Arturo O. answered • 01/02/17

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Prashant,

I do not know if your physics class is with calculus or not. I will give you an answer using calculus.

You are probably already familiar with potential energy U as a function of height h:

U = mgh

But this formula is valid only for a constant acceleration of gravity g. The variation in height above the surface of the earth has to be extremely small so we can say that g ≅ constant.

In general, allowing for variability over a wide distance range, the force of gravitation between the earth M and a mass m is

F = -GMm/r

^{2}where r is the distance between the centers of the earth and the mass m.

Since F = -∂U/∂r, and using ∞ as our reference point,

U(r) = -∫

_{∞}^{r}F(r')dr' = -∫_{∞}^{r}-GMm/r'^{2}dr' = GMm∫dr'/r'^{2}U(r) = -GMm(1/r - 1/∞) = -GMm/rSo you can see that in the more general case, gravitational potential energy is negative, increases as you move away from the earth, and is zero at infinity.

Francisco P.

I would like to add to Arturo's fine explanation to say that the gravitational force (negative gradient of the gravitational potential energy) in both cases is directed to the center of the gravitating body (Earth) and is opposite to the direction of the increase in distance between the center of the two bodies (r).

You can think of the direction of the force as the negative of the slope of the U

_{g}vs. r graph at any position for both the exact potential (U = -GMm/r) and the linearized potential energy near Earth's surface (U = mgz) where z = r - R with R ≡ radius of Earth. In both cases, U increases as r increases for r > 0. The slope is positive, so the force is negative.
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01/02/17

Arturo O.

_{∞}^{r}dr'/r'^{2}= -GMm(1/r - 1/∞) = -GMm/r01/02/17