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 $\frac{m}{1 0}$ so that subsequently the satellite moves in a circular orbit. The kinetic energy of the rocket is ( $G$ is the gravitational constant; $M$ is the mass of the earth) :

Question

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Answer 1

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 $V$ .
$V_{A} = \frac{- G M m}{R}, V_{B} = \frac{- G M m}{2 R}, K E_{A} = \frac{1}{2} m u^{2}$

$K E_{B} = \frac{1}{2} m v^{2}$

$\Rightarrow U_{A} + K E_{A} = U_{B} + K E_{B}$

$\Rightarrow - \frac{- G M m}{R} + \frac{1}{2} m u^{2} = - \frac{G M m}{2 R} + \frac{1}{2} m v^{2}$

$\Rightarrow - \frac{- G M m}{2 R} + \frac{1}{2} m u^{2} = + \frac{1}{2} m v^{2}$

$\Rightarrow v^{2} = u^{2} - \frac{G M}{R} \Rightarrow v = u^{2} - \frac{G M}{R}$ (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.

$\Rightarrow 0 = \frac{- m}{1 0} V_{T} + \frac{9}{1 0} m V_{s a t e l l i t e}$ (at radius 'r' in circular orbit)

$V_{s a t e l l i t e}$ and $V_{T}$ will have opposite direction so $V_{T}$ is negative
$\Rightarrow \frac{9 m}{1 0} V_{s a t e l l i t e} = \frac{m}{1 0} V_{T}$
By KE at radius $2 R$ $V_{s a t e l l i t e} = \frac{G M}{R}$
$\Rightarrow \frac{9 m}{1 0} \frac{4 M}{2 R} = \frac{m}{1 0} V_{T} \Rightarrow V_{T} = 9 \frac{G M}{2 R}$
$\Rightarrow V_{T}^{2} = 81 (\frac{G M}{2 R})$

Now applying conservation of momentum in radial direction.

$\Rightarrow 0 + \frac{m}{1 0} V_{r} = m u^{2} - \frac{G M}{R} \Rightarrow V_{r}^{2} = 100 (u^{2} - \frac{G M}{R})$

Net KE of rocket $= \frac{1}{2} (\frac{m}{1 0}) (V_{r}^{2} + V_{T}^{2})$

$= \frac{1}{2} (\frac{m}{1 0}) (100 (u^{2} - \frac{G M}{R}) + \frac{8 1 G M}{2 R}) = \frac{1 0 0 m}{2 0} (u^{2} - \frac{1 1 9 G M}{2 0 0 R})$

$K E = 5 m (u^{2} - \frac{1 1 9 G M}{2 0 0 R})$

Answer 2

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} = \frac{- G M m}{R}, V_{B} = \frac{- G M m}{2 R}, K E_{A} = \frac{1}{2} m u^{2}

K E_{B} = \frac{1}{2} m v^{2}

\Rightarrow U_{A} + K E_{A} = U_{B} + K E_{B}

\Rightarrow - \frac{- G M m}{R} + \frac{1}{2} m u^{2} = - \frac{G M m}{2 R} + \frac{1}{2} m v^{2}

\Rightarrow - \frac{- G M m}{2 R} + \frac{1}{2} m u^{2} = + \frac{1}{2} m v^{2}

\Rightarrow v^{2} = u^{2} - \frac{G M}{R} \Rightarrow v = u^{2} - \frac{G M}{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 = \frac{- m}{1 0} V_{T} + \frac{9}{1 0} m V_{s a t e l l i t e}

(at radius 'r' in circular orbit)

V_{s a t e l l i t e}

and

V_{T}

will have opposite direction so

V_{T}

is negative

\Rightarrow \frac{9 m}{1 0} V_{s a t e l l i t e} = \frac{m}{1 0} V_{T}

By KE at radius

2 R

V_{s a t e l l i t e} = \frac{G M}{R}

\Rightarrow \frac{9 m}{1 0} \frac{4 M}{2 R} = \frac{m}{1 0} V_{T} \Rightarrow V_{T} = 9 \frac{G M}{2 R}

\Rightarrow V_{T}^{2} = 81 (\frac{G M}{2 R})

Now applying conservation of momentum in radial direction.

\Rightarrow 0 + \frac{m}{1 0} V_{r} = m u^{2} - \frac{G M}{R} \Rightarrow V_{r}^{2} = 100 (u^{2} - \frac{G M}{R})

Net KE of rocket

= \frac{1}{2} (\frac{m}{1 0}) (V_{r}^{2} + V_{T}^{2})

= \frac{1}{2} (\frac{m}{1 0}) (100 (u^{2} - \frac{G M}{R}) + \frac{8 1 G M}{2 R}) = \frac{1 0 0 m}{2 0} (u^{2} - \frac{1 1 9 G M}{2 0 0 R})

K E = 5 m (u^{2} - \frac{1 1 9 G M}{2 0 0 R})

$\frac{m}{2 0} (u^{2} + \frac{1 1 3}{2 0 0} \frac{G M}{R})$

$5 m (u^{2} - \frac{1 1 9}{2 0 0} \frac{G M}{R})$

$\frac{m}{2 0} (u - \frac{2 G M}{3 R})^{2}$

$\frac{3 m}{8} (u + \frac{5 G M}{6 R})^{2}$

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