Answer :
As gravitational force is given by Gm1×m2
r²
F=Gm1×m2/r²⇒G=Fr²/m1×m2
G=[MLT-²] [L²]/[M2] as force dimensional formula is M1L1T-2
G=[M-1L³T-2]
r²
F=Gm1×m2/r²⇒G=Fr²/m1×m2
G=[MLT-²] [L²]/[M2] as force dimensional formula is M1L1T-2
G=[M-1L³T-2]
multiply both the side of the equation by r∧2
Fr∧2= G m1 m2
divided by side m1 m2
of force is measured in newtons ; r in meters ; m1 m2 in kg
[units of G] = Nm∧2/kg∧2
Newton isn't a base unit, so let's pull out the equation from Newtons' Second Law:
FORCE=MASS * ACCELERATION
N= Kg * m/s∧2
substituting that equation for N
[units of G] =(kg.m/s∧2)(m∧2/kg∧2)
[units of G] = m∧3/(kg.s∧2)
[units of G]=m∧3.kg∧-1s∧-2
in term of [M],[L] AND[T]
DIMENSTIONS OF GRAVITATION CONSTANT=G=[L]∧3.[M]∧-1.[T]∧-2
Fr∧2= G m1 m2
divided by side m1 m2
of force is measured in newtons ; r in meters ; m1 m2 in kg
[units of G] = Nm∧2/kg∧2
Newton isn't a base unit, so let's pull out the equation from Newtons' Second Law:
FORCE=MASS * ACCELERATION
N= Kg * m/s∧2
substituting that equation for N
[units of G] =(kg.m/s∧2)(m∧2/kg∧2)
[units of G] = m∧3/(kg.s∧2)
[units of G]=m∧3.kg∧-1s∧-2
in term of [M],[L] AND[T]
DIMENSTIONS OF GRAVITATION CONSTANT=G=[L]∧3.[M]∧-1.[T]∧-2