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 given by: 

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

The mechanical energy on the surface of the Earth is:

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

The mechanical energy at an altitude is:

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

Simplifying further:

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

Now, we calculate the change in total mechanical energy:

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

Substituting the values:

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

Thus, we get:

\( x = 11 \)

Answer 2

Step 1: Data and formula for mechanical energy The total mechanical energy at the Earth’s surface is:

\( T.E_i = -\frac{GM_e m}{R_e}, \)

where:

• \( R_e = 6370 \, \text{km} \) (radius of Earth),
• \( G \) is the gravitational constant,
• \( M_e \) is the mass of Earth.

The altitude of the orbit is given as \( h = 318.5 \, \text{km} \). Approximate:

\( h \approx \frac{R_e}{20}. \)

The total mechanical energy in the orbit is:

\( T.E_f = -\frac{GM_e m}{2(R_e + h)}. \)

Step 2: Substitute \( h \approx \frac{R_e}{20} \)

\( T.E_f = -\frac{GM_e m}{2 \left( R_e + \frac{R_e}{20} \right)} = -\frac{GM_e m}{2 \left( \frac{21R_e}{20} \right)}. \)

Simplify:

\( T.E_f = -\frac{10GM_e m}{21R_e}. \)

Step 3: Change in mechanical energy The change in total mechanical energy is:

\( \Delta E = T.E_f - T.E_i. \)

Substitute:

\( \Delta E = \left( -\frac{10GM_e m}{21R_e} \right) - \left( -\frac{GM_e m}{R_e} \right). \)

Simplify:

\( \Delta E = -\frac{10GM_e m}{21R_e} + \frac{21GM_e m}{21R_e}. \)

\( \Delta E = \frac{11GM_e m}{21R_e}. \)

Thus, \( x = 11 \).

Final Answer: \( x = 11 \).