Question:
For a planet having mass equal to mass of the earth but radius is one fourth of radius of the earth.
Then escape velocity for this planet will be
For a planet having mass equal to mass of the earth but radius is one fourth of radius of the earth.
Then escape velocity for this planet will be
Updated On: Jul 20, 2024
- $11.2 \,km/sec$
- $22.4\, km/sec$
- $5.6\, km/sec$
- $44.8\, km/sec$
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The Correct Option is B
Solution and Explanation
$ v_e = \sqrt{2g R } = \sqrt{ \frac{2GM}{ R}}$
if R is 1/4th then $ v_e =2 v_{e - earth}$
$= 2 \times 11.2 = 22.4\, km/sec $
if R is 1/4th then $ v_e =2 v_{e - earth}$
$= 2 \times 11.2 = 22.4\, km/sec $
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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].