A body weighs 72 N on the surface of the earth. What is the gravitational force on it, at a height equal to half the radius of the earth?
- 48 N
- 32 N
- 30 N
- 24 N
The Correct Option is B
Solution and Explanation
\(w_{s}=mg_{s}=72\,N\)
\(w_{h}=mg_{h}=\frac{mg_{s}}{\left(1+\frac{h}{R}\right)^{2}}\)
\(=\frac{72\,N}{\left(1+\frac{\frac{R }{2}}{2}\right)}\)
\(=\frac{72}{\frac{9 }{4}}\)
\(w_{h}=32\,N\)
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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].