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my question concerns what the correct form of newton's law of universal law of gravity is when my vector r goes from the center of the body acted on towards the center of the body acting on it. In images:enter image description here

I'm interested in the force F of the bigger body on the smaller one. Since it's in the direction of my unit vector r, I can write

$$ \mathbf F = \frac{Gm_1m_2}{R^2}\mathbf r $$

Now, here is why this whole business seems awfully fishy to me. Let's admit that work is the integral of the force with respect to distance, ie:

$$ W = \int \mathbf F \cdot d\mathbf s $$

Since we're moving along the radius, we can write that as: $$ W = \int \mathbf F \cdot d\mathbf r $$

And if we go ahead and compute the integral:

$$ W = -Gm_1m_2\left(\frac{1}{r_f}-\frac{1}{r_i}\right) $$

The reason why I feel like this has to be wrong is because this implies that if we move from a distance a to a distance b, where a > b, the work done by the force is negative, we gain potential energy, and we lose kinetic energy. Which is all... backwards really!

So what's wrong with the reasoning above?

EDIT: Rob's answer made me think about the consequences of choosing my unit vector r going downwards and I have come to the following diagram (sorry about the quality of the drawing):

enter image description here

The seemingly fixes everything, but is it the right way to think about it?

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It's really impossible to answer this without seeing how you did the integral, but I'll take a guess.

To get the correct final sign we need to make sure we get the sign of the force correct, and the sign of the displacement correct. You have chosen a coordinate system that makes that tricky. If $r$ increases going down, then $r_f>r_i$ when the object moves down.

It's almost always easier to 1.) lay it out horizontally and 2.) choose $x$ to increase going to the right. Our brains seem to work better with that. Work the problem that way, and you'll have no problems.

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  • $\begingroup$ Isn't R, the distance between the centers of the two objects,a scalar quantity? $\endgroup$ – Quantaliinuxite Nov 7 '16 at 15:36
  • $\begingroup$ It is not a scalar. A scalar is invariant under rotations. It is a component of a vector. You have hoodwinked yourself by using the letter $r$, which is usually the magnitude of a vector. You are actually integrating over a cartesian coordinate direction, and for that you need a cartesian variable: a vector component in that direction. $\endgroup$ – garyp Nov 7 '16 at 15:36
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The $\vec{r}$ points from the center of the large object outward. But the gravitational force is in the opposite direction. Therefore $\vec{F}=-\frac{Gm_1m_2}{R^2}\vec{r}$. When you carry the negative sign through, it works out properly.

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  • $\begingroup$ I want r to point the same way it does on the diagram! @Michael $\endgroup$ – Quantaliinuxite Nov 7 '16 at 15:29
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You're using $\hat{\mathbf r}$, a downward-pointing unit vector, and $d\mathbf r = dx\hat{\mathbf x} + dy\hat{\mathbf y} + dz\hat{\mathbf z}$, an infinitesimal displacement, in two different ways, and becoming confused because you have given them similar symbols. The coordinate radius, $r = \sqrt{x^2 + y^2 + z^2}$, acquires a larger value as you move away from the origin of your coordinate system. So your scalar radius becomes larger as you move in the $(-\hat{\mathbf r})$ direction, according to your definition, and your potential energy is off by a sign.

Most people choose $\hat{\mathbf r}$ to point in the outward direction so that the scalar $r$ becomes larger in the $+\hat{\mathbf r}$ direction. This is like choosing your $+\hat{\mathbf x}$-axis in the direction that the coordinate $x$ increases, which is how your calculus intuition was developed.

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  • $\begingroup$ I don't understand why U should be off by a sign, after all the direction of our unit vector is just a convention and the math should work out just the same. Are you saying that if I chose r downward, I should flip the integral? (Which would give us the right everything) $\endgroup$ – Quantaliinuxite Nov 7 '16 at 15:45
  • $\begingroup$ I made an edit, did I interpret the implications of choosing r downwards correctly? $\endgroup$ – Quantaliinuxite Nov 7 '16 at 16:05

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