At what height above the earth's surface is the acceleration due to gravity 1% less than its value at the surface ? [R = 6400 km]

Options

A. 16 km
B. 32 km
C. 64 km
D. `32sqrt()2 km`

Answer

To find the height above the Earth's surface where the acceleration due to gravity is 1% less than its value at the surface, we can follow these steps:

### Step 1: Understand the relationship between gravity at height and at the surface
The acceleration due to gravity at a height \( h \) above the Earth's surface can be expressed as:
\[ g_h = \frac{g_0}{(1 + \frac{h}{R})^2} \]
where:
- \( g_h \) = acceleration due to gravity at height \( h \)
- \( g_0 \) = acceleration due to gravity at the surface of the Earth
- \( R \) = radius of the Earth (given as 6400 km)

### Step 2: Set up the equation for 1% less gravity
According to the problem, we want \( g_h \) to be 1% less than \( g_0 \):
\[ g_h = 0.99 g_0 \]

### Step 3: Substitute \( g_h \) in the equation
Substituting \( g_h \) in the equation, we get:
\[ 0.99 g_0 = \frac{g_0}{(1 + \frac{h}{R})^2} \]

### Step 4: Cancel \( g_0 \) from both sides
Assuming \( g_0 \neq 0 \), we can cancel \( g_0 \) from both sides:
\[ 0.99 = \frac{1}{(1 + \frac{h}{R})^2} \]

### Step 5: Rearrange the equation
Taking the reciprocal gives:
\[ (1 + \frac{h}{R})^2 = \frac{1}{0.99} \]

### Step 6: Take the square root
Taking the square root of both sides:
\[ 1 + \frac{h}{R} = \sqrt{\frac{1}{0.99}} \]

### Step 7: Isolate \( h \)
Subtracting 1 from both sides:
\[ \frac{h}{R} = \sqrt{\frac{1}{0.99}} - 1 \]
Multiplying both sides by \( R \):
\[ h = R \left(\sqrt{\frac{1}{0.99}} - 1\right) \]

### Step 8: Substitute the value of \( R \)
Substituting \( R = 6400 \) km:
\[ h = 6400 \left(\sqrt{\frac{1}{0.99}} - 1\right) \]

### Step 9: Calculate \( \sqrt{\frac{1}{0.99}} \)
Calculating \( \sqrt{\frac{1}{0.99}} \):
\[ \sqrt{\frac{1}{0.99}} \approx 1.005 \]

### Step 10: Final calculation for \( h \)
Substituting this back into the equation:
\[ h = 6400 \times (1.005 - 1) \approx 6400 \times 0.005 = 32 \text{ km} \]

### Final Answer
The height above the Earth's surface where the acceleration due to gravity is 1% less than its value at the surface is approximately **32 km**.

---