Question

You are asked to verify Kepler's Laws of Planetary Motion. For these exercises, assume that each planet moves in an orbit given by the vector- valued function \(\mathrm{r}\). Let \(r=\|\mathbf{r}\|,\) let \(G\) represent the universal gravitational constant, let \(M\) represent the mass of the sun, and let \(m\) represent the mass of the planet. Prove Kepler's Third Law: The square of the period of a planet's orbit is proportional to the cube of the mean distance between the planet and the sun.

Short answer

The proof shows Kepler's Third Law, that is, the square of the period of a planet's orbit \(T^2\) is proportional to the cube of the mean distance from the sun to the planet \((4\pi^2/GM)*r^3\).

Step-by-step solution

Understand the force of gravity between the sun and a planet

The force of gravity acting on the planet due to the sun is \(-GMm/r^2\), where M is the solar mass, m is the planet mass, r is the distance between the planet and the sun, and G is the gravitational constant. This force supplies the centripetal force required for the planet to stay in orbit, which is \(mv^2/r\), with v being the velocity of the planet.

Relate the force of gravity to the centripetal force

Set the force of gravity equal to the centripetal force: \(-GMm/r^2 = mv^2/r\). Simplify this equation by canceling out the common factors to get: \(-GM/r = v^2\).

Determine the speed of the planet

The speed of the planet, v, in terms of its orbital radius r and orbital period T (the time it takes for one complete revolution around the sun) is: \(v=2\pi r/T\). Substitute this expression for \(v\) in the equation from the previous step.

Use the speed of the planet to find its period

Substituting the speed gives: \(-GM/r = (2\pi r/T)^2\). This simplifies to: \(-GM/T^2 = 4\pi^2r^2\). Rearrange this to confirm Kepler’s third law: \(T^2 = (4\pi^2/GM)*r^3\). This proves that the square of the period is proportional to the cube of the distance. Thus the Kepler's Third Law is confirmed.