Correct Answer - Option 2 :
\(\frac{v}{\sqrt 6}\)
Concept:
Escape Velocity (v):
V = √(2gR) -- (1)
Calcuation:
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 \(\frac{v}{\sqrt 6}\)
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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.