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304 CHAPTER 9 Gravitation

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4. Since the acceleration at a planet’s surface is a GM/R ,a 3. According to Kepler’s Third Law, the period must be exactly
larger mass M and a smaller gravitational acceleration a are one year. This is so because both the Earth’s orbit (nearly cir-
possible only because the radius R of Uranus is sufficiently cular; the semimajor axis of a circle is its radius) and the
larger than that of the Earth. comet’s orbit have the same semimajor axis, and both orbit the
same central body, the Sun.
5. At the exact center of the Earth, a particle would be equally
attracted in all directions, and so would experience zero net 4. (D) 4. Kepler’s Third Law states that the square of the period
force. is proportional to the cube of the semimajor axis of the orbit,
so to make the period 8 times as large as the Earth’s period
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6. (B) g. The acceleration at the surface is a GM R , so a would make the cube of the semimajor axis 64 times as large;
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E
E
doubled radius would result in an acceleration one-fourth as 1/3
thus the semimajor axis would be 64 4 times as large as
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large, or g. the Earth–Sun distance.
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Checkup 9.2
Checkup 9.5
1. To determine G by measuring the force between the Earth
1. For a circular orbit, we found that the magnitude of the (nega-
and some known mass, we would also have to know the mass
tive) potential energy is twice the size of the kinetic energy.
of the Earth; we have no independent way of determining the
Thus the potential energy decreases so much for the lower
mass of the Earth.
orbit (it becomes more negative) that the kinetic energy can
2. (A) Yes. If we knew the mass of the mountain (and the spatial increase and energy can be lost to friction.
distribution of such mass), then we could determine the gravi-
2. Yes—our derivation of the law depended only on the central
tational force from the plumb bob’s deflection, and thus G.
nature of the force, not on any particular type of orbit (or even
Checkup 9.3 any particular form of the central force).
3. If we ignore air friction (and the body does not encounter any
1. An orbit that is a circle at the latitude of San Francisco is
obstacles), then the body will escape the Earth’s influence in a
impossible, since the center of every orbit must coincide with
parabolic “orbit,” since the escape velocity provides for zero
the center of the Earth.
net energy. The orbit would be similarly parabolic if we
2. The period is proportional to the 3/2 power of the radius of launched the body at any angle (except straight up, although
the orbit, so for a doubled radius, the period of the Moon that resulting linear path can be considered a special case of
would become 2 3/2 27 days 76 days. the parabola). Ultimately, far from the Earth’s influence, the
3. As in Eq. (9.13), we need only know the period and radius of path would be modified by the Sun.
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the moon’s orbit to determine the mass of the planet. 4. No. The gravitational acceleration is g GM/R , whereas the
4. (C) 30 yr. The period is proportional to the 3/2 power of the escape velocity depends on the gravitational potential energy,
radius of the orbit, so the period of Saturn’s motion is which is proportional to M/R. For example, a body with twice
10 3/2 1 yr 30 yr. the mass and twice the radius of the Earth would have half the
gravitational acceleration at the surface, but would have the
Checkup 9.4 same escape velocity.
5. (C) Hyperbolic; elliptical. Recall that a parabolic orbit is a
1. Kepler’s Second Law would remain valid, since it depends only
zero-energy orbit, where the comet can just barely escape to
on the central nature of the force, and otherwise not on any par-
infinity. The energy of comet II must be positive, since it has
ticular form of the force. Kepler’s Third Law, however, like the
a larger speed (a greater kinetic energy, but the same potential
law of periods, Eq. (9.13), depends on the inverse-square nature
energy as it crosses the Earth’s orbit); we found that a posi-
of the force. If we were to perform a similar derivation to that
tive-energy orbit is a hyperbola. Similarly, the energy of
preceding Eq. (9.13) for an inverse-cube force, we would find
comet III must be negative, since it has a smaller speed; nega-
that the period was proportional to the square of the radius.
tive-energy orbits are ellipses, with a semimajor axis given by
2. As in Eq. (9.18), the speeds vary inversely with the distances,
Eq. (9.24).
so for an aphelion distance twice as large as the perihelion dis-
tance, the speed at aphelion will be half as large as the speed at
perihelion, or will be 20 km/s.