Gravitational acceleration on the surface of a planet is 611 g, where g is the gravitational acceleration on the surface of the earth. The average mass density of the planet is 23 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

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

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

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

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

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

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

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

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

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

Let's analyze the problem step by step:

Given:

Gravitational acceleration on planet: $g_{p} = \frac{\sqrt{6}}{11} g$
Average mass density of planet: $ρ_{p} = \frac{2}{3} ρ_{e}$
Escape speed on Earth: $v_{e} = 11 km / s$

Step 1: Recall the formula for gravitational acceleration

For a spherical body: $g = \frac{G M}{R^{2}}$

So for Earth: $g = \frac{G M_{e}}{R_{e}^{2}}$

For planet: $g_{p} = \frac{G M_{p}}{R_{p}^{2}}$

Step 2: Recall the formula for escape speed

Escape speed: $v = \sqrt{\frac{2 G M}{R}}$

So for Earth: $v_{e} = \sqrt{\frac{2 G M_{e}}{R_{e}}}$

For planet: $v_{p} = \sqrt{\frac{2 G M_{p}}{R_{p}}}$

Step 3: Express mass in terms of density

Mass $M = ρ \cdot \frac{4}{3} π R^{3}$

So for Earth: $M_{e} = \frac{4}{3} π ρ_{e} R_{e}^{3}$

For planet: $M_{p} = \frac{4}{3} π ρ_{p} R_{p}^{3}$

Step 4: Find the ratio of planetary radius to Earth's radius

From gravitational acceleration:

\frac{g_{p}}{g} = \frac{\frac{G M_{p}}{R_{p}^{2}}}{\frac{G M_{e}}{R_{e}^{2}}} = \frac{M_{p} R_{e}^{2}}{M_{e} R_{p}^{2}}

Substitute the mass expressions:

\frac{\sqrt{6}}{11} = \frac{(\frac{4}{3} π ρ_{p} R_{p}^{3}) R_{e}^{2}}{(\frac{4}{3} π ρ_{e} R_{e}^{3}) R_{p}^{2}} = \frac{ρ_{p} R_{p}}{ρ_{e} R_{e}}

Given $ρ_{p} = \frac{2}{3} ρ_{e}$ :

\frac{\sqrt{6}}{11} = \frac{(\frac{2}{3} ρ_{e}) R_{p}}{ρ_{e} R_{e}} = \frac{2}{3} \cdot \frac{R_{p}}{R_{e}}

Solving for the radius ratio:

\frac{R_{p}}{R_{e}} = \frac{3}{2} \cdot \frac{\sqrt{6}}{11} = \frac{3 \sqrt{6}}{22}

Step 5: Find the ratio of planetary mass to Earth's mass

\frac{M_{p}}{M_{e}} = \frac{\frac{4}{3} π ρ_{p} R_{p}^{3}}{\frac{4}{3} π ρ_{e} R_{e}^{3}} = \frac{ρ_{p}}{ρ_{e}} \cdot {(\frac{R_{p}}{R_{e}})}^{3}

Substitute known values:

\frac{M_{p}}{M_{e}} = \frac{2}{3} \cdot {(\frac{3 \sqrt{6}}{22})}^{3}

Step 6: Find the escape speed ratio

\frac{v_{p}}{v_{e}} = \frac{\sqrt{\frac{2 G M_{p}}{R_{p}}}}{\sqrt{\frac{2 G M_{e}}{R_{e}}}} = \sqrt{\frac{M_{p} R_{e}}{M_{e} R_{p}}}

Substitute the mass and radius ratios:

\frac{v_{p}}{v_{e}} = \sqrt{[\frac{2}{3} \cdot {(\frac{3 \sqrt{6}}{22})}^{3}] \cdot [\frac{22}{3 \sqrt{6}}]}

Simplify the expression:

\frac{v_{p}}{v_{e}} = \sqrt{\frac{2}{3} \cdot \frac{{(3 \sqrt{6})}^{3}}{22^{3}} \cdot \frac{22}{3 \sqrt{6}}} = \sqrt{\frac{2}{3} \cdot \frac{27 \cdot 6 \sqrt{6}}{22^{3}} \cdot \frac{22}{3 \sqrt{6}}}

Notice that $\sqrt{6}$ terms cancel:

\frac{v_{p}}{v_{e}} = \sqrt{\frac{2}{3} \cdot \frac{27 \cdot 6}{22^{3}} \cdot \frac{22}{3}} = \sqrt{\frac{2}{3} \cdot \frac{27 \cdot 6 \cdot 22}{22^{3} \cdot 3}}

Simplify further:

\frac{v_{p}}{v_{e}} = \sqrt{\frac{2 \cdot 27 \cdot 6 \cdot 22}{3 \cdot 22^{3} \cdot 3}} = \sqrt{\frac{2 \cdot 27 \cdot 6}{9 \cdot 22^{2}}}

Calculate numerical values:

\frac{v_{p}}{v_{e}} = \sqrt{\frac{2 \cdot 27 \cdot 6}{9 \cdot 484}} = \sqrt{\frac{324}{4356}} = \sqrt{\frac{1}{13.444...}} = \sqrt{\frac{1}{\frac{121}{9}}} = \sqrt{\frac{9}{121}} = \frac{3}{11}

Step 7: Calculate the escape speed on the planet

v_{p} = \frac{3}{11} \cdot v_{e} = \frac{3}{11} \cdot 11 = 3 km / s

Final Answer: The escape speed on the surface of the planet is 3 km/s.

Key Formulae

Gravitational acceleration: $g = \frac{G M}{R^{2}}$
Escape speed: $v = \sqrt{\frac{2 G M}{R}}$
Mass of spherical body: $M = ρ \cdot \frac{4}{3} π R^{3}$

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