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 use the formula for gravitational acceleration at a height h above the Earth's surface.

### Step-by-Step Solution:

1. **Understanding the relationship between g and g':**
The acceleration due to gravity at a distance r from the center of the Earth is given by the formula:
\[ g' = \frac{GM}{(r + h)^2} \]
where:
- \( G \) is the universal gravitational constant,
- \( M \) is the mass of the Earth,
- \( r \) is the radius of the Earth,
- \( h \) is the height above the Earth's surface.

2. **Using the value of g at the Earth's surface:**
The acceleration due to gravity at the Earth's surface is:
\[ g = \frac{GM}{r^2} \]

3. **Setting up the equation:**
We want to find h such that:
\[ g' = \frac{1}{4}g \]
Substituting the expressions for g and g':
\[ \frac{GM}{(r + h)^2} = \frac{1}{4} \cdot \frac{GM}{r^2} \]

4. **Canceling GM from both sides:**
Since GM is common on both sides, we can cancel it out:
\[ \frac{1}{(r + h)^2} = \frac{1}{4r^2} \]

5. **Cross-multiplying:**
Cross-multiplying gives:
\[ 4r^2 = (r + h)^2 \]

6. **Expanding the right side:**
Expanding the right side:
\[ 4r^2 = r^2 + 2rh + h^2 \]

7. **Rearranging the equation:**
Rearranging gives:
\[ 4r^2 - r^2 = 2rh + h^2 \]
\[ 3r^2 = 2rh + h^2 \]

8. **Rearranging to form a quadratic equation:**
Rearranging further, we get:
\[ h^2 + 2rh - 3r^2 = 0 \]

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

10. **Calculating the possible values of h:**
This gives us two possible solutions:
\[ h = \frac{2r}{2} = r \quad \text{(taking the positive root)} \]
\[ h = \frac{-6r}{2} = -3r \quad \text{(not physically meaningful)} \]

Thus, the altitude \( h \) above the Earth's surface where the acceleration due to gravity is one-fourth of its value at the surface is:
\[ h = r \]
where \( r \) is the radius of the Earth.

### Final Answer:
The altitude \( h \) is equal to the radius of the Earth.

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