Gravitational acceleration on the surface of a planet is \frac{\sqrt{6}}{11} g, where g is the gravitational acceleration on the surface of the earth. The average mass density of the planet is \frac{2}{3} times that of the earth. If the escape speed on the surface of the earth is taken to be 11 kms–1, the escape speed on the surface of the planet in kms

Engineering

Physics

Gravitational Acceleration

Keplers Law

Center of Mass

Question

Gravitational acceleration on the surface of a planet is $\frac{\sqrt{6}}{11}$ g, where g is the gravitational acceleration on the surface of the earth. The average mass density of the planet is $\frac{2}{3}$ times that of the earth. If the escape speed on the surface of the earth is taken to be 11 kms^–1, the escape speed on the surface of the planet in kms^–1 will be

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$\frac{V_{C 2}}{V_{eρ}} = \frac{ρ_{2} R_{2}}{ρ_{1} R_{1}}$

$= \sqrt{\frac{ρ_{2} R_{2}}{ρ_{1} R_{1}}}$

$\frac{g_{2}}{g_{1}} = (\frac{M_{2}}{M_{1}}) {(\frac{R_{1}}{R_{2}})}^{2}$

$\frac{ρ_{2}}{ρ_{1}} = (\frac{M_{2}}{M_{1}}) {(\frac{R_{1}}{R_{2}})}^{3}$

$\frac{g_{2} / g_{1}}{ρ_{2} / ρ_{1}} = (\frac{R 2}{R_{1}})$

$\frac{V_{C_{2}}}{V_{C_{1}}} = \sqrt{(\frac{g_{2}}{g_{1}}) (\frac{g_{2}}{g_{1}}) / \frac{ρ_{2}}{ρ_{1}}}$

$= \frac{g_{2}}{g_{1}} \sqrt{\frac{ρ_{2}}{ρ_{1}}} = \frac{\sqrt{6}}{11}$

$\sqrt{\frac{3}{2}}$