Question

Question #becfb

Answer 1

Both have the same. Because escape velocity does not depend on mass.

Escape velocity is the minimum initial velocity that any body(object) must have into order to escape to infinity, that is; to 'escape' from the earth's gravitational field.

Derivation:
When a body is thrown up and left to free fall, work is done at the expense of it initial Kinetic Energy and this Kinetic Energy is equivalent to the change in Potential Energy(the Potential difference),
`DeltaKE = DeltaPE`
`1/2mv^2 - 1/2mu^2 = int_(r_1)^(r_2)F_G dr`
`= int_(r_1)^(r_2)GMm^2`
Integrating that gives,
`=> 1/2mu^2 = GMm(1/(r_1) - 1/(r_2)`
For a body to escape to 'infinty' its final position, `r_2 = oo`
So we have,

`=> 1/2mu^2 = GMm(1/(r_1) - 0`
`=> 1/2mu^2 = GMm(1/(r_1)` In this case `r_1` = radius of earth = `r_E`

`=> 1/2mu^2 = GMm(1/(r_E)`
`=> 1/2u^2 = GM(1/(r_E)`
`=> u^2 = 2GM/(r_E)`
`=> u = sqrt(2GM/(r_E))`

This shows that escape velocity is constant for all masses!