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

Answer 1

#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 provides

#(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)#
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Answer 2

The SI units of the universal gravity constant (G) are N m²/kg².

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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