Question about variation of Law of Gravitation

Question

In the book, Fundamentals of Astrodynamics, the author provides a different form of the Law of Gravitation $$\ F_g = - \frac {GMm} {r^2} \frac {\mathbf r}{r} $$

So the book says that the ${\mathbf r} $ is the distance vector, but what does that make non-bold r, the scalar distance between the two masses?

score 2 · Accepted Answer · edited Mar 12 '17 at 16:47

2

Yes, $r$ is the scalar distance as usual. The equation that is handwritten into the picture is the scalar formulation. The printed version is the vector formulation. Notice the $\mathbf{F}$ is bolded.

Sometimes

$$\frac{\mathbf{r}}{r^3}$$

is written as

$$\frac{\mathbf{\hat r}}{r^2}$$

where $\mathbf{\hat r}$ is the unit vector in the $\mathbf{r}$ direction.

edited Mar 12 '17 at 16:47

uhoh

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answered Mar 12 '17 at 16:01

Organic Marble

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Thanks for the answer and the edit. So just to be sure, a distance vector divided by a scalar distance produces only the direction, right? – user78103 Mar 12 '17 at 19:07
@user78103 You will get the direction (a unit vector) only if you're dividing a vector by its magnitude. In this case, $r$ and $\bf{r}$ should have the same magnitude, so you will get the unit vector. – Phiteros Mar 12 '17 at 19:55
2

...and of course the minus sign is for that purpose too: as the distance vector is measured from the origin to the mass in question, the gravity acts towards the origin. – SF. Mar 13 '17 at 12:07
@SF. The vector is drawn from the source of the gravity to the object experiencing it's force. Origin schmorigin - origin has nothing to do with it. That's why we subtract vectors, so we don't have to think about where the origin is. Yes the source seems to be at the origin in the image, but that's not doing anybody any favors in my opinion. $\mathbf{r}= \mathbf{x}m- \mathbf{x}_M$ or $\mathbf{r}= \mathbf{x}{subject}- \mathbf{x}_{source}$ – uhoh Mar 14 '17 at 03:48
@uhoh: Ok, still, $\mathbf{\hat F} = - \mathbf{\hat r}$ – SF. Mar 14 '17 at 09:24
@SF. I don't think I've never put a caret directly above a vector with units as far as I remember, but maybe that's OK. I'd write $\mathbf{\hat n_F}=-\mathbf{\hat r}$ to avoid the quagmire of evaporating newtons. – uhoh Mar 14 '17 at 09:38
@uhoh: since $ \mathbf{\hat x} = { \mathbf{x} \over |x| }$ and both $\mathbf{x}$ and $|x|$ are the same unit, $\mathbf{\hat x} $ is dimensionless. The Newton evaporation is completely justified. – SF. Mar 14 '17 at 09:49
@SF. Ya OK you are right. I still don't think I do it (put carets on top of force vectors), but you are right there doesn't appear to be anything at all wrong with it. My head is here today where everything is unitless. fyi I'm almost there on the proton-shooting particle accelerator ship. Cats can certainly rotate without net angular momentum, so I guess advanced alien races (which cats actually are) can too. https://youtu.be/RtWbpyjJqrU give me one more day. – uhoh Mar 14 '17 at 10:03
@uhoh: That cat uses tail as a reaction wheel! – SF. Mar 14 '17 at 10:13
@SF. Destin says it's not the tail: watch from 02:44 https://youtu.be/RtWbpyjJqrU?t=164 – uhoh Mar 14 '17 at 10:16

Question about variation of Law of Gravitation

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