I tried using dimensional analysis to deduce Newton's law of gravity but I wasn't able to do so as one of the equations were $0=-2$ which is a contradiction. But I thought that we can't do that because the constant of gravitation has some dimensions which make such deduction not possible.

Is that the reason? If yes, Is there any possible way to deduce Newton's law using dimensional analysis (a very good trick for example)?

Also, When does dimensional analysis fail to give us a correct relation? what are its limitations?

Added: Here is my trial on deducing newton law:

First of all, $F$ is propotional to $M_1 , M_2 , r$.

So, $F=K M_1^a M_2^b r^c$ where $K$ is a constant and $a,b,c$ are numbers.

Now, $[F]=[M^1L^1T^{-2}]$ and $R.H.S = [M]^a[M]^b[L]^c = [M^{a+b}L^cT^0]$

So, we have: $[M^{a+b}L^cT^0]=[M^1L^1T^{-2}]$. equating both sides we get, $a+b=1, c=1 , 0=-2$, a contradiction. What is the problem?

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    $\begingroup$ I don't think you can really deduce deduce any fundamental laws with dimensional analysis. In the end you'd be defining at least one of the quantities. $\endgroup$
    – DanielSank
    Oct 25, 2014 at 4:56
  • $\begingroup$ @DanielSank, What do you mean by defining one of the quantities? $\endgroup$
    – FNH
    Oct 25, 2014 at 4:58
  • $\begingroup$ Newton's law of gravitation should never end up with incorrect units, you should always end up with newtons if you do the conversion properly. Did you make sure to convert the km to m in "G?" $\endgroup$
    – Gödel
    Oct 25, 2014 at 5:05
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    $\begingroup$ In addition, his law isn't quite that fundamental, because it is time independent (which says that if the sun "exploded," we would feel the effects immediately, even though it would actually take 8 minutes for us to know, due to the axioms of relativity). $\endgroup$
    – Gödel
    Oct 25, 2014 at 5:07
  • $\begingroup$ I'm not really certain of this because this is subtle, but I don't think you can derive something like $F=ma$ without essentially defining $F$ to be the product of "the amount of stuff" ($m$) and the second derivative of the position ($a$). Furthermore, in a sense, the physical content of $F=ma$ is that for a given amount of "push" (like a rocket or something) the product of the amount of stuff and the acceleration is the same for all objects. That statement can't be deduced from dimensional analysis. $\endgroup$
    – DanielSank
    Oct 25, 2014 at 5:11

3 Answers 3


A major limitation of dimensional analysis is that you must know some of the physics behind the concept that you are attempting to analyze, and since you have chosen to make the proportionality constant dimensionless, it has skewed your results.

However, we can easily deduce the dimensional properties of this constant by rearranging the equation you were attempting to analyze :

$F= \dfrac{K * m_1*m_2}{r^{2}}$

$K= \dfrac{r^2*F}{ m_1*m_2}=\dfrac{m*v*r^2}{ m_1*m_2*t}$

In dimensional form:$\implies \dfrac{[M][L^1][T^{-1}L^{-2}]}{[T][M][M]}={[M^{-1}][L^3][T}^{-2}]$, which is exactly what we would expect for the dimensions of the gravitational proportionality constant. However, to get to this point, you still have to know that $F=m*a$.

  • $\begingroup$ Here is a great Stanford document that exploits both the limits and benefits of dimensional analysis: web.stanford.edu/~cantwell/AA200_Course_Material/… $\endgroup$
    – Gödel
    Oct 25, 2014 at 6:48
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    $\begingroup$ Unfortunately, it doesn't include your exact question, but it does analyze Kepler's third law, which is very similar, and I think it would help you quite a bit. $\endgroup$
    – Gödel
    Oct 25, 2014 at 6:49
  • $\begingroup$ The link seems no longer work... would you mind adding a bit limits of dimensional analysis to your answer? $\endgroup$
    – Shing
    Nov 8, 2016 at 4:04

Also if you have noticed 'c' is coming out to be 1, while it should be -2 . This clearly indicates that constant also has some dimensions. Hence in this case you cannot use dimension analysis to find out the formula. Only way out is the way SIR newton did it.


first we find the dimentions of gravitational constant by gravitational formula. because G is dimentional constant. Then by putting its value in formula of force of gravitation. then formula is balanced perfectly.

  • $\begingroup$ Well, that was th problem - finding the right dimensions... $\endgroup$
    – Jon Custer
    May 31, 2016 at 17:22

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