A spaceship is launched into a circular orbit close to the Earth's surface. What additional velocity has to be imparted to the spaceship to overcome the gravitational pull?

Options

A. `11.2 km s^(-1)`
B. `8 km s^(-1)`
C. `3.2 km s^(-1)`
D. `1.5 km s^(-1)`

Answer

To solve the problem of determining the additional velocity that needs to be imparted to a spaceship in a circular orbit close to the Earth's surface to overcome gravitational pull, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concepts**:
- **Orbital Velocity (v_o)**: The velocity required for an object to maintain a stable orbit around a planet. For a circular orbit close to the Earth's surface, the formula is:
\[ v_o = \sqrt{\frac{GM}{R}} \]
- **Escape Velocity (v_e)**: The minimum velocity needed for an object to break free from the gravitational attraction of a planet. The formula for escape velocity is:
\[ v_e = \sqrt{\frac{2GM}{R}} \]
- Here, \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

2. **Calculate the Additional Velocity**:
- The additional velocity (\( v_{add} \)) required to overcome gravitational pull can be found by subtracting the orbital velocity from the escape velocity:
\[ v_{add} = v_e - v_o \]

3. **Substituting the Formulas**:
- Substitute the formulas for \( v_e \) and \( v_o \):
\[ v_{add} = \sqrt{\frac{2GM}{R}} - \sqrt{\frac{GM}{R}} \]

4. **Factor Out Common Terms**:
- Factor out \( \sqrt{\frac{GM}{R}} \):
\[ v_{add} = \sqrt{\frac{GM}{R}} \left( \sqrt{2} - 1 \right) \]

5. **Express in Terms of Gravitational Acceleration**:
- Since \( g = \frac{GM}{R} \), we can express the additional velocity as:
\[ v_{add} = \sqrt{gR} \left( \sqrt{2} - 1 \right) \]

6. **Final Expression**:
- Therefore, the additional velocity that needs to be imparted to the spaceship to overcome the gravitational pull is:
\[ v_{add} = gR \left( \sqrt{2} - 1 \right) \]

### Final Answer:
The additional velocity required to overcome the gravitational pull is \( gR \left( \sqrt{2} - 1 \right) \).