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Chapter 12

Gravity

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Outline Newton’s law of universal

gravitation Gravitational attraction of spherical

bodies Force of Gravity on a mass m on

the surface of the Earth Force of Gravity on a mass m above

the surface of the Earth

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Newton’s Law of universal gravitation

Newton’s law of universal gravitation: The force of gravity between any two point objects of mass m1 and m2 is attractive and of magnitude

r = the distance between the masses. G = the universal gravitation constant = 6.6710-11 N·m2/kg2.

Direction of the force of gravity: directed along the line connecting the masses.

F Gm1m2

r 2

The force of gravity between two objects forms an action-reaction pair.

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Exercise A man takes his dog for a walk on

a deserted beach. Treating people and dogs as point objects for the moment, find the force of gravity between the 105-kg man and his 11.2-kg dog when they are separated by a distance of (a) 1.00 m and (b) 10.0m.

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Example As part of a daring rescue

attempt, the Millennium Eagle passes between a pair of twin asteroids, as shown. If the mass of the spaceship is 2.50 107 kg and the mass of each asteroid is 3.50 1011 kg, find the net gravitational force exerted on the Millennium Eagle (a) when it is at location A and (b) when it is at location B. Treat the spaceship and the asteroids as if they were point objects.

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Gravitational attraction of spherical bodies (1)

(1) Between a uniform sphere and a point mass: The net force exerted by the sphere on the mass m is the same as if all the mass of the sphere were concentrated at its center. The magnitude of the force

between the mass m and the sphere of mass M is

2r

mMGF

Gravitational attraction of spherical bodies (2)

(2) Between two uniform spheres: The force of gravity between two spheres of finite size is the same as if both were replaced by point masses. For example, the

gravitational force between the Earth and the Moon is

r: center-to-center distance

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2r

MMGF ME

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Force of gravity for a mass m on the surface of the Earth

Force of gravity for a mass m on the surface of the Earth is

Also, force of gravity for a mass m on the surface of the Earth is F = mg.

So, the acceleration due to gravity, g, is

See Appendix C and the back cover of the textbook for Solar System Data.

F GmME

RE2 m

GME

RE2

gGME

RE2 9.81 m/s2

treehugger.com

RE

ME = 5.97x1024 kgRE = 6.37x106 m

m

ME

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Force of gravity for a mass m above the Earth’s surface

If you climb to the top of Mt. Everest, you will be about 5.50 mi above sea level. What is the acceleration due to gravity at this altitude?

1 mi = 1609 m. Answer:

gh GME

RE h 2 9.78 m/s2

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Homework See online homework on

www.masteringphysics.com Hand-written homework:

Ch. 12, p. 409, Problems: #12. Note: For astronomical data, see

the inside of the back cover of your textbook and Appendix C.