Topic 6.6 Motion of Orbiting Satellites: derive the speed and period of a circular orbit, find the orbital energy, and apply conservation of energy and angular momentum to elliptical orbits and Kepler's laws.

What this topic is asking

The College Board (Topic 6.6) wants you to treat gravity as the centripetal force for a circular orbit, to derive the orbital speed and period (Kepler's third law), to find the total orbital energy, and to apply conservation of energy and angular momentum to elliptical orbits. This topic unites the gravitation of Unit 2 with the energy and momentum tools of Unit 6, and it is a frequent FRQ subject.

Circular orbits: speed and period

Solving the centripetal-force balance gives the orbital speed

which depends only on the central mass and the orbital radius, not on the satellite's mass. A higher orbit (larger $r$) is a slower orbit. Substituting into $v = 2\pi r/T$ gives the period

so the square of the period is proportional to the cube of the radius, which is Kepler's third law. This relation lets you compare orbits (a satellite at twice the radius has a period $2^{3/2} \approx 2.8$ times longer) and underlies geostationary-orbit calculations.

Orbital energy

The total mechanical energy of an orbit combines the kinetic and gravitational potential energies. Using $\tfrac{1}{2}mv^2 = \dfrac{GMm}{2r}$ (from the speed) and $U = -\dfrac{GMm}{r}$:

The total energy is negative, the signature of a bound orbit: you would need to add energy to free the satellite. Notice the kinetic energy is exactly half the magnitude of the potential energy, and the total equals minus the kinetic energy. The escape speed, the speed at which the total energy is zero (just barely unbound), is $v_{esc} = \sqrt{\dfrac{2GM}{r}} = \sqrt{2}\,v_{orbit}$, about $41\%$ faster than the circular orbital speed.

Elliptical orbits

Real orbits are generally elliptical, with the central body at one focus (Kepler's first law). Two conservation laws govern them. Gravity is conservative, so mechanical energy is conserved: the satellite trades potential for kinetic energy as it moves in and out. Gravity is also a central force (always along the line to the center), so it exerts no torque about the central body, and angular momentum is conserved. Conservation of angular momentum gives Kepler's second law (equal areas in equal times) and, between the extreme points, $v_p r_p = v_a r_a$: the satellite moves fastest at perigee (closest approach) and slowest at apogee (farthest). Combining energy and angular momentum conservation lets you find both speeds.

Try this

Q1. A satellite orbits at radius $r$ with speed $v$. State its speed at radius $4r$ (same planet). [2 points]

• Cue. $v \propto 1/\sqrt{r}$, so at $4r$ the speed is $v/2$.

Q2. State the total mechanical energy of a circular orbit and explain why it is negative. [2 points]

• Cue. $E = -GMm/2r$; it is negative because the orbit is bound, energy must be added to reach the zero-energy escape state.