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 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. **Understand the Problem**: We need to find the height \( h \) where the acceleration due to gravity \( g' \) is 25% of the surface gravity \( g \). This means:
\[ g' = g \times (1 - 0.75) = \frac{g}{4} \]

2. **Write the Formula for Acceleration due to Gravity at Height \( h \)**: 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. **Write the Formula for Acceleration due to Gravity at the Surface**: The acceleration due to gravity at the surface of the Earth is:
\[ g = \frac{GM}{R^2} \]

4. **Set Up the Equation**: We know from our earlier step that \( g' = \frac{g}{4} \). Thus, we can set up the equation:
\[ \frac{GM}{(R + h)^2} = \frac{1}{4} \cdot \frac{GM}{R^2} \]

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

6. **Cross-Multiply**: This gives us:
\[ 4R^2 = (R + h)^2 \]

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

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

9. **Rearranging Further**: This can be rearranged to form a quadratic equation:
\[ h^2 + 2Rh - 3R^2 = 0 \]

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

11. **Calculate the Values**: This gives us two possible solutions:
- \( h = \frac{2R}{2} = R \) (valid)
- \( h = \frac{-6R}{2} = -3R \) (not valid)

12. **Final Result**: Thus, the height \( h \) at which the acceleration due to gravity decreases by 75% is:
\[ h = R \]
Given that the radius of the Earth \( R \) is approximately \( 6400 \, \text{km} \), the height is:
\[ h \approx 6400 \, \text{km} \]