If the mass of earth is 80 times of that of a planet and diameter is double that of planet and ‘ g ’ on earth is `9.8 m//s^(2)` , then the value of ‘ g ’ on that planet is

Options

A. `4.9 m//s^(2)`
B. `0.98 m//s^(2)`
C. `0.49 m//s^(2)`
D. `49 m//s^(2)`

Answer

To find the value of 'g' on the planet, we can use the formula for acceleration due to gravity:

\[ g = \frac{GM}{R^2} \]

where:
- \( G \) is the universal gravitational constant,
- \( M \) is the mass of the celestial body,
- \( R \) is the radius of the celestial body.

### Step-by-Step Solution:

1. **Identify the mass of the planet:**
- Given that the mass of the Earth is \( M_E \) and the mass of the planet is \( M_P = \frac{M_E}{80} \).

2. **Identify the radius of the planet:**
- Given that the diameter of the planet is half that of the Earth, the radius of the planet is \( R_P = \frac{R_E}{2} \), where \( R_E \) is the radius of the Earth.

3. **Substituting values into the formula for 'g':**
- The formula for 'g' on the planet becomes:
\[ g_P = \frac{G M_P}{R_P^2} \]
- Substitute \( M_P \) and \( R_P \):
\[ g_P = \frac{G \left(\frac{M_E}{80}\right)}{\left(\frac{R_E}{2}\right)^2} \]

4. **Simplifying the expression:**
- The denominator becomes:
\[ \left(\frac{R_E}{2}\right)^2 = \frac{R_E^2}{4} \]
- Thus, we have:
\[ g_P = \frac{G \left(\frac{M_E}{80}\right)}{\frac{R_E^2}{4}} = \frac{4 G M_E}{80 R_E^2} \]
- This simplifies to:
\[ g_P = \frac{G M_E}{20 R_E^2} \]

5. **Using the known value of 'g' on Earth:**
- We know that:
\[ g_E = \frac{G M_E}{R_E^2} = 9.8 \, \text{m/s}^2 \]
- Therefore, we can substitute this into our equation for \( g_P \):
\[ g_P = \frac{9.8}{20} \]

6. **Calculating the final value:**
- Performing the division:
\[ g_P = 0.49 \, \text{m/s}^2 \]

### Final Answer:
The value of 'g' on that planet is \( 0.49 \, \text{m/s}^2 \).

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