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:
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:
11.2 km/s
8.4 km/s
4.2 km/s
7.9 km/s
The Correct Option is D
Solution and Explanation
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.
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