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The Escape Velocity of any object on surface of Earth is v. The Escape velocity at surface having acceleration due to gravity as one sixth of earth and the radius same as earth will be
1. \(\sqrt \frac{v}{3}\)
2. \(\frac{v}{\sqrt 6}\)
3. \(\frac{v}{\sqrt 2}\)
4. v √ 6 

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Correct Answer - Option 2 : \(\frac{v}{\sqrt 6}\)


Escape Velocity (v): 

  • The minimum velocity required to escape the gravity of the earth is called escape velocity.
  • Escape velocity (ve) for a body on earth is given as 
  • \(V_e = \sqrt {\frac{2GM}{R}}\)

    G is the universal gravitational constant, M is the mass of earth, R is the radius of the earth.

  • Escape velocity can also be expressed as

V = √(2gR) -- (1)


We know that the acceleration due to gravity on the planet is one-sixth of acceleration due to gravity on Earth. 

So, acceleration due to the gravity of the planet is one-sixth as that of 

\(g' = \frac{g}{6}\) --- (2)

So, Escape velocity on that planet  is given as

v' = √(2g'R) -- (3)

Radius is Same

Now, putting (2) in (3)

\(v' = \sqrt {2\frac{g'}{6}.r} = \frac{\sqrt {2gr}}{\sqrt{6}}\)

Putting (1) in above equation

\(\implies v' = \frac{v}{\sqrt 6}\)

So, the correct option is \(\frac{v}{\sqrt 6}\)

  • Gravitational Potential Energy: The potential energy is energy stored in an object by going against the gravity of the earth.
  • The gravitational potential energy due to earth is

\(U = -\frac{GMm}{R}\)

  • For escaping the earth's gravitational influence the kinetic energy should overcome this potential energy. The kinetic energy is given as 

\(K = \frac{1}{2}mv^2\)

  • For the body to escape earth's gravity, 

\(\frac{1}{2}mv^2 = \frac{GMm}{R}\)

⇒ \(v = \sqrt{\frac{2GM}{R}}\)

R is the radius of the earth, M is the mass of earth, G is the universal gravitational constant.

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