At what height the gravitational field reduces by 75 % the gravitational field at the surface of earth ?

Options

A. R
B. 2R
C. 3R
D. 4R

Explanation

To solve the problem of finding the height at which the gravitational field reduces by 75% of the gravitational field at the surface of the Earth, we can follow these steps:

### Step 1: Understand the Problem
We know that the gravitational field strength (g) at the surface of the Earth is given by the formula:
\[ g = \frac{GM}{R^2} \]
where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the Earth,
- \( R \) is the radius of the Earth.

If the gravitational field reduces by 75%, that means only 25% of the original gravitational field strength remains.

### Step 2: Set Up the Equation
If the gravitational field strength at height \( h \) is \( \frac{1}{4}g \), we can express this as:
\[ g_h = \frac{GM}{(R + h)^2} \]
where \( g_h \) is the gravitational field strength at height \( h \).

Since we want \( g_h \) to be \( \frac{1}{4}g \), we can write:
\[ \frac{GM}{(R + h)^2} = \frac{1}{4} \cdot \frac{GM}{R^2} \]

### Step 3: Simplify the Equation
We can cancel \( GM \) from both sides of the equation:
\[ \frac{1}{(R + h)^2} = \frac{1}{4R^2} \]

Now, we can cross-multiply:
\[ 4R^2 = (R + h)^2 \]

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

Rearranging the equation:
\[ 4R^2 - R^2 = 2Rh + h^2 \]
\[ 3R^2 = 2Rh + h^2 \]

### Step 5: Solve the Quadratic Equation
This can be rearranged into a standard quadratic form:
\[ h^2 + 2Rh - 3R^2 = 0 \]

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

### Step 6: Calculate the Values
Calculating the two possible values:
1. \( h = \frac{2R}{2} = R \) (taking the positive root)
2. \( h = \frac{-6R}{2} = -3R \) (not physically meaningful)

Thus, the height \( h \) at which the gravitational field reduces by 75% is:
\[ h = R \]

### Final Answer
The height at which the gravitational field reduces by 75% is equal to the radius of the Earth \( R \).