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Question:
The weight of a body at the surface of earth is $18\, N$ The weight of the body at an altitude of $3200 \,km$ above the earth's surface is (given, radius of earth $R _{ e }=6400 \,km$ ):
The weight of a body at the surface of earth is $18\, N$ The weight of the body at an altitude of $3200 \,km$ above the earth's surface is (given, radius of earth $R _{ e }=6400 \,km$ ):
Updated On: Oct 18, 2024
- 19.6 N
- 9.8 N
- 4.9 N
- 8 N
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The Correct Option is D
Solution and Explanation
Acceleration due to gravity at height \(h\)
\(g'=\frac{g}{[1+\frac{h}{R}]^2}\)
So weight at given height
\(mg'=\frac{mg}{[1+\frac{h}{R}]^2}\)
=\(\frac{18}{[1+\frac{1}{2}]^2}=8N\)
\(\text{Therefore, the correct option is (D):} 8N\)
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Concepts Used:
Gravitation
In mechanics, the universal force of attraction acting between all matter is known as Gravity, also called gravitation, . It is the weakest known force in nature.
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,
- F ∝ (M1M2) . . . . (1)
- (F ∝ 1/r2) . . . . (2)
On combining equations (1) and (2) we get,
F ∝ M1M2/r2
F = G × [M1M2]/r2 . . . . (7)
Or, f(r) = GM1M2/r2
The dimension formula of G is [M-1L3T-2].
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