Question

!!WARNING ONLY FOR GENIUS g is inversly proportion tor^2 prove

Answer 1

We know that if an object has been lifted a distance hhfrom the ground then it has a potential energy change:

ΔU=mghΔU=mgh

so hh is proportional to ΔUΔU.

However, we have also the gravitational potential energy law:

U=−GMmrU=−GMmr

where the distance is inversely proportional to the potential energy.

The formula:

ΔU=mghΔU=mgh

is an approximation that applies when the distance hh is small enough that changes in gg can be ignored. As you say, the expression for UU is:

U=−GMmrU=−GMmr

So the change when moving a distance hhupwards is:

ΔU=GMmr−GMmr+hΔU=GMmr−GMmr+h

We rearrange this to get:

ΔU=GMm(1r−1r+h)=GMmhr2+rh=GMr2mh1+h/r≈GMr2mhΔU=GMm(1r−1r+h)=GMmhr2+rh=GMr2mh1+h/r≈GMr2mh

where the last approximation is because h≪rh≪rso 1+h/r≈11+h/r≈1. And since GM/r2GM/r2 is just the gravitational acceleration gg at a distance rr, we get:

ΔU=gmh


Your first potential energy arises from the approximation that the graviational field is approximately constant for "small heights" , i.e.

GMmr2≈mgGMmr2≈mg

The full law leads to your second formula, the approximation to the first. For heights above the earth, it is justified, as we can see by taylor expanding 1r21r2 around the earth's radius RR:

1r2=1R2−2(r−R)R3+O((r−R)2)1r2=1R2−2(r−R)R3+O((r−R)2)

Here, h=r−Rh=r−R, so

1r2=1R2(1−2hR)+O(h2)1r2=1R2(1−2hR)+O(h2)

The term with hh is certainly neglegible for h≪Rh≪R.