A satellite of mass m is initially revolving in a circular orbit at a height equal to the radius of the Earth ( R_E ) from the Earth's surface. An engine fires and does work equal to mgR_E 8 on the satellite to transfer it to a new, higher circular orbit. The final height of the satellite above the surface of the Earth is: (where g is the acceleration due to gravity on the surface of the Earth)

Step-by-step solution

Let the initial height be h₁ = R_E . The initial orbital radius is r₁ = R_E + h₁ = 2R_E . The initial total energy of the satellite is: E₁ = - GMm 2r₁ = - mgR_E^2 2(2R_E) = - mgR_E 4 Work done on the satellite is W = mgR_E 8 . The final total energy E₂ is: E₂ = E₁ + W = - mgR_E 4 + mgR_E 8 = - mgR_E 8 Let the final orbital radius be r₂ . We know that: E₂ = - GMm 2r₂ = - mgR_E^2 2r₂ Equating the two expressions for E₂ : - mgR_E^2 2r₂ = - mgR_E 8 2r₂ = 8R_E r₂ = 4R_E The final height of the satellite above the Earth'