# Why are the dimensions of escape velocity correct?

How does this formula work, from a dimensional analysis perspective?

$$v_\text{escape} = \sqrt{\frac{2GM}{R}}$$

The way I'm thinking about it is that $G$ is in units $\text{N} \cdot \text{m}^2/\text{kg}^2$. You multiply by a kilogram amount (the mass) to turn $G$ into units $N \cdot \text{m}^2/\text{kg}$. You then divide by the radius of the object to turn $G$ into units $N \cdot \text{m}/\text{kg}$.
However, $v_\text{escape}$ is in units $\text{m}/\text{s}$.
$\sqrt{N \cdot \text{m}/\text{kg}} \neq \text{m}/\text{s}$. Therefore, how does the equation even work if the units on either side aren't equal? Or am I doing this all wrong?

• Looks like an edit changed the meaning of the question, so it seems to answer itself... is a rollback in order? – user191954 Aug 26 '18 at 13:29
• @Chair: I made the edit because I assumed that the problem was the meaning of $N$ as a unit, and that the loss of the square root was just a typo. Perhaps virchau could let us know? – TonyK Aug 26 '18 at 13:31
• @TonyK Yeah, it's best if we wait for some confirmation from virchau. I saw your description for the edit while reviewing it, but I'm inclined to believe that it's relatively hard to forget a square root mathjax command, considering the fact that it's a significant number of characters. That's the sticky thing about such questions... – user191954 Aug 26 '18 at 14:22
• Yeah, agreed. I've put the question on hold until virchau can come back and clarify whether that missing square root was just a typo or if it was at the root of their confusion. (Note: a number of people thought this was a homework-like question, but at least in its current form, revision 5, it doesn't look like one to me.) – David Z Aug 27 '18 at 6:54
• @DavidZ: Sorry, I did indeed forget the square root. My main confusion was that I didn't realize that newtons aren't an SI unit; and Time4Tea's answer solved that confusion for me. This edit doesn't change the meaning of the question. – virchau13 Aug 28 '18 at 13:26

Newton is not a fundamental SI unit:

$$\mathrm N=\frac{\mathrm{kg}\cdot\mathrm m}{\mathrm s^2}.$$

So, in fact:

$$\frac{\mathrm N\cdot\mathrm m}{\mathrm{kg}}=\frac{\mathrm m^2}{\mathrm s^2},$$

the square root of which has the units of velocity.

You forget that $\mathrm{N} = \mathrm{kg}\ \mathrm{m}/\mathrm{s}^2.$

• How does that answer the question? – Peter Mortensen Aug 26 '18 at 11:37
• @PeterMortensen: it answers the question succinctly and accurately. It tells the OP that $\sqrt{N \cdot m/kg}$ is indeed equal to $m/s$. – TonyK Aug 26 '18 at 13:01
• It'd be better if you could expand this hint into a full-fledged answer. – user191954 Aug 26 '18 at 13:27
• Since Time4Tea has said the same thing and that answer has been accepted, I think that there's currently no need to expand my answer. – md2perpe Aug 26 '18 at 14:31
• A few thoughts on context and audience. If a student came to my office hours with this question I would probably give them exactly this answer. Because it is better if they follow the logic through themselves. But I would also sit there paying close attention to their face while they worked on it. Because in the event that they are too confused to see the way forward my job isn't done at that point. I still have to prod or lead them forward until they get it. So whether this answer is complete or not depends on the reader. Just one of the oddities of Stack Exchange. – dmckee --- ex-moderator kitten Aug 27 '18 at 2:27