#### Correct option is B. $$5m(u^2-\dfrac{119}{200}\dfrac{GM}{R})$$

As the satellite is launched vertically, the final velocity of the satellite as a whole will be radial. Now applying energy conservation between $$A$$ and $$B$$. Taking final velocity of the whole satellite at point $$B$$ to be $$V$$.$$V_A=\dfrac{-GMm}{R},V_B=\dfrac{-GMm}{2R},KE_A=\dfrac{1}{2}mu^2$$

$$KE_B=\dfrac{1}{2}mv^2$$

$$\Rightarrow U_A+KE_A=U_B+KE_B$$

$$\Rightarrow -\dfrac{-GMm}{R}+\dfrac{1}{2}mu^2=-\dfrac{GMm}{2R}+\dfrac{1}{2}mv^2$$

$$\Rightarrow -\dfrac{-GMm}{2R}+\dfrac{1}{2}mu^2=+\dfrac{1}{2}mv^2$$

$$\Rightarrow v^2=u^2-\dfrac{GM}{R}\Rightarrow v=\sqrt{u^2-\dfrac{GM}{R}}$$ (Radial)

Now as the satellite reached $$B$$ it has only radial velocity $$v$$, no tangential velocity.

Now, as the satellite ejects the recket the final velocity of satellite becomes only tangential as it starts moving in circular orbit.

Let the rocket has velocity $$V_r$$ in radial direction and $$V_T$$ tangential direction.

Applying momentum conservation in tangential direction.

$$\Rightarrow 0=\dfrac{-m}{10}V_T+\dfrac{9}{10}mV_{satellite}$$ (at radius 'r' in circular orbit)

$$V_{satellite}$$ and $$V_T$$ will have opposite direction so $$V_T$$ is negative

$$\Rightarrow \dfrac{9m}{10}V_{satellite}=\dfrac{m}{10}V_T$$

By KE at radius $$2R$$ $$V_{satellite}=\sqrt{\dfrac{GM}{R}}$$

$$\Rightarrow \dfrac{9m}{10}\sqrt{\dfrac{4M}{2R}}=\dfrac{m}{10}V_T\Rightarrow V_T=9\sqrt{\dfrac{GM}{2R}}$$

$$\Rightarrow V^2_T=81\left(\dfrac{GM}{2R}\right)$$

Now applying conservation of momentum in radial direction.

$$\Rightarrow 0+\dfrac{m}{10}V_r=m\sqrt{u^2-\dfrac{GM}{R}}\Rightarrow V^2_r=100\left(u^2-\dfrac{GM}{R}\right)$$

Net KE of rocket $$=\dfrac{1}{2}\left(\dfrac{m}{10}\right)\left(V^2_r+V^2_T\right)$$

$$=\dfrac{1}{2}\left(\dfrac{m}{10}\right)\left(100(u^2-\dfrac{GM}{R})+\dfrac{81GM}{2R}\right)=\dfrac{100m}{20}\left(u^2-\dfrac{119GM}{200R}\right)$$

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