IIT-JEE 2010 · Paper 1 · Physics Q82

← IIT-JEE 2010 · Paper 1

PhysicsInteger+3Question 82 of 84

Gravitational acceleration on the surface of a planet is $\dfrac{\sqrt6}{11}g$ , where g is the gravitational acceleration on the surface of the earth. The average mass density of the planet is $\dfrac23$ times that of the earth. If the escape speed on the surface of the earth is taken to be 11 km s⁻¹, the escape speed on the surface of the planet in km s⁻¹ will be

Your answer

Answer and worked solution

Official answer: 3

Radius. $g=\frac43\pi G\rho R\propto\rho R$ :
$\frac{R_p}{R_e}=\frac{\sqrt6/11}{2/3}=\frac{3\sqrt6}{22}.$

Escape speed. $v_e=\sqrt{2gR}$ :
$\frac{v_p}{v_e}=\sqrt{\frac{\sqrt6}{11}\cdot\frac{3\sqrt6}{22}}=\sqrt{\frac{18}{242}}=\frac3{11}.$
So $v_p=\frac3{11}\times11=3$ km s⁻¹.

Answer: 3

Topics:Escape velocity Variation of g

← Previous question Next question →

IIT-JEE 2010 · Paper 1, Physics question 82