First slide

Motion of planets and satellites

Question

Gravitational acceleration on the surface of the planet is $\frac{\sqrt{8}}{11} g$ , where g is the acceleration due to gravity on the surface of earth. The average mass density of the planet is 2 times 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).

Moderate

Solution

$We know that, g = \frac{G M}{R^{2}} = \frac{4}{3} π G M ρ$
$\begin{array}{l} ∴ \frac{g_{p}}{g_{e}} = \frac{R_{p} ρ_{p}}{R_{e} ρ_{e}} also v_{e} = \sqrt{2 g R} \\ ∴ \frac{{(v_{e})}_{p}}{{(v_{e})}_{e}} = \sqrt{\frac{g_{p} R_{p}}{g_{e} R_{e}}} = \sqrt{\frac{g_{p}}{g_{e}} \cdot \frac{g_{p} ρ_{e}}{g_{e} ρ_{p}}} = \frac{g_{p}}{g_{e}} \sqrt{\frac{ρ_{e}}{ρ_{p}}} = \frac{\sqrt{8}}{11} \sqrt{2} = \frac{4}{11} \\ ∴ {(v_{r})}_{p} = \frac{4}{11} {(v_{e})}_{e} = \frac{4}{11} (11) = 4 {kms}^{- 1} \end{array}$

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