If earth has a mass nine times and radius twice to the of a planet P . Then v e 3 x m s - 1 will be the minimum velocity required by a rocket to pull out of gravitational force of P , where ve is esca

Question

If earth has a mass nine times and radius twice to the of a planet P . Then v e 3 x m s - 1 will be the minimum velocity required by a rocket to pull out of gravitational force of P , where ve is escape velocity on earth. The value of x is

TestHub · Accepted Answer

The formula to calculate the escape velocity $(v_{e})$ for the Earth is given by

$v_{e} = \sqrt{\frac{2 G M_{e}}{R_{e}}} . . . . . . . . . . . . . . . . . . . . . . . (1)$

where, $G$ is the Universal Gravitational constant, $M_{e}$ is the mass and $R_{e}$ is the radius of Earth.

The formula for the escape velocity $(v_{esc})$ of any planet is given by

$v_{esc} = \sqrt{\frac{2 G M}{R}} . . . . . . . . . . . . . . . . . . . . . . . . (2)$

where, $M$ is the mass and $R$ is the radius of the planet.

Given that the mass of the planet is $9$ times the mass of the Earth and the radius of the planet is $2$ times the radius of the Earth.

Substitute $\frac{M_{e}}{9}$ for $M$ and $\frac{R_{e}}{2}$ for $R$ into equation (2) and simplify to obtain the escape velocity for the planet in terms of that in Earth.

$\begin{array}{rcl} v_{esc} & = & \sqrt{\frac{2 G \frac{M_{e}}{9}}{\frac{R_{e}}{2}}} \\ = & \sqrt{\frac{2}{9} \frac{2 G M_{e}}{R_{e}}} \\ = & \frac{\sqrt{2}}{3} v_{e} . . . . . . . . . . . . . . . . . . . . . (3) \end{array}$

Comparing equation (3) with the given expression, it can be written that

$x = 2$ .

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