If a planet consists of a satellite whose mass and radius were both half that of the earh, then acceleration due to gravity at its surface would be

Options

A. `4.9 "ms"^(-2)`
B. `9.8 "ms"^(-2)`
C. `19.6 "ms"^(-2)`
D. `29.4 "ms"^(-2)`

Explanation

To find the acceleration due to gravity at the surface of a satellite that has both mass and radius 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 (or satellite in this case),
- \( R \) is the radius of the planet (or satellite).

### Step-by-Step Solution:

1. **Identify the known values:**
- Let the mass of Earth be \( M \) and the radius of Earth be \( R \).
- The mass of the satellite \( M' = \frac{M}{2} \) (half of Earth's mass).
- The radius of the satellite \( R' = \frac{R}{2} \) (half of Earth's radius).

2. **Substitute the values into the formula for gravitational acceleration:**
- The acceleration due to gravity at the surface of the satellite \( g' \) can be expressed as:
\[ g' = \frac{G \cdot M'}{(R')^2} \]

3. **Plug in the values for \( M' \) and \( R' \):**
\[ g' = \frac{G \cdot \left(\frac{M}{2}\right)}{\left(\frac{R}{2}\right)^2} \]

4. **Simplify the equation:**
- The denominator becomes:
\[ \left(\frac{R}{2}\right)^2 = \frac{R^2}{4} \]
- Thus, the equation for \( g' \) becomes:
\[ g' = \frac{G \cdot \left(\frac{M}{2}\right)}{\frac{R^2}{4}} = \frac{G \cdot M}{2} \cdot \frac{4}{R^2} \]
- This simplifies to:
\[ g' = \frac{4G \cdot M}{2R^2} = \frac{2G \cdot M}{R^2} \]

5. **Relate it to Earth's gravitational acceleration:**
- We know that the acceleration due to gravity at the surface of Earth is:
\[ g = \frac{G \cdot M}{R^2} \]
- Therefore, we can express \( g' \) in terms of \( g \):
\[ g' = 2g \]

6. **Substitute the value of \( g \):**
- The standard value of \( g \) on Earth is approximately \( 9.8 \, \text{m/s}^2 \).
- Thus:
\[ g' = 2 \cdot 9.8 \, \text{m/s}^2 = 19.6 \, \text{m/s}^2 \]

### Final Answer:
The acceleration due to gravity at the surface of the satellite is \( 19.6 \, \text{m/s}^2 \).

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