Gravitation Questions and Answers

Answer: B

The ratio is \(\sqrt{2}\).

For a body near the earth's surface:

\[ \begin{aligned} v_{e} &= \sqrt{\frac{2GM}{R}}, \qquad v_{o} \\ &= \sqrt{\frac{GM}{R}} \end{aligned} \]

so \(\frac{v_{e}}{v_{o}} = \sqrt{2}\). (About 11.2 km/s against 7.9 km/s.) 2 would be the ratio of the squares, i.e. of the kinetic energies, and \(\frac{1}{\sqrt{2}}\) inverts the ratio.

Answer: A

The total energy is -K.

For a satellite in a circular orbit of radius r, gravity supplies the centripetal force, which gives the kinetic and potential energies:

\[ \begin{aligned} K &= \frac{GMm}{2r}, \quad U = -\frac{GMm}{r} = -2K \\ E &= K + U = -K \end{aligned} \]

The total energy is negative because the satellite is bound to the earth, and its size equals the kinetic energy. The result holds at any height h, with \(r = R + h\). 2K is the size of the potential energy, and \(-\frac{K}{2}\) halves the result.

Answer: B

The force becomes \(\frac{F}{4}\).

Uniform spheres attract as if their masses were at their centres. In contact, the centres are 2r apart; the new separation is 4r, twice as far. Gravitational force varies as the inverse square of the distance:

\[ F' = F\left(\frac{2r}{4r}\right)^{2} = \frac{F}{4} \]

The trap is to take the original distance as r (the radius) instead of 2r, which gives \(\frac{F}{16}\). \(\frac{F}{2}\) forgets to square the ratio.

Answer: B

The potential energy is \(-G\).

The gravitational potential energy of two point masses separated by \(r\), taking zero at infinite separation, is:

\[ \begin{aligned} U &= -\frac{G m_{1} m_{2}}{r} \\ &= -\frac{G \times 1 \times 1}{1} \\ &= -G \end{aligned} \]

The negative sign is essential: gravity is attractive, so work must be done on the pair to separate them to infinity, where \(U = 0\). \(G\) has the right size but the wrong sign; \(-\frac{G}{2}\) and \(-\frac{G}{4}\) would need separations of 2 m and 4 m.

Answer: A

The escape velocity is \(\sqrt{2gr}\).

Escape needs kinetic energy enough to bring the gravitational potential energy up to zero:

\[ \begin{aligned} \tfrac{1}{2}mv_{e}^{2} &= \frac{GMm}{r} \\ v_{e} &= \sqrt{\frac{2GM}{r}} \end{aligned} \]

On the surface \(g = \frac{GM}{r^{2}}\), so \(GM = gr^{2}\) and \(v_{e} = \sqrt{2gr}\). (The planet's mass is called m in the question; the result does not depend on the escaping body's mass.)

\(\sqrt{gr}\) is the orbital speed just above the surface, smaller by \(\sqrt{2}\). \(\sqrt{2g/r}\) has the wrong units, and \(gr^{2}\) and \(2gr^{2}\) are not even speeds.

Answer: B

The astronaut is weightless because both the astronaut and the satellite are in free fall towards earth.

Gravity at the height of an orbiting satellite is still nearly as strong as at the surface. But the satellite and everything in it fall towards the earth with the same acceleration, so the floor exerts no normal force on the astronaut, and it is that contact force which we feel as weight.

So the explanation is not that gravity is small (in any direction) there, nor anything to do with the moon's pull.

Answer: D

The height is 2R.

Gravity at height \(h\) is \(g' = g\left(\frac{R}{R + h}\right)^{2}\).

\[ \begin{aligned} \left(\frac{R}{R + h}\right)^{2} &= \frac{1}{9} \\ \frac{R}{R + h} &= \frac{1}{3} \\ R + h &= 3R \Rightarrow h = 2R \end{aligned} \]

The trap is 3R, which is the distance from the earth's centre, not the height above the surface. The small-height formula \(g' = g(1 - \frac{2h}{R})\) cannot be used here, since \(h\) is not small compared with \(R\).

Answer: C

The ratio is 16 : 27.

Escape velocity is \(v_{e} = \sqrt{2gR}\), so \(g = \frac{v_{e}^{2}}{2R}\).

\[ \begin{aligned} \frac{g_{A}}{g_{B}} &= \left(\frac{v_{A}}{v_{B}}\right)^{2}\frac{R_{B}}{R_{A}} \\ &= \frac{4}{9} \times \frac{4}{3} = \frac{16}{27} \end{aligned} \]

The option 4 : 9 squares the velocity ratio but ignores the radii. Note that the radius ratio enters inverted, because \(g\) is divided by \(R\).

Answer: C

The ratio is 1 : \(\sqrt{2}\).

Escape velocity is \(v = \sqrt{\frac{2GM}{R}}\). With \(M = \frac{4}{3}\pi R^{3} \rho\):

\[ \begin{aligned} v &= R\sqrt{\tfrac{8}{3}\pi G \rho} \propto R\sqrt{\rho} \\ \frac{v_{1}}{v_{2}} &= \frac{1}{2} \times \sqrt{\frac{2}{1}} = \frac{1}{\sqrt{2}} \end{aligned} \]

Using the density ratio without the square root gives 1 : 1, and ignoring the density altogether gives 1 : 2. The velocity depends on \(\sqrt{\rho}\) but on the first power of \(R\), which is where the other ratios go wrong.

Answer: B

The ratio is 3 : 2.

At depth d below the surface, \(g_{d} = g\left(1 - \frac{d}{R}\right)\).

\[ \begin{aligned} g_{R/2} &= g\left(1 - \tfrac{1}{2}\right) = \tfrac{g}{2} \\ g_{2R/3} &= g\left(1 - \tfrac{2}{3}\right) = \tfrac{g}{3} \\ \tfrac{1}{2} : \tfrac{1}{3} &= 3 : 2 \end{aligned} \]

Gravity decreases with depth, so the shallower point has the larger g. 2 : 1 and 3 : 1 compare each value with the surface, not with each other.