Taking the radius of the earth to be 6400 km, by what percentage will the acceleration due to gravity at a height of 100 km from the surface of the earth differ from that on the surface of the earth

Options

A. About 1.5%
B. About 5%
C. About 8%
D. About 3%

Explanation

To solve the problem of how the acceleration due to gravity changes at a height of 100 km above the Earth's surface, we can follow these steps:

### Step 1: Understand the formula for gravitational acceleration
The acceleration due to gravity at a height \( h \) above the Earth's surface can be calculated using the formula:
\[ g' = g \cdot \frac{R^2}{(R + h)^2} \]
where:
- \( g' \) is the acceleration due to gravity at height \( h \),
- \( g \) is the acceleration due to gravity at the Earth's surface (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( R \) is the radius of the Earth (given as 6400 km),
- \( h \) is the height above the surface (given as 100 km).

### Step 2: Convert units
Since the radius of the Earth is given in kilometers, we should convert it to meters for consistency:
\[ R = 6400 \, \text{km} = 6400 \times 1000 \, \text{m} = 6.4 \times 10^6 \, \text{m} \]
\[ h = 100 \, \text{km} = 100 \times 1000 \, \text{m} = 1 \times 10^5 \, \text{m} \]

### Step 3: Calculate \( g' \)
Now, we can substitute the values into the formula:
\[ g' = g \cdot \frac{(6.4 \times 10^6)^2}{(6.4 \times 10^6 + 1 \times 10^5)^2} \]
Calculating \( R + h \):
\[ R + h = 6.4 \times 10^6 + 1 \times 10^5 = 6.5 \times 10^6 \, \text{m} \]
Now substituting back:
\[ g' = g \cdot \frac{(6.4 \times 10^6)^2}{(6.5 \times 10^6)^2} \]

### Step 4: Simplify the expression
The ratio can be simplified:
\[ g' = g \cdot \left(\frac{64}{65}\right)^2 \]

### Step 5: Calculate the percentage change in \( g \)
To find the percentage change in gravitational acceleration, we can use:
\[ \text{Percentage Change} = \frac{g - g'}{g} \times 100 \]
Substituting \( g' \):
\[ \text{Percentage Change} = \frac{g - g \cdot \left(\frac{64}{65}\right)^2}{g} \times 100 \]
This simplifies to:
\[ \text{Percentage Change} = \left(1 - \left(\frac{64}{65}\right)^2\right) \times 100 \]

### Step 6: Calculate \( \left(\frac{64}{65}\right)^2 \)
Calculating \( \left(\frac{64}{65}\right)^2 \):
\[ \left(\frac{64}{65}\right)^2 = \frac{4096}{4225} \approx 0.968 \]

### Step 7: Final calculation
Now substituting back:
\[ \text{Percentage Change} = (1 - 0.968) \times 100 \approx 3.2\% \]

### Conclusion
The acceleration due to gravity at a height of 100 km from the Earth's surface differs from that on the surface of the Earth by approximately **3.2%**.