The gravitational potential is given as $$U(r)=-\frac{GMm}{R}$$ where $G$ is the universal gravitational constant $M$ is the mass of the earth and $m$ is the mass of an arbitrary object and $R$ is the distance of the object from the Center of the earth. and if I’m calculating the potential near the surface of the earth it is $$mgh.$$ $mgh$ is obtained by considering the surface of the earth as zero potential. Can I obtain the same by considering another point or line or entity as reference? I do not understand the concept of a reference point in estimation of potential. Is it like when you talk of potential at a certain point you need to specify what is the point of reference?
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$\begingroup$ Related: physics.stackexchange.com/q/286360/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Aug 6, 2023 at 2:57
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What is the significance of a reference point in calculating the potential?
It is merely a convenience. You can add an arbitrary constant to the energy without changing the physics. The reference point is usually chosen to make some common subset of problems easier.
The gravitational potential is given as $U(r)=-\frac{GMm}{R}$ ... and R is the distance of the object from the Center of the earth ...and if I’m calculating the potential near the surface of the earth it is mgh. mgh is obtained by considering the surface of the earth as zero potential....
Supposing that $R_e$ is the radius of the earth, you can write: $$ R=R_e+h\;, $$ where $h$ is small compared with $R_e$.
You can thus perform a Taylor series expansion in the small parameter $h/R_e$ (small compared to $1$): $$ U=-\frac{GMm}{R_e}\frac{1}{1+h/R_e} \approx -\frac{GMm}{R_e}\left(1 - \frac{h}{R_e}\right) =C + \frac{GMm}{R_e^2}h \equiv C + mgh\;, $$ where $C$ is a constant and $g\equiv \frac{GM}{R_e^2}$.
As you mention, you can always redefine the zero of the potential energy by subtracting off the constant $C$: $$ \tilde U(h) = U - C = mgh $$
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$\begingroup$ I understand what you say….but I just had one more thing to confirm though… so what you are saying is that this whole notion of potential is more of a relative kind….where in you measure the potential energy relative to some reference and not so much an absolute one… $\endgroup$– OrpheusCommented Aug 6, 2023 at 11:16
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$\begingroup$ @Orpheus Yes, it is only the energy difference that matters. $\endgroup$– hftCommented Aug 9, 2023 at 0:52