At what height from the surface of earth will the value of g be reduced by `36%` from the value on the surface? Take radius of earth `R = 6400 km`.

Options

A. `1600km`
B. `1800km`
C. `1400km`
D. `1200km`

Explanation

To solve the problem of finding the height from the surface of the Earth at which the value of \( g \) is reduced by 36%, we can follow these steps:

### Step 1: Understand the relationship between \( g \) and height
The acceleration due to gravity \( g' \) at a height \( h \) above the Earth's surface is given by the formula:
\[ g' = \frac{g}{(1 + \frac{h}{R})^2} \]
where \( g \) is the acceleration due to gravity at the surface of the Earth and \( R \) is the radius of the Earth.

### Step 2: Set up the equation for the reduced value of \( g \)
Since we want \( g' \) to be reduced by 36%, we can express this as:
\[ g' = g \times (1 - 0.36) = 0.64g \]

### Step 3: Substitute \( g' \) into the equation
Substituting \( g' = 0.64g \) into the equation from Step 1 gives us:
\[ 0.64g = \frac{g}{(1 + \frac{h}{R})^2} \]

### Step 4: Cancel \( g \) from both sides
Since \( g \) is common on both sides, we can cancel it out:
\[ 0.64 = \frac{1}{(1 + \frac{h}{R})^2} \]

### Step 5: Rearrange the equation
Taking the reciprocal of both sides gives:
\[ (1 + \frac{h}{R})^2 = \frac{1}{0.64} \]
Calculating \( \frac{1}{0.64} \):
\[ \frac{1}{0.64} = \frac{100}{64} = \frac{25}{16} \]

### Step 6: Take the square root
Taking the square root of both sides:
\[ 1 + \frac{h}{R} = \frac{5}{4} \]

### Step 7: Solve for \( h \)
Now, we can isolate \( h \):
\[ \frac{h}{R} = \frac{5}{4} - 1 = \frac{5}{4} - \frac{4}{4} = \frac{1}{4} \]
Multiplying both sides by \( R \):
\[ h = \frac{R}{4} \]

### Step 8: Substitute the value of \( R \)
Given that \( R = 6400 \) km:
\[ h = \frac{6400}{4} = 1600 \text{ km} \]

### Final Answer
Thus, the height from the surface of the Earth where the value of \( g \) is reduced by 36% is:
\[ \boxed{1600 \text{ km}} \]