Question

With "normal" gravity, we used a potential energy of mgh. Now with the gravity that is more accurate over longer distances we use a different potential energy. In you own words compare and contrast the two situations. What are the important differences?

Answer 1

Answer:

A general solution is $\Delta U=mh\frac{GM}{r^{2}}\frac{r}{r+\Delta h}$ 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 $U=-\int F\cdot dx$, we know that the gravitational force is $F=GmM/r^{2}$, so we could find the potential energy taking the integral of F.

$U=-GmM/r$ (1)

We can find the particular case, just finding the gravitational potential energy difference:

$\Delta U=U_{f}-U_{i}$. Here Uf is the potential evaluated in r+Δh and Ui is the potential evaluated in r.

Using (1) we can calculate ΔU.

$\Delta U=-\frac{GmM}{r+\Delta h}+\frac{GmM}{r}$

Simplifying and combining terms we have a simplified expression.

$\Delta U=mh\frac{GM}{r^{2}}\frac{r}{r+\Delta h}$ (2)

Let's call $g=\frac{GM}{r^{2}}$. 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!