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 $V$.$V_{A}=R−GMm ,V_{B}=2R−GMm ,KE_{A}=21 mu_{2}$

$KE_{B}=21 mv_{2}$

$⇒U_{A}+KE_{A}=U_{B}+KE_{B}$

$⇒−R−GMm +21 mu_{2}=−2RGMm +21 mv_{2}$

$⇒−2R−GMm +21 mu_{2}=+21 mv_{2}$

$⇒v_{2}=u_{2}−RGM ⇒v=u_{2}−RGM $ (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.

$⇒0=10−m V_{T}+109 mV_{satellite}$ (at radius 'r' in circular orbit)

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

$⇒109m V_{satellite}=10m V_{T}$

By KE at radius $2R$ $V_{satellite}=RGM $

$⇒109m 2R4M =10m V_{T}⇒V_{T}=92RGM $

$⇒V_{T}=81(2RGM )$

Now applying conservation of momentum in radial direction.

$⇒0+10m V_{r}=mu_{2}−RGM ⇒V_{r}=100(u_{2}−RGM )$

Net KE of rocket $=21 (10m )(V_{r}+V_{T})$

$=21 (10m )(100(u_{2}−RGM )+2R81GM )=20100m (u_{2}−200R119GM )$

$KE=5m(u_{2}−200R119GM ) $