At what altitude will the acceleration due to gravity be 25% of that at the earth’s surface (given radius of earth is R) ?

Options

A. R/4
B. R
C. 3R/8
D. R/2

Answer

To find the altitude at which the acceleration due to gravity is 25% of that at the Earth's surface, we can follow these steps:

### Step 1: Understanding the relationship between gravity at the surface and at altitude
The acceleration due to gravity at the Earth's surface is given by:
\[ g = \frac{GM}{R^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

At a height \( h \) above the Earth's surface, the acceleration due to gravity \( g' \) is given by:
\[ g' = \frac{GM}{(R + h)^2} \]

### Step 2: Setting up the equation for 25% of surface gravity
We want to find the altitude \( h \) where:
\[ g' = 0.25g \]
Substituting the expression for \( g' \):
\[ \frac{GM}{(R + h)^2} = 0.25 \cdot \frac{GM}{R^2} \]

### Step 3: Simplifying the equation
We can cancel \( GM \) from both sides (since it is non-zero):
\[ \frac{1}{(R + h)^2} = 0.25 \cdot \frac{1}{R^2} \]

### Step 4: Cross-multiplying to eliminate the fraction
Cross-multiplying gives:
\[ R^2 = 0.25(R + h)^2 \]

### Step 5: Expanding the equation
Expanding the right side:
\[ R^2 = 0.25(R^2 + 2Rh + h^2) \]
\[ R^2 = 0.25R^2 + 0.5Rh + 0.25h^2 \]

### Step 6: Rearranging the equation
Bringing all terms to one side:
\[ R^2 - 0.25R^2 - 0.5Rh - 0.25h^2 = 0 \]
\[ 0.75R^2 - 0.5Rh - 0.25h^2 = 0 \]

### Step 7: Multiplying through by 4 to eliminate the fraction
Multiplying the entire equation by 4:
\[ 3R^2 - 2Rh - h^2 = 0 \]

### Step 8: Rearranging into standard quadratic form
Rearranging gives:
\[ h^2 + 2Rh - 3R^2 = 0 \]

### Step 9: Using the quadratic formula
Using the quadratic formula \( h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 2R, c = -3R^2 \):
\[ h = \frac{-2R \pm \sqrt{(2R)^2 - 4 \cdot 1 \cdot (-3R^2)}}{2 \cdot 1} \]
\[ h = \frac{-2R \pm \sqrt{4R^2 + 12R^2}}{2} \]
\[ h = \frac{-2R \pm \sqrt{16R^2}}{2} \]
\[ h = \frac{-2R \pm 4R}{2} \]

### Step 10: Solving for \( h \)
This gives two possible solutions:
1. \( h = \frac{2R}{2} = R \)
2. \( h = \frac{-6R}{2} = -3R \) (not a valid solution as height cannot be negative)

Thus, the altitude \( h \) at which the acceleration due to gravity is 25% of that at the Earth's surface is:
\[ h = R \]

### Final Answer
The altitude at which the acceleration due to gravity is 25% of that at the Earth's surface is equal to the radius of the Earth, \( R \).

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