If earth has a mass nine times and radius twice to that of a planet $P$ Then $\frac{v_e}{3} \sqrt{x} ms ^{-1}$ will be the minimum velocity required by a rocket to pull out of gravitational force of $P$, where $v_e$ is escape velocity on earth The value of $x$ is
- 1
18
3
- 2
The Correct Option is D
Approach Solution - 1
Approach Solution -2
Escape velocity for a planet is given by: \[ V_{\text{escape}} = \sqrt{\frac{2GM}{R}} \] For planet P, \[ V_{\text{escape, P}} = \sqrt{\frac{2G \left(\frac{M}{9} \right)}{\frac{R}{2}}} \] \[ = V_e \frac{\sqrt{2}}{3}, \quad x = 2 \]
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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].