Simplified formula of potential energy giving different expected value of mass

Question

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.

Answer 1

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 Ep=40 J for instance)

This method should have worked, so you probably just made some algebra or arithmetic error.

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 $r$.

Answer 2

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.