What is the minimum energy required to launch a satellite of mass $m$ from the surface of a planet of mass $M$ and radius $R$ in a circular orbit at an altitude of $2\,R$ ?
- $ \frac{ 5GmM}{6 R}$
- $ \frac{ 2GmM}{3 R}$
- $ \frac{ GmM}{2 R}$
- $ \frac{ GmM}{3 R}$
The Correct Option is A
Solution and Explanation
$= -\frac{GMm}{2r} -\bigg( - \frac{GMm}{ R} \bigg)$
Here, r = R + 2R = 2>R
Substituting in about equation we get. $E = \frac{ 5GMm}{ 6R}$
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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].