Why can the potential energy at any point be chosen to have any value?

[rudransh verma]

Not quite right. You can choose the potential to be zero anywhere you want. That does not affect the difference in potential energy when an object is displaced by a certain amount. Perhaps you are under the impression that potential energy can only be positive. It can be positive or negative.

Example: You have a well of depth $h$ below the ground. You have a rock at ground level which you drop into the well.
Case I: Choose the zero of potential energy at the bottom of the well.
Initial potential energy: $U_i=+mgh$
Final potential energy: $U_{\!f}=0$
Change in potential energy = $\Delta U=U_{\!f}-U_i=0-mgh=-mgh.$

Case II: Choose the zero of potential energy at ground level.
Initial potential energy: $U_i=0$
Final potential energy: $U_{\!f}=-mgh$
Change in potential energy = $\Delta U=U_{\!f}-U_i=-mgh-0=-mgh.$

It's the change that matters.

Sorry guys.

 

[phinds]

I would never ignore anyone if it’s a fact

So, you think we've just been giving you opinions instead of facts? Why are you here if you don't believe we know what we're talking about?

 

[rudransh verma]

So, you think we've just been giving you opinions instead of facts? Why are you here if you don't believe we know what we're talking about?

I do believe you are right but let it make sense to me.
Actually in case of spring it has PE=0 in relaxed state and its PE increases no matter where it moves. So I thought this must be the case for ball Earth system too. Somewhere I strongly believed in the back of my mind that this applies everywhere.

 

[russ_watters]

Actually in case of spring it has PE=0 in relaxed state and its PE increases no matter where it moves. So I thought this must be the case for ball Earth system too. Somewhere I strongly believed in the back of my mind that this applies everywhere.

It isn't true of the spring either. As I said before, it is common to choose easy/convenient zero reference, but what is easy/convenient can vary.

A common alternative scenario for a spring is when it is constrained or pre-loaded. That's what the "tare" button on a scale is for.

One of the problems with your posture here is that you have blinders on. You are locked into preconceived logic/intuition and thus unable to really listen to/absorb what people are telling you. It would help you if you could generalize the things you are learning and get off wayward paths faster.

 

[rudransh verma]

One of the problems with your posture here is that you have blinders on. You are locked into preconceived logic/intuition and thus unable to really listen to/absorb what people are telling you. It would help you if you could generalize the things you are learning and get off wayward paths faster.

Good point. I will try.

 

[kuruman]

I do believe you are right but let it make sense to me.
Actually in case of spring it has PE=0 in relaxed state and its PE increases no matter where it moves. So I thought this must be the case for ball Earth system too. Somewhere I strongly believed in the back of my mind that this applies everywhere.

What applies everywhere is that the potential energy at any point can have any value you choose because you can choose the zero of potential energy anywhere you choose. This is true even for a spring. Here is how it works for a spring which I assumed has one end fixed whilst the other is allowed to move and stretch or compress the spring.

When you write $PE=\frac{1}{2}kx^2$ the zero of potential energy is at the free end of the spring when the spring is relaxed. Also zero is the position of the free end of the spring. In other words, there are two zeroes implicitly defined at the end of the spring: (a) the zero of potential energy and (b) the origin of coordinates. The latter choice makes it possible to make the position of the free end equal to the displacement of the spring from the relaxed position. This is useful because the change in potential energy of the spring (remember the change is what counts) which is always $\Delta U = \frac{1}{2}k(\Delta x)^2$ can be written as $\Delta U = \frac{1}{2}k x^2$ because we chose the origin of coordinates so that $\Delta x=x.$

To help you digest this, consider what would happen if you measured coordinate $x$ from the fixed end and kept the zero of potential energy at the free end of the relaxed spring. Let $L$ be the relaxed end of the spring. Then the force exerted by the spring is written as $F=-k(x-L)$ and the force is negative when the spring is stretched ($x>L$) and positive when the spring is compressed ($x<L$).

First I will find the change in potential energy when the free end of the spring moves from $x_1$ to $x_2$. That change is the negative of the work done by the spring force: $$\Delta U=U_2-U_1=-\int_{x_1}^{x_2}Fdx=+\int_{x_1}^{x_2}k(x-L)dx=\frac{1}{2}k\left(x_2^2-x_1^2\right)-kL(x_2-x_1).$$ This, I repeat, is the change in potential energy when the spring moves from $x_1$ to $x_2.$

One can get the potential energy function $U(x)$ by choosing the potential energy to be zero at the starting or reference point $x_1=x_{\text{ref}}$ in which case $U_1=0$ in the above equation. Also the end point in the equation can be any $x$ because the subscript $2$ is no longer necessary. Then $$\Delta U=U_2-0= U(x)=\frac{1}{2}k\left(x^2-x_{\text{ref}}^2\right)-kL(x-x_{\text{ref}}).$$This is the most general expression for the potential energy of a spring of length $L$ where $x$ is the position of the free end relative to the fixed end. Note that it is zero at $x=x_{\text{ref}}$.

If you choose the potential to be zero at the free end, i.e. $x_{\text{ref}}=L$, then the potential simplifies to $$U(x)=\frac{1}{2}k\left(x^2-L^2\right)-kL(x-L)=\frac{1}{2}k(x-L)^2=\frac{1}{2}k(\Delta x)^2.$$Finally, if you choose the zero of coordinates to be at the free end, $\Delta x=x$ and you end up with the familiar $$U(x)=\frac{1}{2}kx^2.$$Please read this carefully and digest it. See how you can adjust the zeroes to get different values for the potential energy at a given point. It will liberate your mind.