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278 CHAPTER 9 Gravitation

✔ Checkup 9.2

QUESTION 1: Why don’t we determine G by measuring the (fairly large) force between
the Earth and a mass of, say, 1 kg?
QUESTION 2: Large mountains produce a (small) deflection of a plumb bob suspended
nearby. Could we use this effect to determine G?
(A) Yes (B) No

Online 9.3 CIRCULAR ORBITS
11
Concept
Tutorial The gravitational force is responsible for holding the Solar System together; it makes
the planets orbit around the Sun, and it makes the satellites orbit around the planets.
Although the mutual gravitational forces of the Sun on a planet and of the planet on the
Sun are of equal magnitudes, the mass of the Sun is more than a thousand times as
large as the mass of even the largest planet, and hence its acceleration is much smaller.
Gravitational force provides
the centripetal acceleration. It is therefore an excellent approximation to regard the Sun as fixed and immovable,
and it then only remains to investigate the motion of the planet. If we designate the
m
masses of the Sun and the planet by M and m, respectively, and their center-to-center
S
v F separation by r, then the magnitude of the gravitational force on the planet is
GM m
S
r F 2 (9.7)
r
M S
This force points toward the center of the Sun; that is, the center of the Sun is the
center of force (see Fig. 9.6). For a particle moving under the influence of such a cen-
tral force, the simplest orbital motion is uniform circular motion, with the gravita-
tional force acting as centripetal force.The motion of the planets in our Solar System
The much more massive is somewhat more complicated than that—as we will see in the next section, the plan-
Sun stays essentially fixed.
ets move along ellipses, instead of circles. However, none of these planetary ellipses
FIGURE 9.6 Circular orbit of a planet deviates very much from a circle, and as a first approximation we can pretend that the
around the Sun. planetary orbits are circles.
By combining the expression (9.7) for the centripetal force with Newton’s Second
Law we can find a relation between the radius of the circular orbit and the speed. If the
2
speed of the planet is v, then the centripetal acceleration is v /r [see Eq. (4.49)], and
the equation of motion, ma F, becomes
mv 2
F (9.8)
r
Consequently,
mv 2 GM m
S (9.9)
r 2
r
We can cancel a factor of m and a factor of 1/r, in this equation, and we obtain

GM S
2
v
r
or
GM S
speed for circular orbit v (9.10)
B r