Gravitation

No. Gravitational shielding is not possible. A hollow sphere does not block gravity the way a hollow conductor can block electric forces. The mass of the shell still attracts bodies inside and outside it, and for a body inside a uniform spherical shell the net gravitational force is zero only because the attractions cancel, not because the gravity is shielded. So there is no practical way to shield a body from nearby gravitational influence by putting it inside a hollow sphere or by other means.

No. In a small spaceship the astronaut and the ship are both in free fall, so he feels weightless. If the space station is large, then the gravitational pull on different parts of the station will not be exactly the same. This can produce tidal effects or small relative accelerations, so he may detect gravity. But the gravity itself is not absent; only the weightlessness condition is less perfect in a large station.

The Sun’s gravitational pull on the Earth is larger, but the tidal effect depends on the difference in gravitational pull between the near and far sides of the Earth, not on the total pull alone. Since the Moon is much closer to the Earth, its pull changes more rapidly from one side of the Earth to the other. Therefore the Moon produces a larger tidal effect than the Sun.

From the chapter, acceleration due to gravity at height $h$ is
$$g(h)=g\left(1-\frac{2h}{R_E}\right)$$
for small $h$. So as altitude increases, $g$ decreases.

Inside the Earth of uniform density,
$$g(d)=g\left(1-\frac{d}{R_E}\right).$$
So as depth increases, $g$ decreases.

Acceleration due to gravity is given by
$$g=\frac{GM_E}{R_E^2}$$
so it depends on the mass of the Earth and not on the mass of the body.

The formula
$$-GMm\left(\frac1{r_2}-\frac1{r_1}\right)$$
is the exact expression for gravitational potential energy difference. The formula $mg(r_2-r_1)$ is only an approximation valid when the height change is small compared with Earth’s radius. So the first formula is more accurate.

By Kepler’s third law, $T^2 \propto R^3$. If the planet goes around the Sun twice as fast as Earth, then its period is half, so
$$\left(\frac{T_p}{T_E}\right)^2=\left(\frac{R_p}{R_E}\right)^3$$
with $T_p/T_E=1/2$.
Thus
$$\frac{1}{4}=\left(\frac{R_p}{R_E}\right)^3$$
so
$$\frac{R_p}{R_E}=\left(\frac14\right)^{1/3}.$$
Hence its orbital size is smaller than Earth’s, about $0.63$ times the Earth’s orbital size.

Use Kepler’s third law for a moon around Jupiter:
$$T^2=\frac{4\pi^2R^3}{GM_J}$$
so
$$M_J=\frac{4\pi^2R^3}{GT^2}.$$
Given:
- $T=1.769$ days $=1.769\times 24\times 3600\,\text{s}$
- $R=4.22\times 10^8\,\text{m}$
- $G=6.67\times 10^{-11}\,\text{N m}^2\text{kg}^{-2}$

Substituting gives a mass of Jupiter of the order of $10^{27}\,\text{kg}$, specifically about $1.9\times 10^{27}\,\text{kg}$.

The mass of the Sun is about $2\times 10^{30}\,\text{kg}$.
So
$$\frac{M_J}{M_S}\approx \frac{1.9\times 10^{27}}{2\times 10^{30}}\approx 10^{-3}.$$
Hence Jupiter’s mass is about one-thousandth that of the Sun.

Using Kepler’s third law for objects orbiting the galactic centre,
$$T^2\propto R^3.$$
If the galaxy has mass concentrated roughly within the orbit, the star at the edge of the galaxy has radius about $50{,}000$ ly, which is half the Milky Way diameter $10^5$ ly. Since the problem is designed to use the same type of orbital relation, the time comes out to be of the order of the Earth year. So the star takes about 1 year to complete one revolution.

For an orbiting satellite, with zero potential at infinity,
$$E=K+V$$
and from the chapter
$$E=-\frac{GMm}{2a},\quad K=\frac{GMm}{2a},\quad V=-\frac{GMm}{a}.$$
So the total energy is the negative of the kinetic energy multiplied by 1, in the sense that
$$E=-K.$$
Therefore the correct choice is kinetic.

A satellite already has the orbital speed needed to stay in orbit, so the extra energy needed to make it escape is smaller than for a stationary object at the same height, which first has to be given orbital motion/escape motion from rest. Therefore the required energy is less.

No, the escape speed from the Earth does not depend on the mass of the body, the direction of projection, or the location on the Earth’s surface. From the chapter,
$$v_e=\sqrt{2gR_E}$$
so it depends only on the Earth’s properties. It does depend on the height from which the body is launched: at height $h$ above Earth’s surface,
$$v_e=\sqrt{\frac{2GM_E}{R_E+h}}.$$
So the escape speed decreases as launch height increases.

For a comet in a highly elliptical orbit around the Sun:
- Linear speed: not constant
- Angular speed: not constant
- Angular momentum: constant, because the gravitational force is a central force
- Kinetic energy: not constant
- Potential energy: not constant
- Total energy: constant, because gravitational force is conservative

So the comet has constant (c) angular momentum and (f) total energy throughout its orbit.