Acceleration due to gravity is ‘ g ’ on the surface of the earth. The value of acceleration due to gravity at a height of 32 km above earth’s surface is (Radius of the earth = 6400 km )

Options

A. 0.99 g
B. 0.8g
C. 1.01 g
D. 0.9 g

Answer

To find the acceleration due to gravity at a height of 32 km above the Earth's surface, we can use the formula for gravitational acceleration at a height \( h \) above the Earth's surface:

\[ g' = \frac{g}{(1 + \frac{h}{r})^2} \]

Where:
- \( g' \) is the acceleration due to gravity at height \( h \)
- \( g \) is the acceleration due to gravity at the Earth's surface (approximately \( 9.81 \, \text{m/s}^2 \))
- \( h \) is the height above the Earth's surface (32 km)
- \( r \) is the radius of the Earth (6400 km)

### Step 1: Convert units
First, we need to ensure that the height and radius are in the same units. Since both are given in kilometers, we can use them directly.

### Step 2: Substitute the values into the formula
We know:
- \( h = 32 \, \text{km} \)
- \( r = 6400 \, \text{km} \)

Now substituting these values into the formula:

\[ g' = \frac{g}{(1 + \frac{32}{6400})^2} \]

### Step 3: Simplify the fraction
Calculate \( \frac{32}{6400} \):

\[ \frac{32}{6400} = \frac{1}{200} \]

Now, substitute this back into the equation:

\[ g' = \frac{g}{(1 + \frac{1}{200})^2} \]

### Step 4: Calculate \( 1 + \frac{1}{200} \)
Now, calculate \( 1 + \frac{1}{200} \):

\[ 1 + \frac{1}{200} = \frac{200 + 1}{200} = \frac{201}{200} \]

### Step 5: Square the result
Now, square the result:

\[ \left(\frac{201}{200}\right)^2 = \frac{201^2}{200^2} = \frac{40401}{40000} \]

### Step 6: Substitute back into the equation
Now we can substitute this back into the equation for \( g' \):

\[ g' = \frac{g}{\frac{40401}{40000}} = g \cdot \frac{40000}{40401} \]

### Step 7: Calculate the approximate value
For practical purposes, we can approximate:

\[ g' \approx g \cdot (1 - \frac{1}{200}) = g \cdot 0.995 \]

Thus, the acceleration due to gravity at a height of 32 km above the Earth's surface is approximately:

\[ g' \approx 0.995g \]

### Final Result
If we take \( g \approx 9.81 \, \text{m/s}^2 \):

\[ g' \approx 0.995 \times 9.81 \approx 9.76 \, \text{m/s}^2 \]

### Conclusion
The value of acceleration due to gravity at a height of 32 km above Earth's surface is approximately \( 9.76 \, \text{m/s}^2 \).

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