Newton universal gravitation formula, how f is dimensionally derived?

[yeoG]

If f dimensions are ml/t^2, where does t^2 come from in the equation of
F = G*m1*m2/r^2 where I believe G to be a constant, m1 and m2 to be masses and r to be the distance between two masses - so length. To dimensionally analyse this then, where would the dimension time come from if I were to check if the equation is dimensionally correct and I can't see where it comes from? I do think that it may come from G but i have seen various answers on what the dimensions of G are so I can't cancel the dimensions out to check if it is the same as F? Help would be appreciated, many thanks.

 

[A.T.]

http://en.wikipedia.org/wiki/Gravitational_constant#Dimensions.2C_units.2C_and_magnitude

 

[Integral]

According to Newton

F=ma since acceleration has units l t^-2 the units of force must be m Lt^-2.

Using that you can figure out the units G needs to make the given equation dimensionally correct. Time is factored into the constant.

 

[yeoG]

http://en.wikipedia.org/wiki/Gravitational_constant#Dimensions.2C_units.2C_and_magnitude

It says N (m/kg)^2
so that in dimensions are m x l/t^2(N = Newton = kg x m/s^2??)l^2/M^2)(m/kg^2)
so that would mean F = m x l/t^2 l^2/m^2/l^2 how would this cancel out to F which is ml/t^2? Have I got this concept wrong or is there a mistake I am making?

 

[yeoG]

According to Newton

F=ma since acceleration has units l t^-2 the units of force must be m Lt^-2.

Using that you can figure out the units G needs to make the given equation dimensionally correct. Time is factored into the constant.

Hi thanks for the help however I focusing on F instead of G giving that G = N m^2 kg^-2 or what G is?

 

[nrqed]

It says N (m/kg)^2
so that in dimensions are m x l/t^2(N = Newton = kg x m/s^2??)l^2/M^2)(m/kg^2)
so that would mean F = m x l/t^2 l^2/m^2/l^2 how would this cancel out to F which is ml/t^2? Have I got this concept wrong or is there a mistake I am making?

Isolating, $G= F$ $r^2 / (m_1 m_2 )$. You can tell right away the units of G.

 

[yeoG]

ah I think I understand so by working out the dimensions of G you can show that F must equal to Gxm1xm2 because it will balance the equation to = F?

 

[A.T.]

so by working out the dimensions of G you can show that F must equal to Gxm1xm2

No. Observation shows that F equals Gxm1xm2/r^2. The dimension of G are defined based on this. So there is no point in checking if the equation is dimensionally correct, because it was defined to be correct.