Question

What is the acceleration due to gravity (m/s²) on the surface of a planet if its radius is 1/4th that of earth and its mass is 1/80th that of earth" Assume that the gravity on the surface of the earth is 10 m/s².

Answer 1

Correct Answer: 2

To determine the acceleration due to gravity on the surface of a planet, we use the formula \( g = \frac{G \cdot M}{R^2} \), where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is its radius. For Earth, the gravity \( g_{\text{Earth}} \) is given as 10 m/s². Let \( M_{\text{Earth}} \) and \( R_{\text{Earth}} \) be the mass and radius of Earth, respectively. On the new planet:

• The mass \( M \) is \(\frac{1}{80}\) of Earth, i.e., \( M = \frac{M_{\text{Earth}}}{80} \).
• The radius \( R \) is \(\frac{1}{4}\) of Earth, i.e., \( R = \frac{R_{\text{Earth}}}{4} \).

We substitute these into the formula for gravity: \( g = \frac{G \cdot (M_{\text{Earth}}/80)}{(\frac{R_{\text{Earth}}}{4})^2} = \frac{G \cdot M_{\text{Earth}}}{80 \cdot \frac{R_{\text{Earth}}^2}{16}} = \frac{16 \cdot G \cdot M_{\text{Earth}}}{80 \cdot R_{\text{Earth}}^2} = \frac{1}{5} \cdot \frac{G \cdot M_{\text{Earth}}}{R_{\text{Earth}}^2} = \frac{1}{5} \cdot g_{\text{Earth}} \). Since \( g_{\text{Earth}} = 10 \, \text{m/s}^2 \), it follows that \( g = \frac{1}{5} \times 10 = 2 \, \text{m/s}^2 \).