What is the escape velocity for a body on the surface of a planet on which the acceleration due to gravity is $(3.1)^2 ms^{-2}$ and whose radius is 8100 km?
- $2790\, km-s^{-1}$
- $27.9\,km-s^{-1}$
- $\frac{27.9}{\sqrt5}\,km^{-1}$
- $27.9\sqrt5 \,km-s^{-1}$
The Correct Option is C
Solution and Explanation
Given, $g=(3.1)^2ms^{-2},$
$R = 8100km = 8 1 00 x 10^3 m.$
= $v_e=\sqrt{2\times(3.1)^2\times8100\times10^3}$
$v_e=12.5\times10^3ms^{-1}$
$\Rightarrow v_e=12.5 kms^{-1}$
$\approx\frac{27.9}{\sqrt5}kms^{-1}$
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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].