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 on the new planet.
- \( G \) is the universal gravitational constant.
- \( M' \) is the mass of the new planet.
- \( R' \) is the radius of the new planet.

Given:
- The mass of the new planet \( M' = \frac{1}{2} M \) (where \( M \) is the mass of Earth).
- The radius of the new planet \( R' = \frac{1}{2} R \) (where \( R \) is the radius of Earth).
- The acceleration due to gravity at the surface of Earth \( g = 10 \, \text{m/s}^2 \).

### Step 1: Substitute the values into the formula

We can express \( g' \) in terms of \( g \):

\[ g' = \frac{G \cdot \left(\frac{1}{2} M\right)}{\left(\frac{1}{2} R\right)^2} \]

### Step 2: Simplify the equation

Now, simplifying the denominator:

\[ \left(\frac{1}{2} R\right)^2 = \frac{1}{4} R^2 \]

So we can rewrite \( g' \):

\[ g' = \frac{G \cdot \left(\frac{1}{2} M\right)}{\frac{1}{4} R^2} \]

### Step 3: Further simplify

This can be simplified further:

\[ g' = \frac{G \cdot M}{R^2} \cdot \frac{1/2}{1/4} \]

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

### Step 4: Substitute \( g \)

Since \( \frac{G \cdot M}{R^2} = g \):

\[ g' = 2g \]

### Step 5: Calculate \( g' \)

Now substituting the value of \( g \):

\[ g' = 2 \cdot 10 \, \text{m/s}^2 = 20 \, \text{m/s}^2 \]

### Final Answer

Thus, the acceleration due to gravity at the surface of the planet is:

\[ \boxed{20 \, \text{m/s}^2} \]

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