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

  the unit of the measure =

explanation:

  the property that is quantified by the energy needed per of material to raise the temperature by one.

  si unit of energy = joule (j)

  si unit of temperature = kelvin (k)

  si unit of mass of material = kilogram (kg)

the property that is quantified by the energy needed per of material to raise the temperature by one is called specific heat.

the unit of specific heat is

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