Question

Using dimensional analysis, construct a constant, with units of length only, out of all three of the following fundamental constants of Nature: ℏ, ????, and c. Here, ℏ is Planck's constant, which has dimensions of [????][????]2[T]−1, ???? is Newton's gravitational constant, which has dimensions of [????]−1[????]3[T]−2, and c is the speed of light, with dimensions [????][T]−1.

Answer 1

The quantity with units of length only is $\sqrt{\frac{hG}{c^3}}$

Explanation:

We have to combine the following constants:

- $h$, Planck constant, with units $[m^2][kg][s^{-1}]$

- G, the Newton's gravitational constant, with units $[m^3][kg^{-1}][s^{-2}]$

- $c$, the speed of light, with units $[m][s^{-1}]$

The combination of these constant should have units of length only, so with meters (m).

First, we notice that $h$ has [kg] in its units, while G has $[kg^{-1}]$ in its units, so in order to make the [kg] disappear, we have to multiply them and they should have same power, so:

$hG = [m^{2+3}][kg^{1-1}][s^{-1-2}]=[m^5][s^{-3}]$

Now we have to make the seconds, [s], disappear. We do that by dividing the new quantity by $c^3$, so that the new units are:

$\frac{hG}{c^3}=\frac{[m^5][s^{-3}]}{([m][s^{-1}])^3}=\frac{[m^5][s^{-3}]}{[m^3][s^{-3}]}=[m^2]$

We are almost done: now the quantity has units of an area, squared meters. Therefore, in order to make it have it units of length, we just take its square root:

$\sqrt{\frac{hG}{c^3}}=\sqrt{[m^2]}=[m]$

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