Question:
The percentage decrease in the weight of a rocket, when taken to a height of 32 km above the surface of earth will, be:
(Radius of earth = 6400 km)
The percentage decrease in the weight of a rocket, when taken to a height of 32 km above the surface of earth will, be:
(Radius of earth = 6400 km)
(Radius of earth = 6400 km)
Updated On: Jul 8, 2024
- 1%
- 3%
- 4%
- 0.50%
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The Correct Option is A
Solution and Explanation
We know that,
\(g=\frac {GM}{r^2}\)
The percentage decrease in the weight of a rocket,
\(⇒\frac {Δg}{g}=2\frac {Δr}{r}\)
\(⇒\frac {Δg}{g}×100=2×\frac {32}{6400}×100%\)
\(⇒\frac {Δg}{g}=1%\)
% decrease in weight = \(1%\)
So, the correct option is (A): \(1%\)
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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].