Sign of potential term in Lagrangian mechanics

[Admiralibr123]

I have heard many times that it does not matter where you put the zero to calculate the potential energy and then $L=T-V$. But mostly what we are doing is taking potential energy negative like in an atom for electron or a mass in gravitational field and then effectively adding it to kinetic energy. What if I take zero at the centre of atom and on the surface of earth, write $V$ positive and subtract if in lagrangian. I try that and EOM is different something like $$F=ma=-mg$$ for a free fall object. If I change it for a charge an electron in atom something similar happens.

Please someone explain, does $$F=-mg \quad \textrm{vs} \quad F=+mg$$ not matter? This is just a single term but if the EOM have multiple terms, this sign would surely make a difference. And in essence to me, it feels like it voids the negative sign in $L=T-V$, if we can choose any sign for V. That must definitely not be the case, so where am I wrong in my reasoning?

 

[Vanadium 50]

Calculate the equations of motion for T - V and T - V + a constant. Any difference?

 

[Dale]

You seem to have the mistaken idea that the arbitrariness of the zero means you can flip the sign of the potential. That is a mistake.

You can add a constant to the potential, you cannot multiply it by a constant. That includes multiplying it by -1

 

[PeroK]

That must definitely not be the case, so where am I wrong in my reasoning?

For GPE you have two conventions. If the object remains close to the surface, then you can write: $$V = mgh$$ where $h$ is the height above the surface.

And, in the general case we write $$V = -\frac{GMm}{r}$$.
The critical point, however, is that in both cases $V$ increases as $r$ or $h$ increases.

As a useful exercise, you could show that for $r = R + h$, where $h << R$ we have:
$$V = -\frac{GMm}{r} = -\frac{GMm}{R + h} \approx V_0 + mgh$$ where $V_0$ is constant. This shows that the two conventions are approximately equivalent up to a constant term (in the case where the object remains close to the surface).