Gravitational acceleration on the surface of the planet is 11 x g . If the average mass density of the planet is 3 2 times that of the earth and the escape velocity from the surface of earth is

Question

Gravitational acceleration on the surface of the planet is 11 x ​ ​ g . If the average mass density of the planet is 3 2 ​ times that of the earth and the escape velocity from the surface of earth is 11 km/s and from the planet is 3 km/s, then the value x is ( g is gravitational acceleration of earth)

TestHub · Accepted Answer

The escape velocity from the surface of a planet is given by $v = 2 g R$ , which gives $R = \frac{v ^{2}}{2 g}$ .

The acceleration due to gravity is $g = \frac{4}{3} π GRρ$ .

Substituting the expression for $R$ into the equation for $g$ :

$g = \frac{4}{3} π G (\frac{v ^{2}}{2 g}) ρ$

$g^{2} = \frac{2}{3} π G v^{2} ρ$

$g = v \frac{2}{3} π Gρ$

This implies that $g \propto v ρ$ .

Taking the ratio of the gravitational acceleration on the planet to that on Earth:

$\frac{g _{p}}{g _{e}} = \frac{v _{p}}{v _{e}} \frac{ρ _{p}}{ρ _{e}}$

Given that $v_{e} = 11$ km/s, $v_{p} = 3$ km/s, and $ρ_{p} = \frac{2}{3} ρ_{e}$ :

$\frac{g _{p}}{g _{e}} = \frac{3}{11} \frac{2}{3}$

$\frac{g _{p}}{g _{e}} = \frac{9 \times \frac{2}{3}}{11} = \frac{6}{11}$

$g_{p} = \frac{6}{11} g_{e}$

Comparing this with the given expression $g_{p} = \frac{x}{11} g$ , we get $x = 6$ .

Answer: $6$

Physics - Gravitation Question with Solution | TestHub

Options:

Doubts & Discussion