Kepler\'s 3rd law: $T^2 \propto R^3$
Taking square root: $T \propto R^{3/2}$
Derivation: $T = 2\pi\sqrt{R^3/GM} \propto R^{3/2}$
Answer: $T \propto R^{3/2}$
1st law (Ellipse law): Each planet orbits the Sun in an ellipse with Sun at one focus. Eccentricity $e$: circle $e=0$, ellipse $0 Circular orbit: $v_o = \sqrt{GM/r}$, $T = 2\pi\sqrt{r^3/GM}$. Near Earth: $T \approx 84$ min, $v \approx 7.9$ km/s. Geostationary: $T = 24$ h, $r = 42,164$ km from Earth centre, $v \approx 3.07$ km/s; used for communication, weather. Polar orbit: passes over poles, can scan entire Earth surface over time. GPS: 24 satellites at altitude 20,200 km, $T = 12$ h. Gravitational field: $g = GM/r^2$ (directed toward mass). Gravitational potential: $V = -GM/r$ (always negative). Relation: $g = -dV/dr$. Gravitational PE: $U = mV = -GMm/r$. Potential energy at surface: $U = -GMm/R$. Total energy of satellite in circular orbit: $E = KE + PE = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}$ (always negative → bound system). Weightlessness in orbit: apparent weight = $m(g - a_{centripetal}) = 0$ when $a_{centripetal} = g$ (satellite is in free fall). Not due to absence of gravity (gravity is still present at orbital altitude). True weightlessness: at very large distances from all masses. Effects: no buoyancy, no convection, no sedimentation → unique environment for science experiments, crystal growth, biology research aboard ISS. The law states $T^2 \propto a^3$, where a is the semi-major axis of the orbit. Because the constant of proportionality depends only on the central mass, the law lets you compare two satellites of the same planet without knowing G or M at all: $(T_1/T_2)^2 = (r_1/r_2)^3$. This is why most exam questions on Kepler are ratio questions — they are solvable in one line once you write both sides as a ratio rather than computing each period separately. For a circular orbit, equating gravitational and centripetal force gives $GMm/r^2 = m\omega^2r$ with $\omega = 2\pi/T$. Rearranging yields $T^2 = \frac{4\pi^2}{GM}r^3$, which is exactly Kepler's third law with the constant now identified. This derivation is worth being able to reproduce, because it shows that Kepler's empirical laws are consequences of Newton's law of gravitation rather than independent facts. Kepler's second law says the line joining a planet to the Sun sweeps equal areas in equal times. The reason is that gravity is a central force, so it exerts no torque about the Sun, and the planet's angular momentum $L = mvr$ is therefore conserved. The areal velocity equals $L/2m$, a constant. This is why a planet moves fastest at perihelion and slowest at aphelion — the same conservation law, seen from a different angle. Using the radius instead of the semi-major axis. For a circular orbit they are the same, but for an elliptical orbit the third law needs the semi-major axis, not the perihelion or aphelion distance. Forgetting that Kepler's laws apply to any central-force orbit. They hold for a satellite around the Earth and a moon around Jupiter just as they do for planets around the Sun — only the constant changes.