Question

The mass of the moon is 1/144 times the mass of a planet and its diameter 1/16 times the diameter of a planet. If the escape velocity on the planet is v, the escape velocity on the moon will be:

• A. $\frac{v}{3}$
• B. $\frac{v}{4}$
• C. $\frac{v}{12}$
• D. $\frac{v}{16}$

Answer 1

Answer: (A)

$\frac{v}{3}$

Answer 2

The escape velocity $v_{\text{escape}}$ for a celestial body is given by:

$v_{\text{escape}} = \sqrt{\frac{2GM}{R}},$

where $G$ is the gravitational constant, $M$ is the mass of the body, and $R$ is its radius.

Given that the escape velocity on the planet is $v$:

$v = \sqrt{\frac{2GM}{R}}.$

For the moon:

• Mass $M_{\text{moon}} = \frac{M}{144}$
• Radius $R_{\text{moon}} = \frac{R}{16}$

Substitute these values into the escape velocity formula for the moon:

$v_{\text{moon}} = \sqrt{\frac{2G \cdot \frac{M}{144}}{\frac{R}{16}}} = \sqrt{\frac{2GM \cdot 16}{144R}} = \sqrt{\frac{2GM}{9R}} = \frac{1}{3} \sqrt{\frac{2GM}{R}} = \frac{v}{3}.$

Thus, the escape velocity on the moon is:

$\frac{v}{3}.$