## In This post you will get answer of Q.

The radius of a planet is twice the radius of earth. Both have almost equal average mass-densities. $v_P$ and $v_E$ are escape velocities of the planet and the earth, respectively, then

Solution:

## Here, $R_P = 2 R_E$ , $rho_E = rho_P$

Escape velocity of the earth,

$V_E = sqrt frac {2GM_E}{R_E}= {sqrt {frac {2G}{R_E} bigg(frac {4}{3} pi R_E^3rho_Ebigg)}}$ $= R_E sqrt {frac {8}{3} pi Grho_E }, , , …(i)$

Escape velocity of the planet

$V_P = sqrt frac{2GM_P}{R_P} = {sqrt {frac {2G}{R_P} bigg(frac {4}{3} pi R_P^3rho_Pbigg)}}$$=R_{P}sqrt{frac{8}{3}pi Grho_{P}}ldotsleft(iiright)$

Divide (i) by (ii), we get

$frac {V_E}{V_P} = frac {R_E}{R_P} sqrt frac {rho_E}{rho_P}$

$frac {V_E}{V_P} = frac {R_E}{2R_E} sqrt frac {rho_E}{rho_E} = frac {1}{2}$

or $V_P = 2V_E$