A rocket has landed on planet x, which has half the radius of earth. An astronaut onboard the rocket weighs twice as much on planet x as on earth. If the escape velocity for the rocket taking off from earth is u0, then its escape velocity on planet x is

(a) 2u0

(b) 2u0

(c) u0

(d) u0 2

(e) u0 4

Option (c) u0

Explanation:

The escape velocity has a formula as:

V = √ (2gR)

Where V is the escape velocity,

g is the acceleration due to gravity

R is the radius of the earth.

Now, from the question, we were told that the escape velocity for the rocket taking off from earth is u0 i. e

V (earth) = u0

V (earth) = √ (2gR)

u0 = √ (2gR) = > For the earth

Now, let us calculate the escape velocity for the rocket taking off from planet x. This is illustrated below below:

g (planet x) = 2g (earth) = > since the weight of the astronaut is twice as much on planet x as on earth

R (planet x) = 1/2 R (earth) = > planet x has half the radius of earth

V (planet x) = ?

Applying the formula V = √ (2gR), the escape velocity on planet x is obtained as follow:

V (planet x) = √ (2g (x) x R (x))

V (planet x) = √ (2 x 2g x 1/2R)

V (planet x) = √ (2 x g x R)

V (planet x) = √ (2gR)

The expression obtained for the escape velocity on planet x i. e V (planet x) = √ (2gR), is exactly the same as that obtained for the earth i. e V (earth) = √ (2gR)

Therefore,

V (planet x) = V (earth) = √ (2gR)

But from the question, V (earth) is u0

Therefore,

V (planet x) = V (earth) = √ (2gR) = u0

So, the escape velocity on planet x is u0