A planet has a mass equal to $(1/10)^{ ext{th}}$ of Earth's mass and a diameter equal to half of Earth's diam

Question

A planet has a mass equal to $(1/10)^{	ext{th}}$ of Earth's mass and a diameter equal to half of Earth's diameter. The acceleration due to gravity on this planet is:

Accepted Answer

1. **Formula:** The acceleration due to gravity on the surface of a spherical body is given by $g = \frac{GM}{R^2}$, where $G$ is the gravitational constant, $M$ is the mass, and $R$ is the radius .
2. **Given Data:**
   - Mass of the planet, $M_p = \frac{1}{10} M_E$.
   - Diameter of the planet is half that of Earth, which implies the radius $R_p = \frac{1}{2} R_E$.
   - Acceleration due to gravity on Earth, $g_E = 9.8 	ext{ m s}^{-2}$.
3. **Ratio Calculation:**
   $$ g_p = \frac{G M_p}{R_p^2} = \frac{G (M_E / 10)}{(R_E / 2)^2} $$
   $$ g_p = \frac{1}{10} 	imes \frac{1}{(1/2)^2} 	imes \frac{G M_E}{R_E^2} $$
   $$ g_p = \frac{1}{10} 	imes 4 	imes g_E = 0.4 g_E $$
4. **Result:**
   $$ g_p = 0.4 	imes 9.8 	ext{ m s}^{-2} = 3.92 	ext{ m s}^{-2} $$

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