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If system consists of earth and ball and ball is dropped from height $h_i$ to $h_f$, then:

$\Delta U = -(W_{earth} + W_{ball})$ ($W_{ball}$ can be neglected since it's small)

$\Delta U = -(-mg(h_f - h_i) = = mgh_f - mgh_i$

So change of potential energy in the system when ball was dropped from $h_i$ to $h_f$ is $mgh_f - mgh_i$.

Q1: Since $\Delta U = U_f - U_i$, would it be correct to say that: $U_f = mgh_f$ and $U_i = mgh_i$ or potential energy is always a difference and $U_i = mgh_i$ is incorrect ? I believe it's incorrect because another formula is given by: $U = -\frac{GMm}{r}$. If the object is at height $h_i$ from ground, then $U = -\frac{GMm}{r + h_i}$, so $U_i = mgh_i$ can NOT be right. Right ?

Q2: If $U = -\frac{GMm}{r}$, what's the point of using $mgh$ at all ?

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  • $\begingroup$ mgh is a linearisation, a direct consequence of GMm/r; the convenience of mgh is undeniable. If you continue to learn more maths for physics, you will learn how to work out this problem in far greater detail and solve these questions yourself. $\endgroup$ Oct 19, 2023 at 16:06
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/286360/2451 and links therein. $\endgroup$
    – Qmechanic
    Oct 19, 2023 at 17:31

1 Answer 1

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For any potential energy function, there is an assumed zero. Potential energy is "unique" only up to an overall constant, because the force is equal to the derivative of the potential, and a uniform shift of the potential will therefore yield the same potential.

For $U = \frac{GMm}{r}$, the assumption is that we are using infinitely far away as our reference, i.e., we are saying that the potential is zero as $r\to\infty$. We don't have to do this, but it's a convenient choice, both mathematically and conceptually.

A small change in wording in the OP makes it clear how we should think about things in the linear case. Instead of saying height $h$, we will use position $y$ (or, more carefully, the vertical position $y$) of the ball. Then, the potential energy $U=mgy$ is a function of the position $y$ of the ball, and so the change in potential energy is $\Delta U = U_f-U_i = mgy_f-mgy_i$. Of course, this requires a choice of coordinate system, i.e., a choice of what $y=0$ means. It's often taken to be the ground, i.e., $y=0$ when the ball is on the ground. However, this is not necessary because not matter what, the change in potential energy is the same: $\Delta U = mg(y_f-y_i) = mg((y_f+y_0) - (y_i + y_0))$ so that if we move the origin by $y_0$, $\Delta U$ is the same. This effectively introduces a constant, physically irrelevant offset to $U$.

Finally, $U=mgy$ is an approximation that works when the ball is near the surface of the Earth. As OP mentions, the more correct expression for the potential is $U=-\frac{GMm}{r}$, where $r$ is the separation between the ball and the center of the Earth. However, near the surface of the Earth, the distance between the ball and the center of the Earth is $$ d = R + y\,, $$ where $y\ll R$, and $R$ is the radius of the Earth. In this case, we can expand $U$ in a Taylor series about $y/R=0$, yielding \begin{align} -\frac{GMm}{R+y} &=-\frac{G m M}{R}+\frac{G m M}{R^2} y -\frac{G m M}{R^3} y^2 + \cdots \\&\approx \mbox{constant} + m g y\,, \end{align} since $g=GM/R^2$. Note that $mgy$ is easier to work with mathematically than $GMm/r$, partly because it leads to a constant force, and hence the equations of motion are easy to solve for (the constant-acceleration kinematic equations!). This is why we use $mgy$ when we can.

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  • $\begingroup$ Thank you @march. I submitted edit for your taylor equations(there were typos). From your analysis, $U_i = mgh_i$ seems to be correct/valid(there's a small difference "constant" you pointed out), but neglecting it makes $U_i = mgh_i$ valid near the surface of earth if I'm not mistaken. $\endgroup$
    – Dimitri
    Oct 19, 2023 at 16:40
  • $\begingroup$ Yeah, it's all good: in my formulation, $y$ is the height above the ground, but it doesn't have to be. The point is that we can choose any zero, and so calling it position rather than height makes this more clear, I think. $\endgroup$
    – march
    Oct 19, 2023 at 17:21
  • $\begingroup$ I think in your taylor series, your constant is: $-gRm$ and overally, $-\frac{GMm}{R+y} = -gRm + mgy$ and $-gRm$ is not something that can be just thrown out which makes me think that saying that potential energy of the system when ball is at $h$ height from ground is $mgh$ is wrong. What makes you throw it out ? $\endgroup$
    – Dimitri
    Oct 19, 2023 at 17:25
  • $\begingroup$ $-GmM/R$ is a constant, because none of those quantities (the masses of the ball and Earth, the radius of the Earth, and the constant $G$) change in the problem. Thus, that constitutes a constant offset for the potential energy which is physically irrelevant because it will cancel out every time you compute a $\Delta U$. $\endgroup$
    – march
    Oct 19, 2023 at 17:30
  • $\begingroup$ Yes I know that in computing $\Delta U$, it won't have any effect. What I was saying is when we say that potential energy of the system is $mgh$, it's clearly wrong, isn't it ? we just showed that even near the surface of the earth, potential energy of the system(earth + ball) is $-gRm + mgy$ which is not the same as $mgy$. I'm not talking about the change, but exact potential energy scalar. $\endgroup$
    – Dimitri
    Oct 19, 2023 at 17:32

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