JEE Main 2013 — Gravitation Question with Solution

Energy of the satellite on the surface of the planet

E_i = K.E + P.E = 0 + $$\left(-\frac{GMm}{R}\right)$$ = $$-\frac{GMm}{R}$$

Energy of the satellite at 2R distance from the surface of the planet while moving with velocity v

E_f = $$\frac{1}{2}m{v}^2$$ + $$\left(-\frac{GMm}{R+2R}\right)$$

In the orbital of planet, the centripetal force is provided by the gravitational force

$$\therefore$$ $$\frac{m{v}^2}{R+2R}=\frac{GMm}{{\left(R+2R\right)}^2}$$

$$\Rightarrow v^2=\frac{GM}{3R}$$

$$\therefore$$ E_f = $$\frac{1}{2}m{v}^2$$ + $$\left(-\frac{GMm}{R+2R}\right)$$

$$=\frac{1}{2}m\frac{GM}{3R}-\frac{GMm}{3R}$$

= $$-\frac{GMm}{6R}$$

$$\therefore$$ Minimum energy required required to launch the satellite

= E_f - E_i

= $$-\frac{GMm}{6R}$$ - $$\left(-\frac{GMm}{R}\right)$$

= $$\frac{5GMm}{6R}$$