Question

For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is

Answer 1

David Headley has correctly referenced the virial theorem. You can look that up, but it’s too long to type out here. However, we can show the result with a simple case: Consider a mass m circling the Earth (of mass M) at radius R. The centripetal force is:F_c = -mv^2/RThe gravitational pull of the Earth on the mass is:F_grav= - GMm/R^2Therefore:- GMm/R^2 = F_grav = F_c = - mv^2/R- GMm/R^2 = - mv^2/RSo the kinetic energy of m is:mv^2/2 = (R/2)*(GMm/R^2) = (GMm/R)/2Since the potential energy U = -GMm/R, in this special case:KE = mv^2/2 = (GMm/R)/2 = - U/2The Virial Theorem shows that this relationship is generally true, not just for circular orbits.

Answer 2

Kinetic Energy of Satellite=½ (m v2 )

Now v=( GM/R)1/2

by substituting the value of v in equation of K.E.

K.E=GMm/ 2R

Potential energy=GMm / R

∴ K/U=1/2