How much above the surface of earth does the acceleration due to gravity reduces by `64%` of its value on the earth. Radius of earth `= 6400 km`.

Options

A. `5.71xx 10^(6)m`
B. `1.22 xx 10^(6)m`
C. `11 xx 10^(6)m`
D. `4.27 xx 10^(6)m`

Explanation

To find out how much above the surface of the Earth the acceleration due to gravity reduces by 64% of its value on the Earth, we can follow these steps:

### Step 1: Understand the Problem
We need to find the height \( H \) above the Earth's surface where the acceleration due to gravity \( g' \) is reduced to 36% of its original value \( g \) (since a reduction of 64% means 36% remains).

### Step 2: Express the New Acceleration Due to Gravity
The new acceleration due to gravity at height \( H \) can be expressed as:
\[ g' = \frac{g \cdot R^2}{(R + H)^2} \]
where \( R \) is the radius of the Earth.

### Step 3: Set Up the Equation
We know that:
\[ g' = 0.36g \]
Substituting this into the equation gives:
\[ 0.36g = \frac{g \cdot R^2}{(R + H)^2} \]

### Step 4: Cancel \( g \) from Both Sides
Since \( g \) is not zero, we can cancel it out:
\[ 0.36 = \frac{R^2}{(R + H)^2} \]

### Step 5: Cross Multiply
Cross-multiplying gives:
\[ 0.36(R + H)^2 = R^2 \]

### Step 6: Expand the Equation
Expanding the left side:
\[ 0.36(R^2 + 2RH + H^2) = R^2 \]
This simplifies to:
\[ 0.36R^2 + 0.72RH + 0.36H^2 = R^2 \]

### Step 7: Rearrange the Equation
Rearranging gives:
\[ 0.36H^2 + 0.72RH - 0.64R^2 = 0 \]

### Step 8: Use the Quadratic Formula
This is a quadratic equation in \( H \). We can use the quadratic formula \( H = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where:
- \( a = 0.36 \)
- \( b = 0.72R \)
- \( c = -0.64R^2 \)

Calculating the discriminant:
\[ b^2 - 4ac = (0.72R)^2 - 4(0.36)(-0.64R^2) \]

### Step 9: Substitute Values
Substituting \( R = 6400 \, \text{km} = 6.4 \times 10^6 \, \text{m} \):
\[ b^2 = (0.72 \cdot 6.4 \times 10^6)^2 \]
\[ 4ac = 4 \cdot 0.36 \cdot 0.64 \cdot (6.4 \times 10^6)^2 \]

### Step 10: Solve for \( H \)
After calculating the discriminant and substituting back into the quadratic formula, we can find the height \( H \).

### Final Calculation
After performing the calculations, we find:
\[ H \approx 4.27 \times 10^6 \, \text{m} = 4270 \, \text{km} \]

### Conclusion
Thus, the height above the surface of the Earth where the acceleration due to gravity reduces by 64% of its value on the Earth is approximately \( 4270 \, \text{km} \).

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