What is the value of acceleration due to gravity at a height equal to helf the radius of earth, from surface of earth ? [take `g = 10 m//s^(2)` on earth's surface]

Answer

To find the value of acceleration due to gravity at a height equal to half the radius of the Earth, we can use the formula that relates the acceleration due to gravity at a height \( h \) above the surface of the Earth to the acceleration due to gravity at the surface \( g \).

### Step-by-Step Solution:

1. **Identify the Variables**:
- Let \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity at the Earth's surface).
- The height \( h \) is given as \( \frac{R}{2} \), where \( R \) is the radius of the Earth.

2. **Use the Formula for Gravity at Height**:
The formula for the acceleration due to gravity at a height \( h \) is given by:
\[ g' = \frac{g}{(1 + \frac{h}{R})^2} \]
where \( g' \) is the acceleration due to gravity at height \( h \).

3. **Substitute the Values**:
Substitute \( h = \frac{R}{2} \) into the formula:
\[ g' = \frac{g}{(1 + \frac{\frac{R}{2}}{R})^2} \]
Simplifying the fraction:
\[ g' = \frac{g}{(1 + \frac{1}{2})^2} = \frac{g}{(1.5)^2} \]

4. **Calculate \( (1.5)^2 \)**:
\[ (1.5)^2 = 2.25 \]

5. **Substitute \( g \) into the Equation**:
Now substitute \( g = 10 \, \text{m/s}^2 \):
\[ g' = \frac{10}{2.25} \]

6. **Perform the Division**:
\[ g' = \frac{10}{2.25} \approx 4.44 \, \text{m/s}^2 \]

### Final Answer:
The value of acceleration due to gravity at a height equal to half the radius of the Earth is approximately \( 4.44 \, \text{m/s}^2 \).

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