Question

If the weight of an object is 200 Newtons, then the weight of the object at the midpoint between the Earth's surface and the Earth's center will be:

• A. $ 50 \, \text{N} $
• B. $ 100 \, \text{N} $
• C. $ 150 \, \text{N} $
• D. $ 200 \, \text{N} $

Hint:

Key points to remember:
- Inside Earth, gravity decreases linearly with depth
- At center, weight becomes zero
- Assumes Earth has uniform density (actual variation is more complex)
- Weight at half-depth = half surface weight

Answer 1

The Correct Option is B

Step 1: Understand weight variation with depth
Weight \( W \) varies with distance \( r \) from Earth's center as: \[ W \propto \frac{1}{r^2} \] at surface and above, but follows different rule inside Earth.
Step 2: Gravitational force inside Earth
For points inside Earth (assuming uniform density): \[ F_g \propto r \] where \( r \) is distance from center.
Step 3: Calculate at midpoint
At midpoint \( r = R/2 \) (where \( R \) is Earth's radius): \[ F_g = \frac{1}{2} \times \text{surface value} \] Given surface weight = 200 N: \[ W_{\text{midpoint}} = \frac{200}{2} = 100 \, \text{N} \]
Step 4: Verify the proportionality
Using the complete formula for gravitational force inside Earth: \[ F_g = \frac{GMm}{R^3}r \] At surface (\( r = R \)): \( F_g = 200 \, \text{N} \)
At midpoint (\( r = R/2 \)): \( F_g = 100 \, \text{N} \)