Question

An astronomer discovers a new planet with radius RR with a satellite at a distance 10R10R from the center of the planet. The satellite orbits the planet with period TT. Using this information, derive an expression for the density of the planet rhorho.

a) rho=3/4πG(1/T)^2
b) rho=3/2πG(1/T)^2
c) rho=3/πG(1/T)^2
d) rho=3/8πG(1/T)^2

Answer 1

The correct expression for the density of the planet in terms of its orbital period and radius is given by rho = 3/4πG(1/T)², which is derived using Newton's version of Kepler's third law and the volume equation for a sphere.

Step-by-step explanation:

To derive an expression for the density of the planet (rho), we can use the formula related to the period (T) and radius of orbit (r). The question implies that the satellite orbits the planet at a distance of 10R from the center of the planet, and we are given the satellite's period of orbit is T.

According to Newton's version of Kepler's third law, the relationship between the orbital period and the radius of orbit for a satellite is given by r³/T² = GM/4π², where r is the radius of the orbit, G is the gravitational constant, and M is the mass of the planet.

To express the mass M in terms of the planet's density (rho) and radius (R), we use the volume of a sphere V = 4/3πR³, and thus M = rho V = rho (4/3πR³). Substituting and rearranging for rho in the initial equation, we get:

rho = 3 / (4πG(1/T)²)

Therefore, the correct expression for the density of the planet is option A: rho = 3/4πG(1/T)².