Page 121 - Schaum's Outline of Theory and Problems of Applied Physics

106 CIRCULAR MOTION AND GRAVITATION [CHAP. 9

GRAVITATION
According to Newton’s law of universal gravitation, every body in the universe attracts every other body with a
force that is directly proportional to each of their masses and inversely proportional to the square of the distance
between them. In equation form,

m 1 m 2
Gravitational force = F = G
r 2
where m 1 and m 2 are the masses of any two bodies, r is the distance between them, and G is a constant whose
values in SI and British units are, respectively,

2
SI units: G = 6.67 × 10 −11 N·m /kg 2
2
British units: G = 3.44 × 10 −8 lb·ft /slug 2
A spherical body behaves gravitationally as though its entire mass were concentrated at its center.

SOLVED PROBLEM 9.12

What gravitational force does a 1000-kg lead sphere exert on an identical sphere3maway?

2
2
3
3
(6.67 × 10 −11 N·m /kg )(10 kg)(10 kg)
m 1 m 2 −4
F = G = = 7.4 × 10 N
r 2 (3m) 2
This is less than the force that would result from blowing gently on one of the spheres. Gravitational forces are
usually significant only when at least one of the bodies has a very large mass.
SOLVED PROBLEM 9.13
A girl weighs 128 lb on the earth’s surface. (a) What would she weigh at a height above the earth’s surface
of one earth radius? (b) What would her mass be there?

2
(a) Since the gravitational force the earth exerts on an object a distance r from its center varies as 1/r , the
gravitational force on the girl relative to its value mg at the earth’s surface (where r = r e )is
2

r e
F = mg
r
When the girl is a distance r e above the earth, r = 2r e , and
2 2

r e r e 1
F = mg = w = ( )(128 lb) = 32 lb
r 2r e 4
(b) Her mass of
w 128 lb
m = = = 4 slugs
g 32 ft/s 2
is the same everywhere.

SOLVED PROBLEM 9.14
Find the acceleration of gravity at an altitude of 1000 km.
2
The gravitational force Gm e m/r of the earth on an object of mass m at the distance r from the earth’s center
is equal to the object’s weight mg at that distance, where m e is the earth’s mass. Hence
Gm e m Gm e
= mg g =
r 2 r 2