An astronaut takes a ball of mass m from earth to space. He throws the ball into a circular orbit about earth at an altitude of 318.5 km. From earth's surface to the orbit, the change in total mechanical energy of the ball is The value of is (take = 637

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Solution

Q.
An astronaut takes a ball of mass m from earth to space. He throws the ball into a circular orbit about earth at an altitude of 318.5 km. From earth's surface to the orbit, the change in total mechanical energy of the ball is $x \frac{G M_{e} m}{21 R_{e}} .$ The value of $x$ is (take $R_{e}$ = 6370 km): [2024]

1 11

2 9

3 12

4 10

Ans.
(1)

   $h = 318.5 \approx (\frac{R_{e}}{20})$

   $T \cdot E_{i} = \frac{- G M_{e} m}{R_{e}}$

   $T \cdot E_{f} = \frac{- G M_{e} m}{2 (R_{e} + h)} = \frac{- G M_{e} m}{2 (R_{e} + \frac{R_{e}}{20})}$

Change in total mechanical energy = $T E_{f} - T E_{i}$

$= \frac{G M_{e} m}{R_{e}} [1 - \frac{10}{21}] = \frac{11 G M_{e} m}{21 R_{e}} n$

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