1.State Newton's Universal Law of Gravitation.
Sol: $F = G \frac{m_1 m_2}{r^2}$, where $G = 6.67 \times 10^{-11} \text{ N}\cdot\text{m}^2/\text{kg}^2$.
2.Define escape velocity and write its expression for Earth.
Sol: Minimum velocity to escape Earth's gravitational pull: $v_e = \sqrt{2gR} \approx 11.2 \text{ km/s}$.
3.State Kepler's Three Laws of Planetary Motion.
Sol: 1. Law of Orbits (Ellipses), 2. Law of Areas (Constant areal velocity), 3. Law of Periods ($T^2 \propto a^3$).
4.What is a geostationary satellite? State its height and time period.
Sol: Appears stationary over equator. Period $T = 24 \text{ hours}$, altitude $h \approx 36,000 \text{ km}$.
5.Calculate acceleration due to gravity at height $h = R$ above Earth's surface ($g = 9.8\text{ m/s}^2$).
Sol: $g' = g \left(\frac{R}{R+h}\right)^2 = g \left(\frac{R}{2R}\right)^2 = \frac{g}{4} = 2.45 \text{ m/s}^2$.
6.At what depth below Earth's surface does value of $g$ reduce to half its value at surface?
Sol: $g_d = g \left(1 - \frac{d}{R}\right) = \frac{g}{2} \implies 1 - \frac{d}{R} = \frac{1}{2} \implies d = \frac{R}{2} = 3200 \text{ km}$.
7.Find orbital velocity of a satellite orbiting close to Earth's surface ($R = 6400 \text{ km}, g = 9.8 \text{ m/s}^2$).
Sol: $v_o = \sqrt{gR} = \sqrt{9.8 \times 6.4 \times 10^6} \approx 7.92 \text{ km/s}$.
8.Calculate gravitational potential energy of a $100 \text{ kg}$ mass at distance $2R$ from Earth's center.
Sol: $U = -\frac{GMm}{r} = -\frac{g R^2 m}{2R} = -\frac{1}{2} m g R = -\frac{1}{2}(100)(9.8)(6.4 \times 10^6) = -3.136 \times 10^9 \text{ J}$.
