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What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R ?
(a) 5GmM/6R
(b) 2GmM/3R
(c) GmM/2R
(d) GmM/2R

Lokesh kumar , 11 Years ago
Grade 6
anser 2 Answers
Deepak Patra

Last Activity: 11 Years ago

(a) 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
shimi xavier

Last Activity: 8 Years ago

The kinetic energy at altitude 2 R = G M m ÷ 6 R
The gravitational potential energy at altitude 2 R = – G M m ÷ 3 R
Total energy = Kinetic energy + potential energy
                     = G M m ÷ 6 R – G M m ÷ 3 R
                     = – G M m ÷ 6 R
Potential energy at the surfce is – G M m ÷ R
Therefore, required kinetic energy = G M m ÷ R – G M m ÷ 6 R
                                                       = 5 G M m ÷ 6 R
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