Question

If earth has a mass 9 times and radius twice of a planetvl Mars calculate the minimum velocity required by a rocket to pull out of gravitational force of mars. Take the escape velocity on the surface of earth to be 11.2 km/s

Answer 1

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Answer 2

The minimum velocity required by a rocket to pull out of gravitational force of mars is 5.3 km/s

Escape velocity:

The minimum velocity required to escape the gravitational field of earth is called escape velocity. It is denoted by $V_{e}$

The escape velocity on earth is given by:

$V_{e}=\sqrt{\frac{2 G M}{R}}=11.2 \mathrm{~km} / \mathrm{s}$

Where $\mathrm{G}$ is universal gravitational constant, $\mathrm{M}$ is mass of the earth and $\mathrm{R}$ is radius of the earth.

Given:

$R_{e}=2 R_{m}, v_{e}$

Escape speed on surface of Earth) $=11.2$ km/s

Let$\mathrm{V}_{\mathrm{m}}$ be the speed required to pull out of the gravitational force of mars.

Step by step solution

Formula for the escape velocity

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

$v_{m}=\sqrt{\frac{2 G M_{m}}{R_{m}}}$

Substitute the given values

$=\sqrt{\frac{2 G M_{m}}{R_{m}} \times \frac{R_{e}}{2 G M_{e}}} \\&\frac{v_{m}}{v_{e}}$     $=\sqrt{\frac{M_{m}}{M_{e}} \times \frac{R_{e}}{R_{m}}}=\sqrt{\frac{1}{9} \times 2}=\frac{\sqrt{2}}{3} \\&v_{m}$    $=\frac{\sqrt{2}}{3}(11.2)=5.3 \mathrm{~km} / \mathrm{s}$

$&\Rightarrow \mathbf{v}_{\mathbf{m}}=5.3 \mathrm{~km} / \mathrm{s}\end{aligned}$

Hence the minimum velocity required by a rocket to pull out of gravitational force of mars is 5.3 km/s

Learn more about escape velocity here

https://brainly.in/question/3018919?msp_poc_exp=4

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