First slide

Motion of planets and satellites

Question

Gravitational acceleration on the surface of the planet is $\frac{\sqrt{6}}{11} g$ , where g is the acceleration due to gravity on the surface of earth. The average mass density of the planet
is 2/3 time that of the earth. If the escape speed on the surface of the earth is taken to be 11 kms^-1, then find the escape speed on the surface of the planet (in kms^-1).

Difficult

Solution

$\begin{array}{l} We know that, g = \frac{G M}{R^{2}} = \frac{4}{3} π G R ρ \\ ∴ \frac{g_{p}}{g_{e}} = \frac{R_{p} ρ_{p}}{R_{e} ρ_{e}} also v_{e} = \sqrt{2 g R} \\ ∴ \frac{{(v_{e})}_{p}}{{(v_{e})}_{e}} = \sqrt{\frac{g_{p} R_{p}}{g_{e} R_{e}}} = \sqrt{\frac{g_{p}}{g_{e}} \cdot \frac{g_{p} ρ_{e}}{g_{e} ρ_{p}}} = \frac{g_{p}}{g_{e}} \sqrt{\frac{ρ_{e}}{ρ_{p}}} = \frac{\sqrt{6}}{11} \sqrt{\frac{3}{2}} = \frac{3}{11} \\ ∴ {(v_{e})}_{p} = \frac{3}{11} {(v_{e})}_{e} = \frac{3}{11} (11) = 3 {kms}^{- 1} \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App