- The Hubble Space Telescope (HST) is on a circular,
low-Earth orbit, at an elevation h = 600 km above the
Earth’s surface. What is its orbital period in minutes (to
within 0.1 minute)?
- Kepler's third law
$$
\frac{a^3}{P^2} = \frac{G M}{4 \pi^2}
$$

$$
\frac{(6 \times 10^5+ 6.371 \times 10^8)^3}{P^2} =
\frac{6.674 \times 10^{-11}
6 \times 10^{24}}{4 \pi^2}
$$

$$
P = \frac{1}{60}
\sqrt{\frac{(6 \times 10^{5} + 6.371 \times
10^8)^3 4 \pi^2}{6.674 \times 10^{-11}
5.972 \times 10^{24}}}~\mathrm{min}
$$

$$
= 96.3~\mathrm{minutes}
$$

- For an observer who sees HST pass through the zenith,
how long is HST above the horizon during each orbit in
minutes (to within 0.1 minutes)?

(This problem is a bit more obtuse than I had
intended, so I'll give everyone credit for this
one and it won't be on the exam.)

- Ceres is a dwarf planet orbiting in the Asteroid belt
2.77 AU from the Sun. Like all members of the Asteroid
belt, it is periodically buffeted by the passage of
Jupiter orbiting at 5.20 AU. Use MCeres = 9.43×1020
kilograms, RCeres = 4.77 × 105 meters, MJupiter = 1.90 ×
1027 kilograms and RJupiter = 6.99 × 107 meters. You may
assume that both Ceres and Jupiter follow circular orbits.
Calculate the gravitational acceleration at the surface of
Ceres due to its mass.

$$
g = \frac{G m_{\mathrm{Ceres}}}{r_{\mathrm{Ceres}}^2}
- \frac{G m_{\mathrm{Jupiter}}}{r_{\mathrm{J-C~orbit}}^2}
$$

$$
\frac{6.67 \times 10^{-11} 9.43 \times 10^{20}}
{(4.77 \times 10^{5})^2} -
\frac{6.67 \times 10^{-11} 1.90 \times 10^{27}}
{(7.78 \times 10^{11})^2}
$$

$$
(2.76 \times 10^{-1} - 2.09 \times 10^{-7})~
\mathrm{m}/\mathrm{s}^2
$$

- From question 3.2 in Ryden and Peterson. The asteroid
Eros is seen in opposition from the Earth once every 847
days. What is the sidereal orbital period of Eros in days
(to within 0.1 days)?
$$
P_{\mathrm{Eros}}^{-1} = \frac{1}{365.24} - \frac{1}{847}
$$