At what height above the surface of the earth will the acceleration due to gravity be `25%` of its value on the surface of the earth ? Assume that the radius of the earth is 6400 km .

Options

A. `(R)/(4)`
B. R
C. `(3R)/(8)`
D. `(R)/(2)`

Answer

To find the height above the surface of the Earth where the acceleration due to gravity is 25% of its value on the surface, we can follow these steps:

### Step 1: Understand the relationship of gravitational acceleration
The acceleration due to gravity \( g \) at a distance \( r \) from the center of the Earth is given by the formula:
\[ g = \frac{GM}{r^2} \]
where \( G \) is the gravitational constant and \( M \) is the mass of the Earth.

### Step 2: Set up the equation for height \( h \)
We want to find the height \( h \) above the Earth's surface where the acceleration due to gravity \( g' \) is \( \frac{g}{4} \). At height \( h \), the distance from the center of the Earth becomes \( R + h \), where \( R \) is the radius of the Earth (6400 km).

Thus, we can express \( g' \) as:
\[ g' = \frac{GM}{(R + h)^2} \]

### Step 3: Set the equation for \( g' \)
Since we want \( g' = \frac{g}{4} \), we can write:
\[ \frac{GM}{(R + h)^2} = \frac{1}{4} \cdot \frac{GM}{R^2} \]

### Step 4: Cancel \( GM \) from both sides
We can cancel \( GM \) from both sides of the equation:
\[ \frac{1}{(R + h)^2} = \frac{1}{4R^2} \]

### Step 5: Cross-multiply to solve for \( R + h \)
Cross-multiplying gives us:
\[ 4R^2 = (R + h)^2 \]

### Step 6: Expand the right side
Expanding the right side:
\[ 4R^2 = R^2 + 2Rh + h^2 \]

### Step 7: Rearrange the equation
Rearranging gives:
\[ 4R^2 - R^2 = 2Rh + h^2 \]
\[ 3R^2 = 2Rh + h^2 \]

### Step 8: Rearrange to form a quadratic equation
This can be rearranged to form a quadratic equation in terms of \( h \):
\[ h^2 + 2Rh - 3R^2 = 0 \]

### Step 9: Use the quadratic formula
Using the quadratic formula \( h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 2R, c = -3R^2 \):
\[ h = \frac{-2R \pm \sqrt{(2R)^2 - 4 \cdot 1 \cdot (-3R^2)}}{2 \cdot 1} \]
\[ h = \frac{-2R \pm \sqrt{4R^2 + 12R^2}}{2} \]
\[ h = \frac{-2R \pm \sqrt{16R^2}}{2} \]
\[ h = \frac{-2R \pm 4R}{2} \]

### Step 10: Solve for \( h \)
This gives us two possible solutions:
\[ h = \frac{2R}{2} = R \quad \text{(taking the positive root)} \]
or
\[ h = \frac{-6R}{2} = -3R \quad \text{(not physically meaningful)} \]

Thus, the height \( h \) above the surface of the Earth is equal to the radius of the Earth:
\[ h = R = 6400 \text{ km} \]

### Final Answer
The height above the surface of the Earth where the acceleration due to gravity is 25% of its value on the surface is **6400 km**.

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