Question about variation of Law of Gravitation

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? 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.

• 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? 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. Mar 12 '17 at 19:55
• ...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 3:48
• @uhoh: Ok, still, $\mathbf{\hat F} = - \mathbf{\hat r}$
– SF.
Mar 14 '17 at 9:24