Problems for Chapter 3

The past problems in the earlier sections of this chapter each used one or two ideas at a time. The three in this section join them up. The first asks how long Mars's moon Phobos stays up in the Martian sky, which takes a circular orbit, the turning of the planet and a little geometry. The second finds the height of a satellite from one sighting of it. The third asks for the mass of a body whose density falls with distance from its centre, from its escape speed.

3.11.1From textbook to olympiad

In the first problem below, Phobos goes round Mars faster than Mars turns, and we are asked how long it stays above the horizon for an observer on Mars's equator. The method of Section S2 takes us from the chapter to the solution in these steps.

1. We start by drawing it. Seen from above Mars's North Pole, the observer is a point on the equator, the horizon is the line through that point at right angles to the radius, and Phobos's orbit is a larger circle round Mars. Phobos is above the horizon on the part of its orbit beyond that line.
2. We name the physics. Phobos's speed round Mars comes from the circular orbit of (S.19). The observer is carried round by Mars's rotation, so what matters is how fast Phobos moves relative to the turning planet, the difference of the two angular speeds.
3. We work backwards from the target. The time above the horizon is the angle of orbit that Phobos must cover relative to the observer, divided by that relative angular speed. The drawing gives the angle, since the horizon line cuts Phobos's orbit where $\cos \alpha =R_{\text{M}}/R_{\text{P}}$.
4. We use every given. Mars's mass and Phobos's distance give Phobos's angular speed. Mars's rotation period gives the observer's.
5. We check a limiting case. If Phobos went round at exactly the rate Mars turns, the relative speed would be zero, and it would hang in the sky for ever, as a satellite does above Earth in a geostationary orbit. Phobos moves faster than Mars turns, so it rises in the west and sets in the east.