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 \( R_e = 6370 \, \text{km} \)):
(take \( R_e = 6370 \, \text{km} \)):
- 11
- 9
- 12
- 10
The Correct Option is A
Approach Solution - 1
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:
\( x = 11 \)
Approach Solution -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: \( x = 11 \).
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