At what altitude (h) above the earth's surface would the accelration due to gravity be one fourth of its value at the earth's surface ?

Options

A. h=R
B. h=4R
C. h = 2R
D. h = 16R

Answer

To find the altitude (h) above the Earth's surface where the acceleration due to gravity (g') is one-fourth of its value at the Earth's surface (g), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between g and g':**
The acceleration due to gravity at a height h above the Earth's surface is given by the formula:
\[ g' = \frac{g}{(1 + \frac{h}{r})^2} \]
where \( r \) is the radius of the Earth.

2. **Set up the equation for the given condition:**
We are given that \( g' = \frac{g}{4} \). Therefore, we can substitute this into the equation:
\[ \frac{g}{4} = \frac{g}{(1 + \frac{h}{r})^2} \]

3. **Eliminate g from both sides:**
Since g is present in both sides of the equation, we can cancel it out (assuming \( g \neq 0 \)):
\[ \frac{1}{4} = \frac{1}{(1 + \frac{h}{r})^2} \]

4. **Cross-multiply to simplify:**
Cross-multiplying gives us:
\[ (1 + \frac{h}{r})^2 = 4 \]

5. **Take the square root of both sides:**
Taking the square root results in:
\[ 1 + \frac{h}{r} = 2 \]

6. **Isolate h:**
Subtracting 1 from both sides gives:
\[ \frac{h}{r} = 1 \]
Multiplying both sides by r results in:
\[ h = r \]

7. **Conclusion:**
Therefore, the altitude \( h \) at which the acceleration due to gravity is one-fourth of its value at the Earth's surface is equal to the radius of the Earth.

### Final Answer:
\[ h = r \]