The value of acceleration due to gravity at Earth's surface is `9.8 ms^(-2)`.The altitude above its surface at which the acceleration due to gravity decreases to `4.9 ms^(-2)`m, is close to: (Radius of earth =`6.4xx10^6 m`)

Options

A. `2.6xx10^6 m`
B. `6.4xx10^6 m`
C. `9.0xx10^6 m`
D. `1.6xx10^6 m`

Answer

To solve the problem, we need to find the altitude \( h \) above the Earth's surface where the acceleration due to gravity decreases from \( 9.8 \, \text{m/s}^2 \) to \( 4.9 \, \text{m/s}^2 \).

### Step-by-Step Solution:

1. **Understand the relationship between gravity at the surface and at height**:
The formula for the acceleration due to gravity at a height \( h \) above the surface of the Earth is given by:
\[ g' = g \cdot \frac{R}{R + h}^2 \]
where:
- \( g' \) is the acceleration due to gravity at height \( h \),
- \( g \) is the acceleration due to gravity at the surface (which is \( 9.8 \, \text{m/s}^2 \)),
- \( R \) is the radius of the Earth (given as \( 6.4 \times 10^6 \, \text{m} \)),
- \( h \) is the height we need to find.

2. **Set up the equation**:
We know that at height \( h \), \( g' = 4.9 \, \text{m/s}^2 \). Plugging in the values, we have:
\[ 4.9 = 9.8 \cdot \frac{6.4 \times 10^6}{(6.4 \times 10^6 + h)^2} \]

3. **Simplify the equation**:
Dividing both sides by \( 9.8 \):
\[ \frac{4.9}{9.8} = \frac{6.4 \times 10^6}{(6.4 \times 10^6 + h)^2} \]
This simplifies to:
\[ \frac{1}{2} = \frac{6.4 \times 10^6}{(6.4 \times 10^6 + h)^2} \]

4. **Cross-multiply**:
\[ (6.4 \times 10^6 + h)^2 = 2 \cdot 6.4 \times 10^6 \]

5. **Take the square root**:
Taking the square root of both sides gives:
\[ 6.4 \times 10^6 + h = \sqrt{2} \cdot 6.4 \times 10^6 \]

6. **Isolate \( h \)**:
Rearranging gives:
\[ h = \sqrt{2} \cdot 6.4 \times 10^6 - 6.4 \times 10^6 \]
Factoring out \( 6.4 \times 10^6 \):
\[ h = ( \sqrt{2} - 1 ) \cdot 6.4 \times 10^6 \]

7. **Calculate \( h \)**:
The value of \( \sqrt{2} \) is approximately \( 1.414 \):
\[ h = (1.414 - 1) \cdot 6.4 \times 10^6 \]
\[ h = 0.414 \cdot 6.4 \times 10^6 \]
\[ h \approx 2.65 \times 10^6 \, \text{m} \]

### Final Answer:
The altitude \( h \) at which the acceleration due to gravity decreases to \( 4.9 \, \text{m/s}^2 \) is approximately \( 2.65 \times 10^6 \, \text{m} \).