Question:

A spaceship is launched into a circular orbit of radius R close to the surface of earth. The additional velocity to be imparted to the spaceship in the orbit to overcome the earths gravitational pull is : (g = acceleration due to gravity)

Updated On: Nov 5, 2024
  • 1.4147 Rg
  • 1.414 $ \sqrt{Rg} $
  • 0.414 Rg
  • 0.414 $ \sqrt{Rg} $
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The Correct Option is D

Solution and Explanation

Let a spaceship is launched in a circular orbit of orbital velocity $ {{v}_{o}}. $ That spaceship should have escape velocity $ {{v}_{es}} $ to overcome the earths gravitational pull. Now suppose v is the additional velocity to be imparted to the spaceship. Then according to above statement $ {{v}_{0}}+v={{v}_{es}} $ $ \left[ \begin{align} & \because \,{{v}_{0}}=\sqrt{Rg} \\ & {{v}_{es}}=\sqrt{2}Rg \\ \end{align} \right] $ or $ v={{v}_{es}}-{{v}_{o}} $ $ v=\sqrt{2}{{v}_{o}}-{{v}_{o}}={{v}_{o}}(\sqrt{2}-1)={{v}_{o}}(1.414-1) $ $ =0.414\sqrt{Rg} $
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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].