Question

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 \( x \frac{GM_em}{21R_e} \). The value of \( x \) is
(take \( R_e = 6370 \, \text{km} \)):

• A. 11
• B. 9
• C. 12
• D. 10

Answer 1

The Correct Option is A

The total mechanical energy is calculated as:

\( \text{Total Mechanical Energy} = \frac{PE}{2} \left( \frac{R_e}{20} \right) = 318.5 \)

The mechanical energy at the Earth's surface is:

\( \text{ME on surface of Earth} = - \frac{GMm}{R_e} \quad (\text{KE on surface} = 0) \)

The mechanical energy at a specific altitude is:

\( \text{ME at altitude} = - \frac{GMm}{R_e + \frac{R_e}{20}} = - \frac{20GMm}{2 \times 21R_e} \)

This simplifies to:

\( \text{ME at altitude} = - \frac{10GMm}{21R_e} \)

The change in total mechanical energy is determined by:

\( \text{Change in Total M.E.} = E_f - E_i \)

Substituting the values yields:

\( \text{Change in Total M.E.} = \frac{10GMm}{21R_e} - \frac{GMm}{R_e} = \frac{-10GMm + 21GMm}{21R_e} = \frac{11GMm}{21R_e} \)

Therefore, the result is:

\( x = 11 \)