Question:
A planet in a distant solar system is 10 times more massive than the earth and its radius is 10 times smaller. Given that the escape velocity from the earth is ${11 \, km \, s^{-1}}$, the escape velocity from the surface of the planet would be
A planet in a distant solar system is 10 times more massive than the earth and its radius is 10 times smaller. Given that the escape velocity from the earth is ${11 \, km \, s^{-1}}$, the escape velocity from the surface of the planet would be
Updated On: Jun 7, 2022
- ${0.11 \, km \, s^{-1}}$
- ${1.1 \, km \, s^{-1}}$
- ${11 \, km \, s^{-1}}$
- ${110 \, km \, s^{-1}}$
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The Correct Option is D
Solution and Explanation
Escape velocity from the surface of the earth is
$\upsilon_e = \sqrt{\frac{2GM}{R}}$
$\upsilon_e = 11 \, km \, s^{-1}$
Mass of the planet = $10M$.
Radius of the planet = $R/10$.
$ \therefore \, \upsilon_p = \sqrt{\frac{2GM \times 10}{R/10}} = 10 \upsilon_e = 10 \times 11$
$ = { 110 \, km \, s^{-1}}$
$\upsilon_e = \sqrt{\frac{2GM}{R}}$
$\upsilon_e = 11 \, km \, s^{-1}$
Mass of the planet = $10M$.
Radius of the planet = $R/10$.
$ \therefore \, \upsilon_p = \sqrt{\frac{2GM \times 10}{R/10}} = 10 \upsilon_e = 10 \times 11$
$ = { 110 \, 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].