Question

Escape velocity of a body from earth is 11.2 km/s. If the radius of a planet be one-third the radius of earth and mass be one-sixth that of earth, the escape velocity from the planet is:

• A. 11.2 km/s
• B. 8.4 km/s
• C. 4.2 km/s
• D. 7.9 km/s

Answer 1

The Correct Option is D

Solution: The escape velocity \( V_e \) from a celestial body is given by the formula:

\[ V_e = \sqrt{\frac{2GM}{R}}, \]

where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the body,
- \( R \) is the radius of the body.

Given Values for Earth: Escape velocity from Earth, \( V_{e,\text{earth}} = 11.2 \, \text{km/s} \). For Earth, let:
Mass \( M_e \) and radius \( R_e \) be constants.

For the Planet: The radius of the planet \( R_p = \frac{1}{3} R_e \). The mass of the planet \( M_p = \frac{1}{6} M_e \).

Calculating Escape Velocity for the Planet: Substituting the values into the escape velocity formula:

\[ V_{e,\text{planet}} = \sqrt{\frac{2G \left( \frac{1}{6} M_e \right)}{\frac{1}{3} R_e}}. \]

Simplifying the expression:

\[ V_{e,\text{planet}} = \sqrt{\frac{2GM_e}{R_e}} \times \sqrt{\frac{1}{6} \times 3}. \]

Thus, it can be expressed as:

\[ V_{e,\text{planet}} = V_{e,\text{earth}} \times \sqrt{\frac{1}{2}}. \]

Substituting \( V_{e,\text{earth}} = 11.2 \, \text{km/s} \):

\[ V_{e,\text{planet}} = 11.2 \times \sqrt{\frac{1}{2}} = 11.2 \times 0.7071 \approx 7.9 \, \text{km/s}. \]

Thus, the escape velocity from the planet is: 7.9 km/s.