Question

what is escape velocity on the moon

Answer 1

hiiiii friend!!
==========================================================
The formula for escape velocity is:

Vesc=(√(2∗G∗M)/r)Vesc=((2∗G∗M)/r)

Where:

G = 6.67384E-11 N*m^2*kg^-2 (Gravitational Constant)
M = 7.34767E22 kg (mass of the moon)
r = 1737400 m (radius of the moon)

Plug those in and we get an escape velocity of 2375.89 m/s
========================================================
i think it may help u!!!

AND.......

Answer 2

$\huge \underline\red{Answer:-} \\ \\ \underline\red{escape \: velocity \: :-} \: \ \: velocity \: which \: is \: \\ needed \: by \: any \: object \: (body) \: to \: escape \: \\ from \: gravitational \: influence \: \\ is \: know \: to \: be \: escape \: velocity . \\ \\ { \red{= > }the \: formula \: of \: escape \: velocity \: we } \\ {found \: that, \: for \: earth, \: } \\ { it \: is \:} \red{ \sqrt{\frac{{2 \: G \: M} }{R} } \: or \: \sqrt{2 \: g \: R} } \: . \\ \\ {\red{V \: =} escape \: velocity}</p><p> \\ {\red{G \: =} \: gravitational \: constant \:} \\ { is \: 6.67408 × {10}^{ - 11} {m}^{3} {kg}^{ - 1} {sec}^{ - 2} }</p><p></p><p> \\ { \red{M \: = }\: mass \: of \: the \: planet}</p><p></p><p> \\ {\red{R \: =} \: radius \: from \: the \: centre \: of \: } \\ {gravity \: } \\ {\red{g \: = } \: is \: the \: acceleration \: due \: } \\ {to \: the \: gravity \: of \: earth}</p><p>\\ \\ \underline\red{The \: escape \: velocity \: of \: earth \: is \: } \\ \underline\red{11 \: kilometers \: per \: second.} \: \\ \\ \underline \red{the \: escape \: velocity \: of \: moon \:} \\ \underline \red{is \: 2.38 \: km/s.}\\ \\ \fcolorbox{red}{white}{i \: hope \: it \: helps \: you.}$