Question

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)

• A. 11.2 (√2 -1) km/s
• B. 7.9 (√2 - 1) km/s
• C. 8(√2 -1) km/s
• D. 7.4(√2 - 1) km/s

Answer 1

The Correct Option is C

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:

\( \Delta V = 8 (\sqrt{2} - 1) \, \text{km/s} \).