1. Newtonian Gravitation

2. F = G m1m r2
In 1686, Newton applied the first law to the moon, surmising that a force must be acting on the moon to keep it in a circular orbit around the earth, otherwise it would move in a liner fashion. The force must be the same as that which keeps apples falling to the earth. Newton’s ideas were used by Halley to predict the return in 1758 of Halley’s comet, and in 1860, astronomers were able to discover Neptune by calculating its effect on Uranus.

3. The mass of an object will act as if located at the center of its radius in terms of calculating the gravitational force.
Basically, any two objects separated in space will exert forces on each other based upon mass and distance between them. These forces will be equal in magnitude but opposite in direction. At the surface of the earth, F = mg will still apply, but when altitude changes dramatically, acceleration will change.

4. A man takes his dog for a walk
A man takes his dog for a walk. Find the force of gravity between the 105 kg man and his 11.2 kg dog when they are separated by 1.00 m and 10.0 m.
F = (6.67x10-11)(105 )(11.2)/12 =
7.84 x 10-8 N
F = 6.67x10-11)(105 )(11.2)/10 2 =
7.84x N
Force exerted by earth on man is 1030 N and dog is 110 N

5. F = mgh = G m1ME = GME (RE + h)2 (RE + h)2
Find the acceleration of gravity in the space shuttle orbiting at 100. km above the earth.
gh = 6.67 x 10-11) ( 5.97 x 1024kg)
(6.37x m)2
= 9.51 m/s2

6. Gravitational PE (U)
As we move further away from the earth, the force of gravitational attraction lessens. If PE equals mgh, how can PE be determined in other than near earth situations.
U = - G m1m2 r
Can be used to find U wrt the earth or any two separated masses.

7. Determine the potential energy of a 12 kg meteorite one earth radius above the earth.
U = - G m1m2 = x 1015
r x 107 m
U = x 108 J
What about at the surface of the earth?
r x 106 m
U = x 108 J (which equals mgh if h is radius of the earth)

8. Energy conservation
Mechanical energy for astronomical systems should follow the same rules as near earth; specifically, energy must be conserved.
Etot = K + U = ½ mv2 – G m1m2 r
We now consider an asteroid with vo = 0 at infinity, falling directly to the earth. Etot = 0

9. Etot = ½ mvf2 – G m1m2 r
where final dist = RE
vf = (2GME/RE)1/2 notice that mass of the object is irrelevant, and the speed will be:
vf = [ (2)(6.67x 10-11)(5.97 x 1024)/6.37x106]1/2
vf = 11,200 m/s (25,000 mi/h)
The converse of this situation, an object leaving the earth and going to infinity where it will have E = 0; would require a speed we call escape velocity, ve = 11,200 m/s

10. Useful data