A space ship of mass 2 × 104 kg is launched into a circular orbit close to the earth surface. The additional velocity to be imparted to the spaceship in the orbit to overcome the gravitational pull will be (if g = 10 m/s2 and radius of earth = 6400 km)
- 11.2 (√2 -1) km/s
- 7.9 (√2 - 1) km/s
- 8(√2 -1) km/s
- 7.4(√2 - 1) km/s
The Correct Option is C
Solution and Explanation
Step 1: Formula for \( \Delta V \)
The change in velocity (\( \Delta V \)) is given by:
\[ \Delta V = \frac{GM}{R} (\sqrt{2} - 1), \]
where:
- \( G \): Gravitational constant
- \( M \): Mass of the central body (e.g., Earth)
- \( R \): Radius of the orbit
Step 2: Simplify the Expression
The term \( \frac{GM}{R} \) can be rewritten using the acceleration due to gravity at the surface of the central body (\( g \)):
\[ g = \frac{GM}{R^2}. \]
Substitute \( gR \) for \( \frac{GM}{R} \):
\[ \Delta V = gR (\sqrt{2} - 1). \]
Step 3: Substitute Numerical Values
We are given \( gR = 8000 \, \text{m/s} \) or \( 8 \, \text{km/s} \). Substitute this into the equation:
\[ \Delta V = 8000 (\sqrt{2} - 1) \, \text{m/s}. \]
Convert to km/s:
\[ \Delta V = 8 (\sqrt{2} - 1) \, \text{km/s}. \]
Final Answer:
\( \Delta V = 8 (\sqrt{2} - 1) \, \text{km/s} \).
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