I was reading about inertial mass and gravitational mass equivalence from this post Answer to https://www.quora.com/Why-is-gravitational-mass-equal-to-inertial-mass/answer/Leo-C-Stein?ch=15&oid=45062&share=c7307f40&srid=iaGyH&target_type=answer

Here, According to equivalence principle inertial mass(Mi) and gravitational charge(Qg) have the same value.

Gravitational force= (Qg)*g = (Mi)*a


from here they have taken Qg/Mi=1, but since (g) contains gravitational constant G then why can't we write Qg/Mi =G or any other constant, so that we can say the value of G =6.67•10^-11 is actually C.G' where G' is the real gravitational constant and C the ratio of inertial and gravitational mass


1 Answer 1


Qg/Mi=G, with $G$ the gravitational constant with the value $G=6.67\cdot 10^{-11} \frac{m^3}{kg}s^2$ is not possible, because the ratio Qg/Mi is dimensionless, whereas $G$ has the dimension as indicated ($\frac{m^3}{kg}s^2$).

Qg/Mi is dimensionless because Qg as well as Mi are masses.

One could imagine that one would use a dimensionless constant C=Qg/Mi instead, but nevertheless it would not change the main observation that all bodies fall with the same acceleration (on earth) which would be in that case:

$a =Qg\cdot g/Mi = C\cdot g$

As $C$ is a constant, $a$ would always have the same value: $Cg$.

Anyway, $a$ is the acceleration experienced by any fallen body observed on earth, therefore there is no reason to distinguish it from $g$ which has exactly this as definition, i.e. $a=g$ (justified by observation -- or better make the experiment yourself). In other words $C=1$.

For charges there is a different observation though, so it would not apply.

  • $\begingroup$ Thank you very much for your answer, I don't know how the value for G is extracted but can the actual value of the G be different? Like G =6.67•10^-11/C then the value of 'a' will not be equal to g but to C•g itself? I agree that still it will give acceleration of different masses as same, but can the ratio of inertial and gravitational mass be a constant that is being multiplied with the real gravitational constant giving us the apparent value of 6.67•10^-11? $\endgroup$
    – veke
    Apr 3, 2022 at 21:49
  • $\begingroup$ @veke I really wonder why you'd like $Qg\neq Mi$. One can define a $g'=\frac{GM}{Cr^2}$, then can set $Qg/Mi=C$. But what for ? So that school kids would first learn that due to Newton's gravity Qg is different from Mi and then on a later stage they will have to learn according to general relativity that it is actually not true because conceptually (and subsequently in terms of value) they are completely identical. And wonder why in special relativity physicists set speed of light $c=1$. Getting rid of useless constants is a simplification. $\endgroup$ Apr 4, 2022 at 8:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.