Question:
Mass of moon is $ 1/81 $ times that of earth and its radius is $ 1/4 $ the radius of earth. If escape velocity on the surface of the earth is $11.2\,km/s $ . Then the value of escape velocity at surface of the moon will be
Mass of moon is $ 1/81 $ times that of earth and its radius is $ 1/4 $ the radius of earth. If escape velocity on the surface of the earth is $11.2\,km/s $ . Then the value of escape velocity at surface of the moon will be
Updated On: Jul 27, 2022
- 5 km/s
- 2.5 km/s
- 0.5 km/sec
- 0.14 km/ sec
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The Correct Option is B
Solution and Explanation
Here : Mass of moon $M_{m}=\frac{M_{e}}{81}$
Radius of moon $R_{m}=\frac{R_{e}}{4}$
Escape velocity on the surface of earth $V_{e s(e)}=11.2\, km / s$
The relation for escape velocity is
$v=\sqrt{\frac{2 G M}{R}} \propto \sqrt{\frac{M}{R}}$
Hence $\frac{v_{e s}(m)}{v_{e s}(e)}=\sqrt{\frac{M_{m}}{R_{m}} \times \frac{R_{e}}{M_{e}}}$
$=\sqrt{\frac{M_{e}}{81} \times \frac{4}{R_{e}} \times \frac{R_{e}}{M_{e}}}=\frac{2}{9}$
$v_{e s(m)}=\frac{2}{9} \times 11.2=2.5\, km / s$
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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].