Question

If the escape speed of a projectile on Earth's surface is \( 11.2 \, \text{km/s} \) and a body is projected out with thrice this speed, then determine the speed of the body far away from the Earth.

• A. 56.63 km/s
• B. 33 km/s
• C. 39 km/s
• D. 31.7 km/s

Hint:

The speed of a body projected at thrice the escape velocity will be three times the escape velocity at infinity.

Answer 1

The Correct Option is D

Given:

• Escape speed at Earth's surface, \( v_e = 11.2 \, \text{km/s} \)
• Projection speed, \( v_0 = 3v_e = 33.6 \, \text{km/s} \)

Step 1: Understand escape speed

The escape speed \( v_e \) is the minimum speed needed to break free from Earth's gravitational pull without additional propulsion. At escape speed, the total energy at infinity is zero.

Step 2: Apply energy conservation

At Earth's surface:

\[ \text{Total Energy} = \text{Kinetic Energy} + \text{Potential Energy} \] \[ E_{\text{total}} = \frac{1}{2}mv_0^2 - \frac{GMm}{R} \]

Where:

• \( G \) = gravitational constant
• \( M \) = Earth's mass
• \( R \) = Earth's radius
• \( m \) = mass of the projectile

At infinity (far from Earth):

\[ E_{\text{total}} = \frac{1}{2}mv_{\infty}^2 \]

Step 3: Relate to escape speed

We know that escape speed relates to potential energy:

\[ \frac{1}{2}mv_e^2 = \frac{GMm}{R} \]

Substituting into the energy equation:

\[ \frac{1}{2}mv_0^2 - \frac{1}{2}mv_e^2 = \frac{1}{2}mv_{\infty}^2 \]

Step 4: Solve for final speed

\[ v_{\infty}^2 = v_0^2 - v_e^2 \] \[ v_{\infty} = \sqrt{(3v_e)^2 - v_e^2} \] \[ v_{\infty} = \sqrt{9v_e^2 - v_e^2} \] \[ v_{\infty} = \sqrt{8v_e^2} \] \[ v_{\infty} = v_e\sqrt{8} \] \[ v_{\infty} = 11.2 \times 2.828 \] \[ v_{\infty} \approx 31.7 \, \text{km/s} \]