Question

The mass of a planet is \(\frac{1}{10}\)th that of the earth and its diameter is half that of the earth. The acceleration due to gravity on that planet is :

• A. 19.6 m s^–2
• B. 9.8 m s^–2
• C. 4.6 m s^–2
• D. 3.92 m s^–2

Answer 1

The Correct Option is D

Step 1: Use the Formula for Acceleration Due to Gravity

The formula for acceleration due to gravity is:

$$ g = \frac{G M}{R^2} $$

Where:

G = Gravitational constant

M = Mass of the planet

R = Radius of the planet

Step 2: Express the Planet’s Mass and Radius in Terms of Earth’s

Let Earth’s mass and radius be:

\( M_e \) = Earth's mass

\( R_e \) = Earth's radius

For the given planet:

$$ M_p = \frac{M_e}{10}, \quad R_p = \frac{R_e}{2} $$

Step 3: Substitute Values into the Formula

Using the gravitational formula for the planet:

$$ g_p = \frac{G M_p}{R_p^2} $$

Substituting the values:

$$ g_p = \frac{G \cdot \frac{M_e}{10}}{\left(\frac{R_e}{2}\right)^2} $$

Rewriting the denominator:

$$ g_p = \frac{G \cdot \frac{M_e}{10}}{\frac{R_e^2}{4}} $$

Since division by a fraction is multiplication by its reciprocal:

$$ g_p = G \cdot M_e \cdot \frac{4}{10 R_e^2} $$

Since we know that:

$$ g_e = \frac{G M_e}{R_e^2} = 9.8 \text{ m/s}^2 $$

We substitute \( g_e \) into the equation:

$$ g_p = \frac{4}{10} \cdot 9.8 $$

Solving:

$$ g_p = 3.92 \text{ m/s}^2 $$

Conclusion

The acceleration due to gravity on the given planet is 3.92 m/s².