Given Newton's gravity formula, #F=G(Mm)/(r^2)#, what are the SI units of the universal gravity constant, #G#?

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Answer 1 · 2016-03-19T12:16:05

#G# has the SI units #m^3/(kg*s^2)#

#F# is the force of gravity: #(kg*m)/s^2#

#M# and #m# are masses, in #kg#

#r# is the distance between the masses, in #m# (meters)

This gives

#(kg*m)/s^2 = G xx (kg^2)/(m^2)#

#m^2*(kg*m)/s^2 = G xx kg^2#
#m^2*(kg*m)/s^2*1/k^2 = G#
#m^2*(cancel(kg)*m)/s^2*1/k^cancel(2) = G#
#(m^3)/s^2*1/k = G#
#G = m^3/(kg*s^2)#

Given Newton's gravity formula, #F=G*(M*m)/(r^2)#, what are the SI units of the universal gravity constant, #G#?