If earth has a mass nine times and radius twice to that of a planet P. Then (ve/3) √x ms -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

Question

If earth has a mass nine times and radius twice to that of a planet P. Then (ve/3) √x ms -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 (A) 1 (B) 18 (C) 3 (D) 2

Tardigrade Admin · Accepted Answer

v text (escape) plant =√(2 G MP/RP) =√(2 G((Me/9))/((Re)2))=(ve √2/3) ∴ x=2

Q. If earth has a mass nine times and radius twice to that of a planet $P$ . Then $\frac{v _{e}}{3} x m s^{- 1}$ will be the minimum velocity required by a rocket to pull out of gravitational force of $P$ , where $v_{e}$ is escape velocity on earth. The value of $x$ is

1

18

3

2

$v_{(escape) plant} = \frac{2 G M _{P}}{R _{P}}$
$= \frac{2 G ( \frac{M _{e}}{9} )}{( \frac{R _{e}}{2} )} = \frac{v _{e} 2}{3}$
$∴ x = 2$

Q. If earth has a mass nine times and radius twice to that of a planet P. Then 3ve​​x​ms−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

1

18

3

2

v(escape) plant ​=RP​2GMP​​​ =(2Re​​)2G(9Me​​)​​=3ve​2​​ ∴x=2

Q. If earth has a mass nine times and radius twice to that of a planet $P$ . Then $\frac{v _{e}}{3} x m s^{- 1}$ will be the minimum velocity required by a rocket to pull out of gravitational force of $P$ , where $v_{e}$ is escape velocity on earth. The value of $x$ is

$v_{(escape) plant} = \frac{2 G M _{P}}{R _{P}}$
$= \frac{2 G ( \frac{M _{e}}{9} )}{( \frac{R _{e}}{2} )} = \frac{v _{e} 2}{3}$
$∴ x = 2$