A body weights 63 N on the surface of the earth At a height h above the surface of Earth, its weight is 28 N While at a depth h below the surface Earth, the weight is 31.5 N. The value of h is

Options

A. 0.4 R
B. 0.5 R
C. 0.8 R
D. R

Answer

To solve the problem, we need to determine the height \( h \) above the surface of the Earth where the weight of the body changes. We will also consider the weight of the body at a depth \( h \) below the surface of the Earth.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Weight on the surface of the Earth, \( W = 63 \, \text{N} \)
- Weight at height \( h \), \( W_h = 28 \, \text{N} \)
- Weight at depth \( h \), \( W_d = 31.5 \, \text{N} \)

2. **Relate Weight to Mass and Gravity:**
- The weight of the body at the surface can be expressed as:
\[ W = mg \]
where \( m \) is the mass of the body and \( g \) is the acceleration due to gravity at the surface of the Earth.

3. **Weight at Height \( h \):**
- The weight at height \( h \) can be expressed as:
\[ W_h = mg_h \]
- The acceleration due to gravity at height \( h \) is given by:
\[ g_h = \frac{g}{(1 + \frac{h}{R})^2} \]
where \( R \) is the radius of the Earth.

4. **Setting Up the Equation for Height \( h \):**
- We can write:
\[ W_h = W \cdot \frac{g_h}{g} \]
- Substituting the values:
\[ 28 = 63 \cdot \frac{g_h}{g} \]
- This simplifies to:
\[ \frac{g_h}{g} = \frac{28}{63} = \frac{4}{9} \]

5. **Using the Gravity Equation:**
- From the gravity equation at height \( h \):
\[ \frac{g_h}{g} = \frac{1}{(1 + \frac{h}{R})^2} \]
- Setting this equal to \( \frac{4}{9} \):
\[ \frac{4}{9} = \frac{1}{(1 + \frac{h}{R})^2} \]
- Taking the reciprocal gives:
\[ (1 + \frac{h}{R})^2 = \frac{9}{4} \]
- Taking the square root:
\[ 1 + \frac{h}{R} = \frac{3}{2} \]
- Rearranging gives:
\[ \frac{h}{R} = \frac{3}{2} - 1 = \frac{1}{2} \]
- Therefore:
\[ h = \frac{1}{2} R \]

6. **Weight at Depth \( h \):**
- The weight at depth \( h \) can be expressed as:
\[ W_d = mg_d \]
- The acceleration due to gravity at depth \( h \) is given by:
\[ g_d = g \left(1 - \frac{h}{R}\right) \]
- Substituting the values:
\[ W_d = W \left(1 - \frac{h}{R}\right) \]
- From earlier, we found \( \frac{h}{R} = \frac{1}{2} \):
\[ W_d = 63 \left(1 - \frac{1}{2}\right) = 63 \cdot \frac{1}{2} = 31.5 \, \text{N} \]
- This confirms our calculation.

### Final Result:
The value of \( h \) is:
\[ h = 0.5 R \]