Derive an expression for the energy needed to launch an object from the surface of Earth to a height h above the surface. Ignoring Earth's rotation, how much energy is needed to get the same object into orbit at height h? Express your answer in terms of some or all of the variables h, mass of the object m, mass of Earth mE, its radius RE, and gravitational constant G.

Question

Serena Carey

Physics

3 November, 02:26

Derive an expression for the energy needed to launch an object from the surface of Earth to a height h above the surface. Ignoring Earth's rotation, how much energy is needed to get the same object into orbit at height h? Express your answer in terms of some or all of the variables h, mass of the object m, mass of Earth mE, its radius RE, and gravitational constant G.

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Dwayne Hays · Answer 1

Potential energy on the surface of the earth

= - GMm / R

Potential at height h

= - GMm / (R+h)

Potential difference

= GMm / R - GMm / (R+h)

= GMm (1/R - 1 / R+h)

= GMmh / R (R + h)

This will be the energy needed to launch an object from the surface of Earth to a height h above the surface.

Extra energy is needed to get the same object into orbit at height h

= Kinetic energy of the orbiting object at height h

= 1/2 x potential energy at height h

= 1/2 x GMm / (R + h)