Satellite moving in an orbit around the earth , the ratio of kinetic energy and potential energy is
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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/R
The gravitational pull of the Earth on the mass is:
F_grav= - GMm/R^2
Therefore:
- GMm/R^2 = F_grav = F_c = - mv^2/R
- GMm/R^2 = - mv^2/R
So the kinetic energy of m is:
mv^2/2 = (R/2)*(GMm/R^2) = (GMm/R)/2
Since the potential energy U = -GMm/R, in this special case:
KE = mv^2/2 = (GMm/R)/2 = - U/2
The Virial Theorem shows that this relationship is generally true, not just for circular orbits.
F_c = -mv^2/R
The gravitational pull of the Earth on the mass is:
F_grav= - GMm/R^2
Therefore:
- GMm/R^2 = F_grav = F_c = - mv^2/R
- GMm/R^2 = - mv^2/R
So the kinetic energy of m is:
mv^2/2 = (R/2)*(GMm/R^2) = (GMm/R)/2
Since the potential energy U = -GMm/R, in this special case:
KE = mv^2/2 = (GMm/R)/2 = - U/2
The Virial Theorem shows that this relationship is generally true, not just for circular orbits.
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