gravitational field. So the field actually gets smaller if you tunnel down below the M surface.

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* 10 Determine the surface . Its mass * 10 Determine the surface gravity of Earth. Its mass is 6.0 x 1024 kg and its radius is 6.4 x 106 m. is 6.0 x 1024 kg and its radius is 6.4 x 106 m. Answer

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11 Determine the surface gravity of Earth's moon. Its 11 Determine the surface gravity of Earth's moon. Its * mass is 7.4 x 1022 kg and its radius is 1.7 x 106 m. * mass is 7.4 x 1022 kg and its radius is 1.7 x 106 m. Answer

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* 12 Determine the surface gravity of Earth's sun. Its * 12 Determine the surface gravity of Earth's sun. Its mass is 2.0 x 1030 kg and its radius is 7.0 x 108 m. mass is 2.0 x 1030 kg and its radius is 7.0 x 108 m. Answer

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13Compute g for the surface of a planet whose radius 13Compute g for the surface of a planet whose radius * is double that of the Earth and whose mass is triple * is double that of the Earth and whose mass is triple that of Earth. that of Earth.

Answer

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* Gravitational field in space

While gravity gets weaker as you get farther from a planet, it never becomes Gravitational Field zero.

in Space There is always some gravitational field present due to every planet, and moon in the universe.

Return to Table of Contents Slide 32 / 57 Slide 33 / 57 Gravitational field in space Gravitational field in space * * The contribution of a planet to the local gravitational field can be The local gravitational field calculated using the same equation we've been using. You just is usually dominated by have to be careful about "r". nearby masses since gravity gets weaker as the If a location, "A", is a height "h" above a planet of radius "R", it is a inverse square of the distance "r" from the planet's center, where r = R + h. distance.

The contribution of a planet R to the local gravitational field can be calculated using the same equation h we've been using. You just M r have to be careful about "r". A

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14 Determine the gravitational field of Earth at a 14 Determine the gravitational field of Earth at a * height of 6.4 x 106 m (1 ). Earth's * height of 6.4 x 106 m (1 Earth radius). Earth's mass is 6.0 x 1024 kg and its radius is 6.4 x 106 m. mass is 6.0 x 1024 kg and its radius is 6.4 x 106 m. Answer

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15 Determine the gravitational field of Earth at a 15 Determine the gravitational field of Earth at a * height 2.88 x 108 m above its surface (the height * height 2.88 x 108 m above its surface (the height of the moon above Earth). of the moon above Earth).

Earth's mass is 6.0 x 1024 kg and its radius is Earth's mass is 6.0 x 1024 kg and its radius is 6.4 x 106 m. 6.4 x 106 m. Answer

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16 The occupants of the International Space Station The International Space Station (ISS) * (ISS) float and appear to be weightless. Determine The International Space Station (ISS) is a research facility, the on- the strength of Earth's gravitational field acting on construction of which began in 1998. The space station is in a astronauts in the ISS. Low Earth Orbit and can be seen from Earth with the naked eye! Earth's mass is 6.0 x 1024 kg and its radius is 6.4 x It orbits at an altitude of approximately 106 m. The ISS is 350km (3.5 x 105m) above the 350 km (190 mi) above the surface of surface of Earth. the Earth, and travels at an average of 27,700 kilometers (17,210 mi) per hour. This means the astronauts see 15 sunrises everyday!

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16 The occupants of the International Space Station 17 How does the gravitational field acting on the * (ISS) float and appear to be weightless. Determine * occupants in the space station compare to the the strength of Earth's gravitational field acting on gravitational field acting on you now? astronauts in the ISS.

Earth's mass is 6.0 x 1024 kg and its radius is 6.4 x 106 m. The ISS is 350km (3.5 x 105m) above the A It's the same surface of Earth. B It's slightly less C It's about half as strong D There is no gravity acting on them Answer

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17 How does the gravitational field acting on the * occupants in the space station compare to the gravitational field acting on you now?

A It's the same B It's slightly less C It's about half as strong D There is no gravity acting on them ** Orbital Motion B Answer

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Return to Table of https://www.njctl.org/video/?v=DaDGZF-rPTY https://www.njctl.org/video/?v=Ah0inrolRgM Contents Slide 40 / 57 Slide 41 / 57 ** Orbital Motion ** Orbital Motion

We've already determined that the The gravitational field will be pointed gravitational field acting on the towards the center of Earth and occupants of the space station, d represents the acceleration that a R and on the space station itself, is mass would experience at that not very different than the force location (regardless of the mass). h r acting on us. Earth ISS a Earth In this case any object would ISS How come they don't fall to Earth? simply fall to Earth.

How could that be avoided? This diagram should look really familiar….

Slide 42 / 57 Slide 43 / 57 Orbital Motion ** Orbital Motion **

If the object has a tangential Here is Newton's own drawing of velocity perpendicular to its a thought experiment where a acceleration, it will go in a cannon on a very high mountain circle. (above the ) shoots a shell with increasing speed, It will keep falling to Earth, but v shown by trajectories for the shell never strike Earth. of D, E, F, and G and finally so a fast that it never falls to earth, but goes into orbit.

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or

v v

a a

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18 Compute g at a distance of 7.3 x 108 m from the 18 Compute g at a distance of 7.3 x 108 m from the ** center of a spherical object whose mass is ** center of a spherical object whose mass is 3.0 x 1027 kg. 3.0 x 1027 kg.

Answer

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19 Use your previous answer to determine the 19 Use your previous answer to determine the ** velocity, both magnitude and direction, for an ** velocity, both magnitude and direction, for an object orbiting at a distance of 7.3 x 108 m from the object orbiting at a distance of 7.3 x 108 m from the center of a spherical object whose mass is center of a spherical object whose mass is 3.0 x 1027 kg. 3.0 x 1027 kg. Answer

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20 Use your previous answer to determine the orbital 20 Use your previous answer to determine the orbital ** period of for an object orbiting at a distance of ** period of for an object orbiting at a distance of 7.3 x 108 m from the center of a spherical object 7.3 x 108 m from the center of a spherical object whose mass is 3.0 x 1027 kg. whose mass is 3.0 x 1027 kg. Answer

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21 Compute g at a height of 59 earth radii above the 21 Compute g at a height of 59 earth radii above the *surface of Earth. *surface of Earth.

Answer

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22 Use your previous answer to determine the 22 Use your previous answer to determine the ** velocity, both magnitude and direction, for an ** velocity, both magnitude and direction, for an

object orbiting at height of 59 RE above the object orbiting at height of 59 RE above the surface of Earth. surface of Earth. Answer

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23 Use your previous answer to determine the 23 Use your previous answer to determine the ** of an object orbiting at height of ** orbital period of an object orbiting at height of

59 RE above the surface of Earth. 59 RE above the surface of Earth. Answer

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Slide 52 / 57 Slide 53 / 57 ** Orbital Motion Now, we can find the relationship between the period, T, and the orbital radius, r, for any orbit.

** Kepler's Third Law of Motion

Return to Table of https://www.njctl.org/video/?v=5vr5aiqDYyY Contents Slide 54 / 57 Slide 55 / 57 ** Kepler's Third Law ** Kepler's Third Law If you know the period (T) of a planet's orbit, you can determine its distance (r) from the sun. Since all planets orbiting the Kepler had noted that the ratio of T2 / r3 yields the same result sun have the same period to for all the planets. That is, the square of the period of any distance ratio, the following is planet's orbit divided by the cube of its distance from the sun true: always yields the same number.

We have now shown why: (4π2 ) / (GM) is a constant; its the same for all orbiting objects, where M is the mass of the object being orbited; it is independent of the object that is orbiting.

T(white)2 T(green)2 = r(white)3 r(green)3 Slide 56 / 57 Slide 56 (Answer) / 57

24 The period of the Moon is 27.3 days and its orbital 24 The period of the Moon is 27.3 days and its orbital ** radius is 3.8 x 108 m. What would be the orbital radius ** radius is 3.8 x 108 m. What would be the orbital radius of an object orbiting Earth with a period of of an object orbiting Earth with a period of 20 days? 20 days? Answer

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25 What is the orbital period (in days) of an 25 What is the orbital period (in days) of an ** unknown object orbiting the sun with an orbital ** unknown object orbiting the sun with an orbital radius of twice that of earth? radius of twice that of earth? Answer

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