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

Options

A. `5 ms ^(-2)`
B. `10 ms ^(-2)`
C. `20 ms ^(-2)`
D. `40 ms ^(-2)`

Answer

To find the acceleration due to gravity at the surface of a planet with half the mass and half the radius 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 the given values:**
- Acceleration due to gravity at Earth's surface, \( g_E = 10 \, \text{m/s}^2 \).
- Mass of the planet, \( M_1 = \frac{1}{2} M_E \) (where \( M_E \) is the mass of Earth).
- Radius of the planet, \( R_1 = \frac{1}{2} R_E \) (where \( R_E \) is the radius of Earth).

2. **Write the formula for gravitational acceleration on the planet:**
\[ g_1 = \frac{G \cdot M_1}{R_1^2} \]

3. **Substitute the values of \( M_1 \) and \( R_1 \):**
\[ g_1 = \frac{G \cdot \left(\frac{1}{2} M_E\right)}{\left(\frac{1}{2} R_E\right)^2} \]

4. **Simplify the denominator:**
\[ \left(\frac{1}{2} R_E\right)^2 = \frac{1}{4} R_E^2 \]
Thus, we can rewrite \( g_1 \):
\[ g_1 = \frac{G \cdot \left(\frac{1}{2} M_E\right)}{\frac{1}{4} R_E^2} \]

5. **Simplify the expression:**
\[ g_1 = \frac{G \cdot M_E}{R_E^2} \cdot \frac{1/2}{1/4} = \frac{G \cdot M_E}{R_E^2} \cdot 2 \]
Therefore:
\[ g_1 = 2 \cdot \frac{G \cdot M_E}{R_E^2} \]

6. **Substitute the value of \( g_E \):**
Since \( \frac{G \cdot M_E}{R_E^2} = g_E = 10 \, \text{m/s}^2 \):
\[ g_1 = 2 \cdot 10 \, \text{m/s}^2 = 20 \, \text{m/s}^2 \]

7. **Conclusion:**
The acceleration due to gravity at the surface of the planet is:
\[ g_1 = 20 \, \text{m/s}^2 \]

### Final Answer:
The value of acceleration due to gravity at the surface of the planet is **20 m/s²**.