JEE Main 2020 — Gravitation Question with Solution

Using energy conservation

K_i + U_i = K_f + U_f

$$\Rightarrow$$ $$\frac{1}{2}m{u}^2-\frac{GmM}{R}$$ = $$\frac{1}{2}m{v}^2-\frac{GmM}{2R}$$

$$\Rightarrow$$ v = $$\sqrt{u^2-\frac{GM}{R}}$$

After ejecting a rocket of mass $$\frac{m}{10}$$ the remaining part of mass $$\frac{9m}{10}$$ will rotate the earth with orbital velocity v₀.

$$\therefore$$ v₀ = $$\sqrt{\frac{GM}{2R}}$$

Applying momentum conservation along radial direction,

Before firing rocket momentum of satelite in radial direction = mv

And after firing rocket momentum of satelite in radial direction = 0 and momentum of rocket in radial direction = $$\frac{m}{10}{v}_2$$

$$\therefore$$ mv = $$\frac{m}{10}{v}_2$$

$$\Rightarrow$$ v₂ = 10v

Now applying momentum conservation along tangential direction we get,

0 = $$\frac{m}{10}{v}_1$$ - $$\frac{9m}{10}{v}_0$$

$$\Rightarrow$$$$\frac{9m}{10}{v}_0$$ = $$\frac{m}{10}{v}_1$$

$$\Rightarrow$$ v₁ = 9v₀

$$\therefore$$Total Kinetic Energy of rocket

= $$\frac{1}{2}\frac{m}{10}\left(v_1^2+v_2^2\right)$$

= $$\frac{1}{2}\frac{m}{10}\left(81v_0^2+100v^2\right)$$

= $$\frac{m}{20}\left(81\left(\frac{GM}{2R}\right)+100\left(u^2-\frac{GM}{R}\right)\right)$$

= $$\frac{m}{20}\left(100u^2+\frac{81GM}{2R}-\frac{100GM}{R}\right)$$

= $$\frac{m}{20}\left(100u^2-\frac{119GM}{2R}\right)$$

= $$5m\left(u^2-\frac{119GM}{200R}\right)$$