To find the additional velocity that should be imparted to the spaceship to overcome Earth's gravitational pull, we need to calculate the difference between the escape velocity and the orbital velocity of the spaceship.
1. **Orbital Velocity (v₀):**
For a circular orbit close to Earth's surface, the orbital velocity can be calculated using:
\[
v_0 = \sqrt{g \cdot R}
\]
where \( g \) is the acceleration due to gravity (9.8 m/s²), and \( R \) is the radius of Earth (6400 km = 6.4 × 10⁶ m).
\[
v_0 = \sqrt{9.8 \, \text{m/s}^2 \times 6.4 \times 10^6 \, \text{m}}
\]
\[
v_0 \approx \sqrt{6.272 \times 10^7} \approx 7900 \, \text{m/s} \approx 7.9 \, \text{km/s}
\]
2. **Escape Velocity (vₑ):**
The escape velocity from the Earth's surface is given by:
\[
v_e = \sqrt{2g \cdot R}
\]
\[
v_e = \sqrt{2 \times 9.8 \, \text{m/s}^2 \times 6.4 \times 10^6 \, \text{m}}
\]
\[
v_e \approx \sqrt{1.2544 \times 10^8} \approx 11.2 \, \text{km/s}
\]
3. **Additional Velocity Needed:**
The additional velocity required to overcome gravitational pull is the difference between the escape velocity and the orbital velocity:
\[
\Delta v = v_e - v_0
\]
\[
\Delta v \approx 11.2 \, \text{km/s} - 7.9 \, \text{km/s} = 3.3 \, \text{km/s}
\]
Rounding to the nearest option provided:
\[
\Delta v \approx 3.2 \, \text{km/s}
\]
So, the correct answer is:
**C) 3.2 km/s**
