At which height from the earth's surface does the acceleration due to gravity decrease by 75 % of its value at earth's surface ?

Answer

To solve the problem of finding the height \( h \) from the Earth's surface at which the acceleration due to gravity decreases by 75% of its value at the Earth's surface, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We need to find the height \( h \) where the acceleration due to gravity \( g' \) is 25% of the acceleration due to gravity at the Earth's surface \( g \).

2. **Using the Formula for Acceleration Due to Gravity**:
The acceleration due to gravity at a height \( h \) above the Earth's surface is given by:
\[ g' = \frac{GM}{(R + h)^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

3. **Acceleration at Earth's Surface**:
The acceleration due to gravity at the Earth's surface is:
\[ g = \frac{GM}{R^2} \]

4. **Setting Up the Equation**:
According to the problem, we want \( g' \) to be 25% of \( g \):
\[ g' = \frac{1}{4} g \]
Substituting the expressions for \( g' \) and \( g \):
\[ \frac{GM}{(R + h)^2} = \frac{1}{4} \cdot \frac{GM}{R^2} \]

5. **Canceling Common Terms**:
Since \( GM \) appears in both sides, we can cancel it:
\[ \frac{1}{(R + h)^2} = \frac{1}{4R^2} \]

6. **Cross Multiplying**:
Cross-multiplying gives:
\[ 4R^2 = (R + h)^2 \]

7. **Expanding the Right Side**:
Expanding the right side:
\[ 4R^2 = R^2 + 2Rh + h^2 \]

8. **Rearranging the Equation**:
Rearranging the equation gives:
\[ 4R^2 - R^2 = 2Rh + h^2 \]
\[ 3R^2 = 2Rh + h^2 \]

9. **Rearranging Further**:
Rearranging further:
\[ h^2 + 2Rh - 3R^2 = 0 \]

10. **Using the Quadratic Formula**:
We can solve this quadratic equation using the quadratic formula:
\[ h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 2R \), and \( c = -3R^2 \):
\[ h = \frac{-2R \pm \sqrt{(2R)^2 - 4 \cdot 1 \cdot (-3R^2)}}{2 \cdot 1} \]
\[ h = \frac{-2R \pm \sqrt{4R^2 + 12R^2}}{2} \]
\[ h = \frac{-2R \pm \sqrt{16R^2}}{2} \]
\[ h = \frac{-2R \pm 4R}{2} \]

11. **Calculating the Values**:
This gives 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, we have:
\[ h = R \]

12. **Substituting the Radius of the Earth**:
Given that the radius of the Earth \( R \) is approximately \( 6400 \) km, we find:
\[ h = 6400 \text{ km} \]

### Final Answer:
The height from the Earth's surface at which the acceleration due to gravity decreases by 75% of its value at the Earth's surface is approximately **6400 km**.