Question:
Escape velocity from Earth is about 11 km/s. The escape velocity from a planet, 4 times the size of the Earth with same mean density is
Escape velocity from Earth is about 11 km/s. The escape velocity from a planet, 4 times the size of the Earth with same mean density is
Updated On: Jul 4, 2024
- 44 km/s
- 22 km/s
- 16.5 km/s
- 88 km/s
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The Correct Option is A
Solution and Explanation
Escape velocity of a planet
$v=\sqrt{\frac{2GM}{R}} = \sqrt{\frac{2G \times\frac{4}{3} \pi R^{3} \times\rho}{R}}$
$v=R\sqrt{\frac{8\pi}{3} G\rho}$
For same $\rho, v \propto R$
$\therefore\:\: \frac{v_{E}}{v_{P}} = \frac{R_{E}}{R_{P}} $
Here, $v_{E} = 11 Km s^{-1} , R_{E} = R, R_{p} = 4R, v_{P } = ? $
$ \therefore\:\:\: \frac{11}{v_{P} } = \frac{R}{4R}$
or, $ v_{P} = 44 km s^{-1}$
$v=\sqrt{\frac{2GM}{R}} = \sqrt{\frac{2G \times\frac{4}{3} \pi R^{3} \times\rho}{R}}$
$v=R\sqrt{\frac{8\pi}{3} G\rho}$
For same $\rho, v \propto R$
$\therefore\:\: \frac{v_{E}}{v_{P}} = \frac{R_{E}}{R_{P}} $
Here, $v_{E} = 11 Km s^{-1} , R_{E} = R, R_{p} = 4R, v_{P } = ? $
$ \therefore\:\:\: \frac{11}{v_{P} } = \frac{R}{4R}$
or, $ v_{P} = 44 km s^{-1}$
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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].