At what height above the surface of earth the value of "g" decreases by 2 % [ radius of the earth is 6400 km ]

Options

A. 32 km
B. 64 km
C. 128 km
D. 1600 km

Answer

To solve the problem of finding the height above the Earth's surface where the value of "g" decreases by 2%, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We know that the acceleration due to gravity at the surface of the Earth (g) is approximately 9.8 m/s². We need to find the height (h) above the Earth's surface where the value of g decreases by 2%.

2. **Calculate the New Value of g**:
A decrease of 2% means that the new value of g (g') is:
\[ g' = g - 0.02g = 0.98g \]

3. **Using the Formula for g at Height h**:
The formula for the acceleration due to gravity at a height h above the Earth's surface is given by:
\[ g' = \frac{g}{(1 + \frac{h}{R})^2} \]
where R is the radius of the Earth. For small heights compared to the radius of the Earth, we can use the binomial approximation:
\[ (1 + x)^{-2} \approx 1 - 2x \quad \text{for small } x \]
Thus, we can approximate:
\[ g' \approx g \left(1 - \frac{2h}{R}\right) \]

4. **Setting Up the Equation**:
We set the approximate expression for g' equal to 0.98g:
\[ 0.98g \approx g \left(1 - \frac{2h}{R}\right) \]

5. **Cancelling g**:
Since g is common on both sides, we can cancel it out:
\[ 0.98 \approx 1 - \frac{2h}{R} \]

6. **Rearranging the Equation**:
Rearranging gives us:
\[ \frac{2h}{R} \approx 1 - 0.98 = 0.02 \]

7. **Solving for h**:
Now we can solve for h:
\[ 2h \approx 0.02R \]
\[ h \approx \frac{0.02R}{2} = 0.01R \]

8. **Substituting the Radius of the Earth**:
Given that the radius of the Earth (R) is 6400 km:
\[ h \approx 0.01 \times 6400 \text{ km} = 64 \text{ km} \]

### Final Answer:
The height above the surface of the Earth at which the value of "g" decreases by 2% is approximately **64 km**.

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