Question

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 . [JEE main 2013,] ​

Answer 1

• Explanation:- From conservation of energy ,

Total energy at the planet = total energy at altitude ,

Let the velocity of launched satellite is Vₐ

-GMm/R + (KE)surface = -GMm /3R + 1/2mvₐ²___(i)

In the orbital of planet , the necessary centripetal force is obtained by gravitational force .

∴ mvₐ²/R+2R = GMm/(R+2R )²

⟹ Vₐ= √GM /3R _______(ii)

From eqs (i) and (ii) we get

(KE )surface = GMm/R -GMm/3R + 1/2m* [√(GM/3R)]² _____ From (i) and (ii)

(KE)surface = 2GmM /3R+ 1/6GMm/6R

⟹ 4GMm + GMm /6R

(KE)surface = 5/6GMm/6R

since , minimum energy required to launch a satellite is 5/6GMm/6R Answer