At what height, the weight of the body is same as that at same depth from the earth's surface (take, earth's radius = R)

Options

A. `(sqrt(3) R - R)/2 `
B. `(sqrt(5))/2 R-R `
C. ` (sqrt(5) R-R)/2`
D. `R/2`

Explanation

To solve the problem of finding the height at which the weight of a body is the same as that at the same depth from the Earth's surface, we can follow these steps:

### Step-by-Step Solution:

1. **Define Variables**:
- Let the height above the Earth's surface be \( h \).
- Let the depth below the Earth's surface be \( d \).
- The radius of the Earth is \( R \).

2. **Weight at Height**:
- The effective gravitational acceleration \( g_h \) at height \( h \) is given by:
\[ g_h = \frac{g}{(1 + \frac{h}{R})^2} \]
- Therefore, the weight of the body at height \( h \) is:
\[ W_h = mg_h = m \cdot \frac{g}{(1 + \frac{h}{R})^2} \]

3. **Weight at Depth**:
- The effective gravitational acceleration \( g_d \) at depth \( d \) is given by:
\[ g_d = g \left(1 - \frac{d}{R}\right) \]
- Therefore, the weight of the body at depth \( d \) is:
\[ W_d = mg_d = mg \left(1 - \frac{d}{R}\right) \]

4. **Set Weights Equal**:
- We need to find the condition where the weights are equal:
\[ mg \left(1 - \frac{d}{R}\right) = m \cdot \frac{g}{(1 + \frac{h}{R})^2} \]
- Cancel \( mg \) from both sides (assuming \( m \neq 0 \)):
\[ 1 - \frac{d}{R} = \frac{1}{(1 + \frac{h}{R})^2} \]

5. **Substitute \( d \) with \( h \)**:
- Since \( d = h \), we can substitute \( d \) in the equation:
\[ 1 - \frac{h}{R} = \frac{1}{(1 + \frac{h}{R})^2} \]

6. **Cross Multiply**:
- Cross multiplying gives:
\[ (1 - \frac{h}{R}) (1 + \frac{h}{R})^2 = 1 \]

7. **Expand and Simplify**:
- Expanding the left-hand side:
\[ (1 - \frac{h}{R}) (1 + \frac{2h}{R} + \frac{h^2}{R^2}) = 1 \]
- This simplifies to:
\[ 1 + 2\frac{h}{R} + \frac{h^2}{R^2} - \frac{h}{R} - 2\frac{h^2}{R^2} - \frac{h^3}{R^3} = 1 \]
- Rearranging gives:
\[ \frac{h^2}{R^2} - \frac{h^3}{R^3} + \frac{h}{R} = 0 \]

8. **Factor Out \( \frac{h}{R} \)**:
- Factoring out \( \frac{h}{R} \):
\[ \frac{h}{R} \left(h - \frac{h^2}{R}\right) = 0 \]
- This gives us \( h = 0 \) or \( h^2 - hR + R^2 = 0 \).

9. **Solve the Quadratic Equation**:
- Using the quadratic formula:
\[ h = \frac{R \pm \sqrt{R^2 + 4R^2}}{2} = \frac{R \pm \sqrt{5R^2}}{2} = \frac{R(1 \pm \sqrt{5})}{2} \]
- Since height cannot be negative, we take:
\[ h = \frac{R(1 + \sqrt{5})}{2} \]

### Final Result:
The height \( h \) at which the weight of the body is the same as that at the same depth is:
\[ h = \frac{R(1 + \sqrt{5})}{2} \]