If the weight of an object on earth's surface is 400 N, then weight of the same particle at a depth \(\frac{R}{2}\) from surface would be (R is radius of earth)
- 100 N
- 300 N
- 200 N
- 250 N
The Correct Option is C
Solution and Explanation
Step 1: Use the formula for weight at depth.- Weight on surface: W = mg = 400N.- At depth d = R/2,
Step 2: Substitute the values.
W' = W(1 - dR)
W' = 400(1 - R/2R)
W' = 400(1 - 12)
W' = 400 $\cdot$ 12 = 200N
Final Answer: Weight at half the earth’s radius = 200N
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Concepts Used:
Newtons Law of Gravitation
Gravitational Force
Gravitational force is a central force that depends only on the position of the test mass from the source mass and always acts along the line joining the centers of the two masses.
Newton’s Law of Gravitation:
According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,
- Directly proportional to the product of their masses i.e. F ∝ (M1M2) . . . . (1)
- Inversely proportional to the square of the distance between their center i.e. (F ∝ 1/r2) . . . . (2)
By combining equations (1) and (2) we get,
F ∝ M1M2/r2
F = G × [M1M2]/r2 . . . . (7)
Or, f(r) = GM1M2/r2 [f(r)is a variable, Non-contact, and conservative force]