A body of mass m is projected with velocity λve in vertically upward direction from the surface of the earth into space. It is given that ve is escape velocity and λ< 1. If air resistance is considered to be negligible, then the maximum height from the centre of earth, to which the body can go, will be:
(R : radius of earth)
(R : radius of earth)
\(\frac{R}{1+λ^2}\)
\(\frac{R}{1-λ^2}\)
\(\frac{R}{1-λ}\)
\(\frac{λ^2R}{1-λ^2}\)
The Correct Option is B
Solution and Explanation
The correct answer is (B): \(\frac{R}{1-λ^2}\)
Using energy conservation
\(-\frac{GM_em}{R_e} + \frac{1}{2}m \left(λ\sqrt{\frac{2GM_e}{R_e}} \right)^2 = -\frac{GM_em}{r}\)
\(\frac{GM_em}{r} = \frac{GM_em}{R_e} - \frac{GM_em}{R_e} λ^2\)
\(r = \frac{R_e}{1-λ^2}\)
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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].