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It is said that since work is done by the gravitational potential itself , so at a finite distance the gravitational potential energy of the body is negative. Could someone explain why?

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I don't find any of the linked answers satisfying or clear.

First a physical argument. Assume potential energy (or potential) is zero at infinity. What happens when I put an object in the gravitational field, initially at rest? It accelerates toward the Earth. That is, its kinetic energy increases. To conserve energy, potential energy must decrease. But PE was zero at infinity. To decrease from zero, the PE at all points has to be negative.

Another argument. This one requires the definition of potential energy. Many presentations are not clear about this. Here is the correct definition of potential energy: $$\Delta U = -W_\mathrm{internal,\, conservative}$$

The system is the Earth-object system; the internal force is gravity. Gravity is conservative.

Continuing,

$$U_r - U_\infty = -W_\mathrm{internal}$$

Since $U_\infty = 0$ we have $$ U_r = -W_\mathrm{internal}$$

The work done by gravity is always positive. We know this because if we release a particle at rest its speeds up. $W_\mathrm{internal} > 0$, so $U_r < 0$

Mathematics. This is what ties people's minds in knots. Try this. Our job is to calculate the potential energy of a body in a gravitational field. We need the work done by gravity in bringing an object from $\infty$ to $r$. Less confusing, but the same thing, is to bring the object from $-\infty$ to $-r$. This avoids suspicious arguments about the sign of $\mathrm{d}r$. The integration moves "from left to right" so there is no argument: $\mathrm{d}r >0$, and the displacement is in the same direction as the force. The work is positive. No minus sign there, either. $$W_{\mathrm{internal}} = \int_{-\infty}^{-r} \frac{1}{r^2}\,\mathrm{d}r $$ All of the other variables are positive definite so they play no role in determining the sign. We leave them off. $$W_{\mathrm{internal}} = \left[-\frac{1}{r}\right]_{-\infty}^{-r} $$ $$ W_{\mathrm{internal}} = \frac{1}{r} - (-\infty) $$ $$ W_{\mathrm{internal}} = \frac{1}{r} $$ and since $$ U_r = -W_\mathrm{internal}$$ we get $$U_r = -\frac{1}{r}$$

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First of all I would like to correct your statement. The work would not be done by potential, work is always done by a force resulting a change in potential energy.

Let us consider two objects (say 1 and 2 and both of them spherical) which form an isolated system, having zero kinetic energy. Hence the their total energy will purely be due to their potential energy. Now, absolute potential can never be defined, i.e. only a reference can be set, which we set at infinity in case of point objects and spheres. Now the objects considered by us, 1 and 2 would attract each other due to their mutual gravitational force of attraction. Hence the work done by an external agent would be negative to bring them together from infinity, as the force would act in opposite direction to that of displacement which in turn is in the direction of gravitational force. And by definition of potential energy we can conclude that the potential energy of the system will be negative and hence the potential due to either object.

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Its not that potential energy is negative for all finite r values... It is that work done in moving from a farther point to a closer point is always negative. Since the potential at $\infty$ is set to zero, all other potentials become negative.

Because potential at r is the work done in bringing a unit mass from $\infty$ (which is far away) to a distance r (which is closer).

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