The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about $10^{-40}$. An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and p

Question

The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about $10^{-40}$. An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction. You will find the answer interesting.

Accepted Answer

> [!TLDR]
> Replacing the Coulomb force $ke^2/r^2$ with the gravitational force $Gm_em_p/r^2$ in the Bohr derivation gives a 'gravitational Bohr radius' of about $1.2	imes10^{29}\ 	ext{m}$ for $n=1$ -- enormously larger than an atom, or even the observable universe.

**Step 1 -- Redo the Bohr derivation with gravity as the central force.**
In the ordinary Bohr model, the centripetal force balance is
$$\frac{mv^2}{r} = \frac{ke^2}{r^2}, \qquad k = \frac{1}{4\pi\varepsilon_0}$$
and angular momentum quantisation gives $mvr = n\dfrac{h}{2\pi}$. Combining these (exactly as in the text's derivation of the Bohr radius) gives
$$r_n = \frac{n^2h^2}{4\pi^2 m k e^2}$$
Nothing about this derivation actually used the fact that the force was *electrical* -- only that it was a $1/r^2$ attractive force between the electron (mass $m_e$) and a much heavier fixed centre. So if the electron and proton were instead bound purely by **gravity**, we simply replace the Coulomb coupling $ke^2$ by the gravitational coupling $Gm_em_p$:
$$r_n^{	ext{grav}} = \frac{n^2h^2}{4\pi^2 m_e\left(Gm_em_pight)} = \frac{n^2h^2}{4\pi^2 Gm_e^2m_p}$$

**Step 2 -- Evaluate for $n = 1$.**
Using $h = 6.63	imes10^{-34}\ 	ext{J s}$, $G = 6.67	imes10^{-11}\ 	ext{N m}^2	ext{kg}^{-2}$, $m_e = 9.11	imes10^{-31}\ 	ext{kg}$, $m_p = 1.67	imes10^{-27}\ 	ext{kg}$:
$$r_1^{	ext{grav}} = \frac{h^2}{4\pi^2 Gm_e^2m_p} \approx 1.2	imes10^{29}\ 	ext{m}$$

**Step 3 -- Put the number in perspective.**
$1.2	imes10^{29}\ 	ext{m}$ is about a thousand times *larger* than the radius of the observable universe ($\sim4	imes10^{26}\ 	ext{m}$). This dramatically illustrates the exercise's opening fact -- gravity is weaker than the Coulomb attraction between an electron and proton by a factor of about $10^{-40}$ -- an atom held together by gravity alone, at the same quantum number, would be unimaginably larger than anything that actually exists.

> [!ANSWER]
> $\boxed{r_1^{	ext{grav}} \approx 1.2	imes10^{29}\ 	ext{m}}

Concept understanding — Bohr Model Quantization

Why does an electron not spiral into the nucleus?

The radical idea: allowed orbits only

What this means physically

What it predicts

The key insight for exams

Why "quantization"?

Why this formula?

Why Angular Momentum is Quantised in the Bohr Model

The core problem Bohr faced

The de Broglie wavelength argument (the cleanest derivation)

Why this fixes the energy levels

The deeper reason: it's not really about orbits

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