# Simplified formula of potential energy giving different expected value of mass

I am trying to calculate the mass of a planet by the following image

I have the mass of the object, $$2$$ kg, and the radius of the planet, $$5000$$ km (also the gravitational constant $$G$$). My first attempt was to calculate the variation of the potential energy with the classical formula, taking any point of the graph ($$h = 10$$ m and $$E_p = 40$$ J for instance)

$$U_G(r) = -\frac{GMm}{r}$$

But trying to get $$M$$ is giving me the wrong result with this method.

Looking to the correct answer, the author is using the simplified formula

$$U_G = mgh$$

But I have no clue of why this works and the normal method not. In my attempt I got a mass of $$1.5 \cdot 10^{18}$$ kg, and the answer is $$7.5 \cdot 10^{23}$$ kg.

• In this graph what is h given relative to? I suspect it's given relative to the 'ground' and not the center of the planet. In this case, $U=mgh$ can be used as a good enough approximation. Also, note that if it was relative to the center (as if someone drilled a hole through the planet) and the planet was uniform, then potential would actually be quadratic in h. – DanDan0101 Sep 1 at 20:30
• @DanDan0101 yes, it's relative to the ground. Could I use the classical formula in this case? – Norhther Sep 1 at 20:35

Since $$U=-\frac{GMm}{r}$$ then $$\frac{dU}{dr} = \frac{GMm}{r^2}$$ which gives us $$M=\frac{dU}{dr} \frac{r^2}{Gm}$$ From the graph $$\frac{dU}{dr}=4 \ \text{J/m}$$ and the rest of the values are known. This gives the correct answer of $$M=7.5 \ 10^{23} \ \text{kg}$$
To me this seems like the easiest way to work this problem. Otherwise you have to remember the formula for $$g$$ in terms of $$G$$, $$M$$, and $$r$$.
I'll be trying to explain why $$U=mgh$$ is a good enough approximation for potential at the surface of the planet.
Let $$R$$ be the radius of the planet. Then, according to the equation for potential, $$U=-\frac{GMm}{R+h}$$ Note that $$R$$ is about 5 orders of magnitude larger than $$h$$. Perform a Maclaurin expansion on the term $$\frac{h}{R}$$, and use the definition $$g=\frac{GM}{R^{2}}$$: $$U=-\frac{GMm}{R}\left(1+\frac{h}{R}\right)^{-1}=-mgR\left(1-\frac{h}{R}+\mathcal{O}\left(\left(\frac{h}{R}\right)^{2}\right)\right)$$ If we now define our potential as relative to the potential on the surface of the planet, which is $$-mgR$$, then: $$U=mgh-\mathcal{O}\left(\frac{h}{R}mgh\right)$$ Usually $$\frac{h}{R}$$ is so small that the term is treated as negligible, leading to the familiar equation $$U=mgh$$. I think you know how to do the problem now using this fact.