A body weight 250N on the surface. Assuming the earth to be a sphere of uniform mass density, how much would it weight half way down to the centre of earth

Options

A. 240 N
B. 210 N
C. 195 N
D. 125 N

Answer

To find the weight of a body halfway down to the center of the Earth, we can follow these steps:

### Step 1: Understand the weight of the body on the surface
The weight of the body on the surface of the Earth is given as 250 N. The weight (W) is related to the mass (m) of the body and the acceleration due to gravity (g) at the surface:
\[ W = mg \]
Where:
- \( W = 250 \, \text{N} \)
- \( g \) at the surface of the Earth is approximately \( 9.81 \, \text{m/s}^2 \).

### Step 2: Calculate the mass of the body
From the weight equation, we can rearrange it to find the mass:
\[ m = \frac{W}{g} \]
Substituting the values:
\[ m = \frac{250 \, \text{N}}{9.81 \, \text{m/s}^2} \approx 25.48 \, \text{kg} \]

### Step 3: Determine the depth halfway to the center of the Earth
The radius of the Earth (R) is approximately 6371 km. Halfway to the center would be:
\[ \text{Depth} = \frac{R}{2} = \frac{6371 \, \text{km}}{2} = 3185.5 \, \text{km} \]

### Step 4: Calculate the new acceleration due to gravity (g') at that depth
The formula for the acceleration due to gravity at a depth (d) is:
\[ g' = g \left(1 - \frac{d}{R}\right) \]
Here, \( d = \frac{R}{2} \), so:
\[ g' = g \left(1 - \frac{R/2}{R}\right) = g \left(1 - \frac{1}{2}\right) = \frac{g}{2} \]
Thus, at half the radius:
\[ g' = \frac{9.81 \, \text{m/s}^2}{2} \approx 4.905 \, \text{m/s}^2 \]

### Step 5: Calculate the weight of the body at that depth
Now, we can find the new weight (W') of the body at this depth:
\[ W' = mg' \]
Substituting the values:
\[ W' = 25.48 \, \text{kg} \times 4.905 \, \text{m/s}^2 \approx 125 \, \text{N} \]

### Final Answer
The weight of the body halfway down to the center of the Earth would be approximately **125 N**.

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