A satellite of mass $$M$$ is launched vertically upwards with an initial speed $$u$$ from the surface of the earth. After it reaches height $$R$$ ($$R$$ = radius of the earth), it ejects a rocket of mass $$\frac{M}{10}$$ so that subsequently the satellite moves in a circular orbit. The kinetic energy of the rocket is ($$G$$ is the gravitational constant; $$M_e$$ is the mass of the earth):

We have a satellite of mass $$M$$ that is fired vertically from the surface of the earth (radius $$R$$) with speed $$u$$. Its total mechanical energy just after launch is

$$E_0=\frac12Mu^{2}-\frac{GM_eM}{R}.$$

When it rises to a height $$R$$ above the surface its distance from the earth’s centre is $$r=R+R=2R$$. At that instant let its speed be $$v$$. Applying conservation of mechanical energy between the point of launch and the point $$r=2R$$, we write

$$\frac12Mu^{2}-\frac{GM_eM}{R}= \frac12Mv^{2}-\frac{GM_eM}{2R}.$$

Simplifying,

$$\frac12Mu^{2}-\frac{GM_eM}{R}+ \frac{GM_eM}{2R}= \frac12Mv^{2}

\;\Longrightarrow\;

\frac12Mv^{2}= \frac12Mu^{2}-\frac{GM_eM}{2R}.$$

Dividing by $$\tfrac12M$$ we get

$$v^{2}=u^{2}-\frac{GM_e}{R}. \quad -(1)$$

Immediately afterwards the satellite ejects a small rocket of mass $$\dfrac{M}{10}$$, so the remaining mass of the satellite becomes

$$M_s=\frac{9M}{10}.$$

The problem states that the satellite (of mass $$\dfrac{9M}{10}$$) now moves in a circular orbit of radius $$2R$$. The speed required for a circular orbit of radius $$r$$ is obtained from the centripetal-gravitational balance

$$\frac{v_c^{2}}{r}=\frac{GM_e}{r^{2}}

\;\Longrightarrow\;

v_c=\sqrt{\frac{GM_e}{r}}

=\sqrt{\frac{GM_e}{2R}}. \quad -(2)$$

During the extremely short interval of ejection, the only external force is gravity, which is radial; its impulse is therefore negligible. Hence the linear momentum of the system (satellite + rocket) is conserved separately in the radial and tangential directions.

Radial direction. Before ejection the entire mass $$M$$ is moving radially outward with speed $$v$$, so the initial radial momentum is

$$P_{r,\,\text{initial}}=Mv.$$

After ejection, the satellite is moving purely tangentially (no radial component), so all the radial momentum must be carried by the rocket (mass $$\dfrac{M}{10}$$). Let $$v_{r,\parallel}$$ be the radial component of the rocket’s velocity. Conservation gives

$$Mv=\frac{M}{10}\,v_{r,\parallel}

\;\Longrightarrow\;

v_{r,\parallel}=10v. \quad -(3)$$

Tangential direction. Initially the tangential momentum is zero (vertical launch). After ejection the satellite (mass $$\dfrac{9M}{10}$$) moves tangentially with speed $$v_c$$. Let $$v_{r,\perp}$$ be the tangential component of the rocket’s velocity. Conservation of tangential momentum gives

$$0=\frac{9M}{10}\,v_c+\frac{M}{10}\,v_{r,\perp}

\;\Longrightarrow\;

v_{r,\perp}=-9v_c. \quad -(4)$$

(The negative sign shows that the rocket’s tangential motion is opposite to that of the satellite.)

Magnitude of the rocket’s velocity. Because the radial and tangential components are perpendicular, the speed of the rocket is

$$v_r=\sqrt{v_{r,\parallel}^{2}+v_{r,\perp}^{2}}

=\sqrt{(10v)^{2}+(-9v_c)^{2}}

=\sqrt{100v^{2}+81v_c^{2}}. \quad -(5)$$

Kinetic energy of the rocket. Using $$K=\tfrac12mv^{2}$$,

$$K_r=\frac12\left(\frac{M}{10}\right)v_r^{2}

=\frac{M}{20}\bigl(100v^{2}+81v_c^{2}\bigr). \quad -(6)$$

Now substitute from (1) and (2):

$$v^{2}=u^{2}-\frac{GM_e}{R}, \qquad

v_c^{2}=\frac{GM_e}{2R}.$$

Putting these into (6),

$$

\begin{aligned}

K_r

&=\frac{M}{20}\Bigl[100\Bigl(u^{2}-\frac{GM_e}{R}\Bigr)

+81\Bigl(\frac{GM_e}{2R}\Bigr)\Bigr] \\

&=\frac{M}{20}\Bigl[100u^{2}-100\frac{GM_e}{R}

+\frac{81}{2}\frac{GM_e}{R}\Bigr] \\

&=\frac{M}{20}\Bigl[100u^{2}-100\frac{GM_e}{R}

+40.5\frac{GM_e}{R}\Bigr] \\

&=\frac{M}{20}\Bigl[100u^{2}-59.5\frac{GM_e}{R}\Bigr].

\end{aligned}

$$

Factorising the common factor $$100$$ inside the bracket,

$$

K_r

=\frac{M}{20}\cdot100\Bigl[u^{2}-0.595\frac{GM_e}{R}\Bigr]

=5M\left(u^{2}-\frac{119}{200}\frac{GM_e}{R}\right). \quad -(7)

$$

This matches Option B exactly.

Hence, the correct answer is Option 2.