Question:
A body of weight 72 N moves from the surface of earth at a height half of the radius of earth, then gravitational force exerted on it will be
A body of weight 72 N moves from the surface of earth at a height half of the radius of earth, then gravitational force exerted on it will be
Updated On: Mar 14, 2024
- 36 N
- 32 N
- 144 N
- 50 N
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The Correct Option is B
Solution and Explanation
$ F_{surface } = G \frac{Mm}{ {R_e}^2}$
$ F_{R_c /2} = G \frac{ Mm}{ (R_e + R_e /2)^2} =\frac{4}{9} \times F_{surface } = \frac{4}{9}\times 72= 32\, N $.
$ F_{R_c /2} = G \frac{ Mm}{ (R_e + R_e /2)^2} =\frac{4}{9} \times F_{surface } = \frac{4}{9}\times 72= 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].