A spaceship is launched into a circular orbit close to the earth's surface . What additional velocity has now to be imparted to the spaceship in the orbit to overcome the gravitational pull. Radius of earth `= 6400 km`, `g = 9.8m//s^(2)`.

Options

A. `3.28 km//s`
B. `12 km//s`
C. `10 km//s`
D. `40 km//s`

Answer

To solve the problem of finding the additional velocity required for a spaceship in a circular orbit close to the Earth's surface to overcome gravitational pull, we can follow these steps:

### Step 1: Understand the parameters
- The radius of the Earth (r) = 6400 km = 6400 × 10^3 m
- The acceleration due to gravity (g) = 9.8 m/s²

### Step 2: Calculate the orbital velocity (V₀)
The orbital velocity (V₀) for an object in a circular orbit close to the Earth's surface can be calculated using the formula:
\[ V₀ = \sqrt{g \cdot r} \]

Substituting the values:
\[ V₀ = \sqrt{9.8 \, \text{m/s}² \cdot (6400 \times 10^3 \, \text{m})} \]

### Step 3: Calculate the escape velocity (Vₑ)
The escape velocity (Vₑ) required to break free from the Earth's gravitational pull is given by:
\[ Vₑ = \sqrt{2 \cdot g \cdot r} \]

Substituting the values:
\[ Vₑ = \sqrt{2 \cdot 9.8 \, \text{m/s}² \cdot (6400 \times 10^3 \, \text{m})} \]

### Step 4: Find the additional velocity (ΔV)
The additional velocity required (ΔV) to overcome the gravitational pull is the difference between the escape velocity and the orbital velocity:
\[ \Delta V = Vₑ - V₀ \]

### Step 5: Substitute the values and simplify
Using the expressions for Vₑ and V₀:
\[ \Delta V = \sqrt{2 \cdot g \cdot r} - \sqrt{g \cdot r} \]

Factoring out \(\sqrt{g \cdot r}\):
\[ \Delta V = \sqrt{g \cdot r} \left( \sqrt{2} - 1 \right) \]

### Step 6: Substitute the known values
Now substituting g = 9.8 m/s² and r = 6400 × 10^3 m:
\[ \Delta V = \sqrt{9.8 \cdot (6400 \times 10^3)} \left( \sqrt{2} - 1 \right) \]

### Step 7: Calculate the numerical value
Calculating \(\sqrt{9.8 \cdot (6400 \times 10^3)}\):
1. Calculate \(9.8 \cdot (6400 \times 10^3) = 62752000\)
2. Calculate \(\sqrt{62752000} \approx 7914.4 \, \text{m/s}\)

Now calculate \(\Delta V\):
\[ \Delta V \approx 7914.4 \left( \sqrt{2} - 1 \right) \]

Using \(\sqrt{2} \approx 1.414\):
\[ \Delta V \approx 7914.4 \cdot (1.414 - 1) \approx 7914.4 \cdot 0.414 \approx 3280.4 \, \text{m/s} \]

### Step 8: Convert to km/s
Converting to kilometers per second:
\[ \Delta V \approx 3.28 \, \text{km/s} \]

### Final Answer
The additional velocity required is approximately **3.28 km/s**.

---