Question

The radius of a planet is twice the radius of earth.
Both have almost equal average mass-densities. If
Vp and Ve are escape velocities of the planet and
the earth, respectively, then
(1) ve = 1.5vp
(2) Vp = 1.5VE
(3) vp = 2V
(4) Ve = 3vp​

Answer 1

Answer:(2)

Explanation:

Answer 2

The escape velocity of planet is two times of escape velocity of earth.

(3) is correct option.

Explanation:

Given that,

The radius of a planet is twice the radius of earth.

$R_{p}=2R_{e}$

Both have almost equal average mass-densities.

$\rho_{p}=\rho_{e}$

We need to calculate the escape velocity

Using formula of escape velocity

$v=\sqrt{\dfrac{2GM}{R}}$

Escape velocity of the earth

$v_{e}=\sqrt{\dfrac{2GM_{e}}{R_{e}}}$

$v_{e}=\sqrt{\dfrac{2G}{R_{e}}\times\dfrac{4}{3}\pi\times R_{e}^3\rho_{e}}$

$v_{e}=R_{e}\sqrt{\dfrac{8}{3}\pi G\rho_{e}}$....(I)

Escape velocity of the planet

$v_{p}=\sqrt{\dfrac{2G}{R_{p}}\times\dfrac{4}{3}\pi\times R_{p}^3\rho_{p}}$

$v_{p}=R_{p}\sqrt{\dfrac{8}{3}\pi G\rho_{p}}$....(II)

Divided equation (I) by equation (II)

$\dfrac{v_{e}}{v_{p}}=\dfrac{R_{e}\sqrt{\dfrac{8}{3}\pi G\rho_{e}}}{R_{p}\sqrt{\dfrac{8}{3}\pi G\rho_{p}}}$

$\dfrac{v_{e}}{v_{p}}=\dfrac{R_{e}}{R_{p}}\sqrt{\dfrac{\rho_{e}}{\rho_{p}}}$

Put the value into the formula

$\dfrac{v_{e}}{v_{p}}=\dfrac{R_{e}}{2R_{e}}$

$\dfrac{v_{e}}{v_{p}}=\dfrac{1}{2}$

$v_{p}=2v_{e}$

Hence, The escape velocity of planet is two times of escape velocity of earth.

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Topic : Escape velocity

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