Q63 / 90 · JEE Main 2023 (13 Apr Shift 1) · Physics

A planet having mass $9 M_{e}$ and radius $4 R_{e},$ where $M_{e}$ and $R_{e}$ are mass and radius of earth respectively, has escape velocity in $km s^{- 1}$ given by: (Given escape velocity on earth $V_{e} = 11.2 \times 10^{3} m s^{- 1}$ )

A
$67.2$
Correct answer
B
$16.8$
C
$11.2$
D
$33.6$

Solution

The escape velocity on earth is given by $V_{e} = \sqrt{\frac{2 G M_{e}}{R_{e}}}$
Escape velocity for the planet is
$V'_{e s c} = \sqrt{\frac{2 G (9 M_{e})}{4 R_{e}}} . . . (i)$
The escape velocity of earth is $V_{e} = 11.2 \times 10^{3} m s^{- 1}$
From equation (i)
$V'_{e s c} = \sqrt{\frac{9}{4} (\frac{2 G M_{e}}{R_{e}})} = \frac{3}{2} \times 11.2 \times 10^{3} m s^{- 1} \Rightarrow V'_{esc} = 16.8 km s^{- 1}$

Hence, the correct answer is option: 'B'.

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A planet having mass 9 Me and radius 4Re, where Me and Re are mass and radius of earth respectively, has escape velocity in km s-1 given by: (Given escape velocity on earth Ve=11.2×103 m s-1)

Solution

Relevant Questions

A planet having mass $9 M_{e}$ and radius $4 R_{e},$ where $M_{e}$ and $R_{e}$ are mass and radius of earth respectively, has escape velocity in $km s^{- 1}$ given by: (Given escape velocity on earth $V_{e} = 11.2 \times 10^{3} m s^{- 1}$ )