Question:
Escape velocity from a planet is $ V_{e}$ . If its mass is increased to $8$ times and its radius is increased to $2$ times, then the new escape velocity would be:
Escape velocity from a planet is $ V_{e}$ . If its mass is increased to $8$ times and its radius is increased to $2$ times, then the new escape velocity would be:
Updated On: Jun 18, 2022
- $V_{e}$
- $ \sqrt{2}v_{e}$
- $ 2v_{e}$
- $ 2\sqrt{2}v_{e}$
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The Correct Option is C
Solution and Explanation
Velocity of projection of body, at which the body goes out or gravitational field of earth or planet and never return, is called escape velocity. $v_{e}=\sqrt{\frac{2 G M}{R}}$
where $G$ is gravitational constant, $M$ is mass of plant and $R_{e}$ is radius.
When $M=8\, M,\, R=2 R$,
then $v_{e}=\sqrt{\frac{2 G(8 M)}{(2 R)}} v_{e}=2 \sqrt{\frac{2 G M}{R}}$ $v_{e}=2 v_{e} .$
Note: Escape velocity is independent of mass of body.
where $G$ is gravitational constant, $M$ is mass of plant and $R_{e}$ is radius.
When $M=8\, M,\, R=2 R$,
then $v_{e}=\sqrt{\frac{2 G(8 M)}{(2 R)}} v_{e}=2 \sqrt{\frac{2 G M}{R}}$ $v_{e}=2 v_{e} .$
Note: Escape velocity is independent of mass of body.
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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].