Orbits and Satellites: Kepler's Laws and Orbital Energy

Circular orbits: gravity is the centripetal force

A satellite in a circular orbit has only one force on it, so the free-body diagram is a single arrow toward the planet's centre. Setting that gravitational force equal to $mv^2/r$:

The mass of the satellite cancels: a screw and a space station in the same orbit move at the same speed. The second result is Kepler's third law — period squared is proportional to orbital radius cubed. Higher orbits are slower: with $GM_E\approx4\times10^{14}$ m³/s², a satellite at $r=6{,}8\times10^6$ m (about $400$ km up) moves at $v=\sqrt{4\times10^{14}/6{,}8\times10^6}\approx7{,}7$ km/s and circles the Earth every $92$ minutes.

Orbital energy

With $U=-\frac{GMm}{r}$ and $K=\tfrac12mv^2=\frac{GMm}{2r}$, the total mechanical energy of a circular orbit is

It is negative because the satellite is bound. Raising a satellite to a higher orbit lowers its speed but raises its total energy — the rocket must do positive work even though the satellite ends up slower. $E=0$ is the escape condition.

Elliptical orbits: two conservation laws

Gravity is a central force: it points at the planet, so it exerts no torque about the planet and angular momentum $L=mvr_\perp$ is conserved. Combined with energy conservation, this explains Kepler's other two laws: the orbit is an ellipse with the planet at one focus, and the satellite is fastest at the closest point (perigee) and slowest at the farthest (apogee), sweeping out equal areas in equal times. Compare closest and farthest points with $mv_pr_p=mv_ar_a$; do not use $v=\sqrt{GM/r}$ there, which holds only for circles.