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 Gravitational Field Formula
The gravitational field \( g \) 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,
- \( M \) is the mass of the Earth,
- \( r \) is the distance from the center of the Earth.

### Step 2: Define the Gravitational Field at the Surface
At the surface of the Earth, the gravitational field \( g_0 \) can be expressed as:
\[ g_0 = \frac{GM}{R^2} \]
where \( R \) is the radius of the Earth.

### Step 3: Determine the Condition for 75% Reduction
If the gravitational field reduces by 75%, it means that 25% of the original gravitational field remains. Therefore, we can write:
\[ g' = 0.25 g_0 \]
Substituting the expression for \( g_0 \):
\[ g' = 0.25 \left(\frac{GM}{R^2}\right) \]

### Step 4: Express the Gravitational Field at Height \( h \)
At a height \( h \) above the Earth's surface, the distance from the center of the Earth becomes \( R + h \). Thus, the gravitational field at height \( h \) is:
\[ g' = \frac{GM}{(R + h)^2} \]

### Step 5: Set Up the Equation
Now we can set the two expressions for \( g' \) equal to each other:
\[ \frac{GM}{(R + h)^2} = 0.25 \left(\frac{GM}{R^2}\right) \]
We can cancel \( GM \) from both sides (assuming \( GM \neq 0 \)):
\[ \frac{1}{(R + h)^2} = 0.25 \cdot \frac{1}{R^2} \]

### Step 6: Simplify the Equation
Rearranging the equation gives:
\[ (R + h)^2 = 4R^2 \]
Taking the square root of both sides:
\[ R + h = 2R \]

### Step 7: Solve for Height \( h \)
Now, we can solve for \( h \):
\[ h = 2R - R = R \]
Thus, the height \( h \) at which the gravitational field reduces by 75% is equal to the radius of the Earth.

### Final Answer
The height at which the gravitational field reduces by 75% of the gravitational field at the surface of the Earth is:
\[ h = R \]
where \( R \) is the radius of the Earth.

---