Question:

A space station is set up in space at a distance equal to earth's radius from earth's surface. Suppose a satellite can be launched from space station. Let $V_1$ and $V_2$ be the escape velocites of the satlhte on earth's surface and space station respectively. Then

Updated On: Jul 6, 2022
  • $V_2=V_1$
  • $V_2$
  • $V_2>V_1$
  • No relation
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The Correct Option is B

Solution and Explanation

For an object to escape from an orbit, its total energy must be zero. For space station: $T . E =\frac{1}{2} mv _{2}^{2}+\left(-\frac{ GMm }{2 R }\right)=0$ $\Longrightarrow v_{2}=\sqrt{\frac{G M}{R}}$ where $R$ is the radius of earth. For earth's surface : $v_{1}=\sqrt{\frac{2 GM }{ R }}$ Hence $v_{1}>v_{2}$
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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].