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

Question

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

TestHub · Accepted Answer

The escape velocity on Earth is given by $v_{e} = \frac{2 G M _{e}}{R _{e}}$

The escape velocity on planet P is given by $v_{p} = \frac{2 G M _{p}}{R _{p}}$

Given that $M_{e} = 9 M_{p} \Rightarrow M_{p} = \frac{M _{e}}{9}$ and $R_{e} = 2 R_{p} \Rightarrow R_{p} = \frac{R _{e}}{2}$

Substituting these values into the expression for $v_{p}$ :

$v_{p} = \frac{2 G ( \frac{M _{e}}{9} )}{\frac{R _{e}}{2}}$

$v_{p} = \frac{2 G M _{e}}{R _{e}} \times \frac{2}{9}$

$v_{p} = \frac{2 G M _{e}}{R _{e}} \times \frac{2}{3}$

$v_{p} = \frac{v _{e}}{3} 2$

Comparing this with the given expression $\frac{v _{e}}{3} x$ , we get $x = 2$ .

Answer: $2$

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