The escape velocity from the earth's surface is v. The escape velocity from the surface of another planet having a radius, four times that of earth and same mass density is [NEET 2021]

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Correct answer: D. (d) We know that, escape velocity from earth's surface, v_e= 2GM_eR_e Also, Density = Mass Volume ⇒ Mass = Density Volume or M_e= V or M_e= 43 R_e^3( ∵V= 43 R_e^3) Substituting the values in the escape velocity expression, we get v_e= 2 Gρ 43 R_e^3R_e v_e= 2 Gρ 43 R_e^2…(i) Given, radius of another planet is 4 times the radius of the earth, ⇒v_p= 2 Gρ 43 R_p^2 v_p= 2 Gρ 43 (4R_e )^2( ∵R_p=4R_e) v_p=4 2 Gρ 43 (R_e )^2 … . . ( ii ) On dividing Eq. (ii) by Eq. (i), we get v_pv_e= 4 2 Gρ 43 (R_e )^2 2 Gρ 43 R_e^2 ⇒v_p=4v_e=4v

$The escape velocity from the earth's surface is v . The escape velocity from the surface of another planet having a radius, four times that of earth and same mass density is [NEET 2021]$

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