Find the escape velocity on the surface of the moon, if the mass and radius of the moon are $7.34 \times 10^{22} k g$ and $1.74 \times 10^{6} m$ respectively.
2.37

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Solution

The correct option is A 2.37
Given: Gravitational constant $G = 6.67 \times 10^{- 11} N m^{2} k g^{- 2}$
Mass of moon, $M = 7.34 \times 10^{22} k g$
Radius of the moon, $R = 1.74 \times 10^{6} m$
We know, escape velocity,
$v_{e s c}$ = $\sqrt{\frac{2 G M}{R}}$
$∴ v_{e s c} = \sqrt{\frac{2 \times 6.67 \times 10^{- 11} \times 7.34 \times 10^{22}}{1.74 \times 10^{6}}}$
$= 2.37 k m s^{- 1}$

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Find the escape velocity on the surface of the moon, if the mass and radius of the moon are 7.34×1022kg and 1.74×106m respectively.2.37

The correct option is A 2.37Given: Gravitational constant G=6.67×10−11Nm2kg−2 Mass of moon, M=7.34×1022 kg Radius of the moon, R=1.74×106m We know, escape velocity, vesc = √2GMR ∴vesc=√2×6.67×10−11×7.34×10221.74×106 =2.37 kms−1

Find the escape velocity on the surface of the moon, if the mass and radius of the moon are $7.34 \times 10^{22} k g$ and $1.74 \times 10^{6} m$ respectively.
2.37