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Anon25 [30]
3 years ago
9

Derive an expression for the energy needed to launch an object from the surface of Earth to a height h above the surface.Ignorin

g 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.
Physics
1 answer:
Sergio [31]3 years ago
8 0

Answer:

Explanation:

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)

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