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 solve the problem of finding the acceleration due to gravity at the surface of a satellite that has half the mass and half the radius of Earth, we can follow these steps:

### Step 1: Understand the formula for acceleration due to gravity
The acceleration due to gravity \( g \) at the surface of a planet or satellite is given by the formula:
\[ g = \frac{GM}{R^2} \]
where \( G \) is the universal gravitational constant, \( M \) is the mass of the body, and \( R \) is the radius of the body.

### Step 2: Define the parameters for the satellite
According to the problem, the mass \( M' \) and radius \( R' \) of the satellite are given as:
\[ M' = \frac{M}{2} \quad \text{and} \quad R' = \frac{R}{2} \]
where \( M \) and \( R \) are the mass and radius of Earth.

### Step 3: Substitute the parameters into the formula
Now, we can substitute \( M' \) and \( R' \) into the formula for \( g \):
\[ g' = \frac{G \cdot M'}{(R')^2} = \frac{G \cdot \left(\frac{M}{2}\right)}{\left(\frac{R}{2}\right)^2} \]

### Step 4: Simplify the expression
Now simplify the expression:
\[ g' = \frac{G \cdot \left(\frac{M}{2}\right)}{\left(\frac{R^2}{4}\right)} = \frac{G \cdot M}{2} \cdot \frac{4}{R^2} \]
This simplifies to:
\[ g' = \frac{4GM}{2R^2} = \frac{2GM}{R^2} \]

### Step 5: Relate it to Earth's gravity
We know that \( g = \frac{GM}{R^2} \), so we can express \( g' \) in terms of \( g \):
\[ g' = 2 \cdot \frac{GM}{R^2} = 2g \]

### Step 6: Calculate the value of \( g' \)
Given that the acceleration due to gravity at the surface of Earth \( g \) is approximately \( 9.8 \, \text{m/s}^2 \):
\[ g' = 2 \cdot 9.8 \, \text{m/s}^2 = 19.6 \, \text{m/s}^2 \]

### Conclusion
Thus, the acceleration due to gravity at the surface of the satellite is:
\[ \boxed{19.6 \, \text{m/s}^2} \]