The minimum energy required to launch a m kg satellite from the earth's surface in a circular orbit at an altitude 2r, where r is the radius of earth is
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Answered by
24
As we know, Gravitational potential energy = - GM /r
and orbital velocity, v0 = vGM/R+h
Ef = ½ mv20 – GMm/3R = ½ m GM/3R – GMm/3R
= GMm/3R (1/2-1) = - GMm/6R
Ei = - GM/R + K
Ei = Ef
Therefore minimum required energy, K = 5GMm/6R
and orbital velocity, v0 = vGM/R+h
Ef = ½ mv20 – GMm/3R = ½ m GM/3R – GMm/3R
= GMm/3R (1/2-1) = - GMm/6R
Ei = - GM/R + K
Ei = Ef
Therefore minimum required energy, K = 5GMm/6R
Answered by
32
Answer:
5/6mgR
Explanation:
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