Acceleration due to gravity at earth's surface is `10ms^(-2)`. The value of acceleration due to gravity at the surface of a planet of mass `(1/5)^(th)` and radius `1/2` of the earth is

Options

A. `4ms^(-2)`
B. `6ms^(-2)`
C. `8ms^(-2)`
D. `12ms^(-2)`

Answer

To find the acceleration due to gravity at the surface of a planet with a mass that is one-fifth that of Earth and a radius that is half that of Earth, we can use the formula for gravitational acceleration:

\[ g = \frac{G \cdot M}{R^2} \]

Where:
- \( g \) is the acceleration due to gravity,
- \( G \) is the universal gravitational constant,
- \( M \) is the mass of the planet,
- \( R \) is the radius of the planet.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Acceleration due to gravity at Earth's surface, \( g_e = 10 \, \text{m/s}^2 \).
- Mass of the planet, \( M_p = \frac{1}{5} M_e \) (where \( M_e \) is the mass of Earth).
- Radius of the planet, \( R_p = \frac{1}{2} R_e \) (where \( R_e \) is the radius of Earth).

2. **Write the Formula for Acceleration due to Gravity for Earth**:
\[ g_e = \frac{G \cdot M_e}{R_e^2} \]

3. **Write the Formula for Acceleration due to Gravity for the Planet**:
\[ g_p = \frac{G \cdot M_p}{R_p^2} \]

4. **Substitute the Mass and Radius of the Planet**:
Substitute \( M_p \) and \( R_p \) into the formula for \( g_p \):
\[ g_p = \frac{G \cdot \left(\frac{1}{5} M_e\right)}{\left(\frac{1}{2} R_e\right)^2} \]

5. **Simplify the Expression**:
\[ g_p = \frac{G \cdot \frac{1}{5} M_e}{\frac{1}{4} R_e^2} \]
\[ g_p = \frac{G \cdot M_e}{R_e^2} \cdot \frac{1}{5} \cdot 4 \]
\[ g_p = g_e \cdot \frac{4}{5} \]

6. **Substitute the Value of \( g_e \)**:
\[ g_p = 10 \cdot \frac{4}{5} \]
\[ g_p = 10 \cdot 0.8 \]
\[ g_p = 8 \, \text{m/s}^2 \]

### Final Answer:
The value of acceleration due to gravity at the surface of the planet is \( 8 \, \text{m/s}^2 \).

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