I am a little bit confused about Taylor series.
I understand how to expand some function around some $x_0$ point using formula
$$f(x)=\sum_{n}\frac{f^{(n)}(x_0)(x-x_0)^n}{n!}$$
Let's assume I have to expand potential energy $E_p$ around point on Earth (around radius $R$). Potential energy is $$E_p=-k/|R+h|,$$ where $h$ is height on Earth.
The result should be $-k/R +kh/R^2+…$
My problem is when I expand potential energy in Taylor series, I should take the value of derivative in $x_0$ point right?, that point is $R$, but when I look online for solution, everyone take $x_0$ as equal to zero?
Why is that?
Best Answer
If we write the potential energy given by gravitational potential, it is given, using your notation, as
$$E_p = E_p(r) = \frac{-k}{r}$$
Where $k$ is a constant real positive value $k>0$. If you set the origin (the origin in physics is one of the most important concepts, there is no sense to talk about results without explicitly mark where is the origin) at the centre of the Earth you'll get the result you pointed out
$$E_p(R+h) = -\frac{k}{R+h}$$
Now, if you want to expand the potential energy what you should do is to expand $E_p$ with respect to $r$
$$E_p(r) = \sum_{n}\frac{E_p^{(n)}(r_0)}{n!}(r-r_0)^n = E_p(r_0) + \frac{\mathrm d}{\mathrm d r}(E_p(r_0)) (r-r_0) + \frac{1}{2}\frac{\mathrm d^2}{\mathrm dr^2}(E_p(r_0))(r - r_0)^2+ \dots$$
What is important so you can expand is that the $E_p$ must be $n$ times differentiable at $r = r_0$. If this is true then you can use the $r_0$ you want.
So if you set $r_0 = R$ you'll get the answer
$$E_p(R) = \frac{-k}{R} + \frac{k}{R^2}(r - R) + \dots$$
and now you just set $r = R + h$ to have what you want: approximate potential at high $h$. It is not possible expand this around zero because this point is not defined. This fractions all have a term $1/r^n$ and these functions cannot have a zero as a point so you cannot expand this function with respect to $r_0 = 0$. You can see that your result is in agree with experience. If you forget the constant term the second term is
$$\frac{GM}{R^2}mh $$
Using that $G = 6.647\times10^{-11}m^3kg^{-1}s^{-2}$, $M = 5.973\times10^{24}kg$ and $R \approx 6.378 \times 10^{6}$ you get that
$$\frac{GM}{R^2} = 0.9799\times10 m/s^2 \approx 9.8 m/s^2 = g$$
so you get that close to the Earth
$$E_p = mgh$$