Answer:
A general solution is [tex]\Delta U=mh\frac{GM}{r^{2}}\frac{r}{r+\Delta h}[/tex] and a particualr case is mgh, it is just to distance around the radius Earth.
Explanation:
We can use a general equation of the potential energy to understand the particular and general case:
The potential energy is defined as [tex]U=-\int F\cdot dx[/tex], we know that the gravitational force is [tex]F=GmM/r^{2}[/tex], so we could find the potential energy taking the integral of F.
[tex]U=-GmM/r[/tex] (1)
We can find the particular case, just finding the gravitational potential energy difference:
[tex]\Delta U=U_{f}-U_{i}[/tex]. Here Uf is the potential evaluated in r+Δh and Ui is the potential evaluated in r.
Using (1) we can calculate ΔU.
[tex]\Delta U=-\frac{GmM}{r+\Delta h}+\frac{GmM}{r}[/tex]
Simplifying and combining terms we have a simplified expression.
[tex]\Delta U=mh\frac{GM}{r^{2}}\frac{r}{r+\Delta h}[/tex] (2)
Let's call [tex]g=\frac{GM}{r^{2}}[/tex]. It is the acceleration due to gravity on the Earth's surface, if r is the radius of Earth and M is the mass of the Earth and we can write (2) as ΔU=mgh, but if we have distance grader than r we should use (2), otherwise, we could get incorrect values of potential energy.
I hope i hleps you!
