The ratio of escape velocity at earth ve to the escape velocity at a planet vp whose radius and mean density are twice as that of earth is :

Question

The ratio of escape velocity at earth ve to the escape velocity at a planet vp whose radius and mean density are twice as that of earth is : (A) 1: 2 √2 (B) 1: 4 (C) 1: √2 (D)  1: 2

Tardigrade Admin · Accepted Answer

Ve = √(2GM/R) = √(2G/R).((4/3) π R3ρ) propto R√ρ ∴ Ratio = 1: 2 √2

Q. The ratio of escape velocity at earth $v_{e}$ to the escape velocity at a planet $v_{p}$ whose radius and mean density are twice as that of earth is :

$1 : 22$

$1 : 4$

$1 : 2$

$1 : 2$

$V e = \frac{2 GM}{R} = \frac{2 G}{R} . (\frac{4}{3} π R^{3} ρ) \propto R ρ$
$∴$ Ratio = $1 : 22$

Q. The ratio of escape velocity at earth ve​ to the escape velocity at a planet vp​ whose radius and mean density are twice as that of earth is :

1:22​

1:4

1:2​

1:2

Ve=R2GM​​=R2G​.(34​πR3ρ)​∝Rρ​ ∴ Ratio = 1:22​

Q. The ratio of escape velocity at earth $v_{e}$ to the escape velocity at a planet $v_{p}$ whose radius and mean density are twice as that of earth is :

$1 : 22$

$1 : 4$

$1 : 2$

$1 : 2$

$V e = \frac{2 GM}{R} = \frac{2 G}{R} . (\frac{4}{3} π R^{3} ρ) \propto R ρ$
$∴$ Ratio = $1 : 22$