-2

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 ?

Qmechanic
  • 201,751
Dimitri
  • 3
  • 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. – naturallyInconsistent Oct 19 '23 at 16:06
  • Possible duplicates: https://physics.stackexchange.com/q/286360/2451 and links therein. – Qmechanic Oct 19 '23 at 17:31

1 Answers1

0

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.

Dimitri
  • 3
march
  • 7,644
  • 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. – Dimitri Oct 19 '23 at 16:40
  • 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. – march Oct 19 '23 at 17:21
  • 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 ? – Dimitri Oct 19 '23 at 17:25
  • $-GmM/R$ 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 $\Delta U$. – march Oct 19 '23 at 17:30
  • 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. – Dimitri Oct 19 '23 at 17:32
  • The point that I'm making is that there is no unique choice for a potential energy function. None of them are wrong, because they all lead to the same physics. If we had started with $U=\frac{GMm}{r} +\frac{GMm}{R}$, where we have chosen a different reference point to set $U=0$ (which is perfectly acceptable to do!), we would have arrived at $U=mg y$ as our potential energy function. So saying that one or the other is wrong is just not something you can say. All that matters is the physics at the end, and each expression leads to the same prediction. – march Oct 19 '23 at 17:51
  • I think my confusion is that I see potential energy as a scalar and direct value and not about the change. Imagine earth mass is 100, ball mass is 1, radius from earth center to earth surface is 10 and G=1 for simplicity and this gives $g = 1$. When ball is put on the surface of earth, potential energy of the system is: 100/10 = 10. when ball is at height $2$ from surface, potential energy of the system is 100/12= 8.3. if you calculate this with $mgy$, you get: $112 = 2$. That's what I meant - $mgy$ can't be the potential energy calculation, it's a change – Dimitri Oct 19 '23 at 18:00
  • Did you read my post carefully though? - $mgy$ is an approximation that only works when $y$ is much smaller than the radius of the Earth; that's the whole context of taking the Taylor series. For example, a ball goes up 30 meters compared to the $6\times10^6$ meter radius of the Earth. You've chosen numbers $R=10$ and $y=2$, but 2 is not much less than 10; it's a full 20% of 10! – march Oct 19 '23 at 18:07
  • Furthermore, "I see potential energy as a scalar and direct value and not about the change". Well, don't do that. The value of the potential energy of a configuration of masses means absolutely nothing on its own. It only becomes meaningful in reference to the value of the potential energy of another configuration. This is why I keep talking about changes or differences in potential energy, because that is the only physically meaningful quantity in the context of potential energy. – march Oct 19 '23 at 18:09
  • I think I get it now. What I was trying to do is match the result of $\frac{GMm}{R+2}$ to $mg2$ and they clearly won't be equal even if I use the actual $G, M, R$ values. It seems that $\frac{GMm}{R}$ formula gives the potential energy of the system in relative to infinite separation. So if $-\frac{GMm}{R}$ gives us the number (-20), that means that I need to give the system 20J energy to make them separated at infinity distance. From this logic(if I'm right), that's how $mgy$ is derived in which case, now on the ground, potential energy is assumed as 0. – Dimitri Oct 19 '23 at 18:27
  • You've essentially got it now. – march Oct 19 '23 at 18:30
  • Thank you very much <3 – Dimitri Oct 19 '23 at 18:53