Big G and small g

PMBC

My physics is rudimentary LC 1970s. At that time I understood g to be the acceleration due to gravity and G the gravitational constant in the equation of te attractive force between two masses. Also that this was the simpler explanation as per Newton that Einstein hjad turned upside down by explaining the 'attractions' as the curvature of space-time which every mass effects.
However, to set my mind at rest and stop my brain puzzling over it, what was the relationship between 'the two 'jees''?

Bit cynical

PMBC wrote: »

My physics is rudimentary LC 1970s. At that time I understood g to be the acceleration due to gravity and G the gravitational constant in the equation of te attractive force between two masses. Also that this was the simpler explanation as per Newton that Einstein hjad turned upside down by explaining the 'attractions' as the curvature of space-time which every mass effects.
However, to set my mind at rest and stop my brain puzzling over it, what was the relationship between 'the two 'jees''?

They are both ways of calculating the force [latex]F[/latex] on a body due to gravity. To calculate the force of a body on the surface of the earth we can either use

[latex]\displaystyle{F=mg}[/latex] or [latex]\displaystyle{F=G\frac{mm_{e}}{r^{2}}}[/Latex],

where [latex]m_e[/latex] is the mass of the earth and [latex]r[/latex] is the radius.

Since these should give the same value we can equate the two formulas:

[latex]\displaystyle{mg=G\frac{mm_{e}}{r^{2}}}[/latex]

The [latex]m[/latex]s cancel out on both sides and we are left with

[latex]\displaystyle{g=\frac{Gm_{e}}{r^{2}}}[/latex].

So that is essentially the relationship. If you have big G, you multiply it by the mass of the earth and divide by the radius squared to get little g. Try it out by plugging in the numbers. Of course the figure will vary from point to point on the earth, but this will give the rough figure.