Escape Speed

The equation of escape velocity is
so,

The escape velocity from earth is given by
...(i)
The orbital velocity of a satellite revolving around earth is given by

where, mass of earth,
radius of earth,
height of satellite from surface of earth.
By the relation
so, ...(ii)
Dividing equation by , we get

Squaring on both side, we get

or i.e.,

The correct option is (B):
The escape velocity of an object is given by the equation .
where is the gravitational constant, is the mass of the planet, and is the distance you're at from the centre of the planet.
In this case, the acceleration due to gravity on the surface of the moon is one sixth that on the surface of earth's and the diameter of the moon is one fourth of that of earth.
Therefore, the escape velocity in the moon will be

.
Hence, the ratio of escape velocity on moon and earth will be .

Step 1: Recall the escape velocity formula. Step 2: When radius is doubled (i.e., ) and mass remains the same: Step 3: Compute new velocity: