Bohr radius of Earth-Sun system

[songoku]

Homework Statement

Let the Sun and Earth are put as part Hydrogen atom. Find the Bohr radius in this case

Relevant Equations

Bohr radius = $\frac{n^2 h^2}{4 \pi^{2}mkq^2}$

When I looked up for Bohr radius, the formula has $q$ in it, which is charge of the object. For this question, the electron and proton are replaced by sun and Earth so it means that I have to know the charge of Earth and sun?

Thanks

 

[PeroK]

Homework Statement:: Let the Sun and Earth are put as part Hydrogen atom. Find the Bohr radius in this case
Relevant Equations:: Bohr radius = $\frac{n^2 h^2}{4 \pi^{2}mkq^2}$

When I looked up for Bohr radius, the formula has $q$ in it, which is charge of the object. For this question, the electron and proton are replaced by sun and Earth so it means that I have to know the charge of Earth and sun?

Thanks

The Earth is kept in orbit by a gravitational force, not by an electromagnetic force!

The Earth and Sun are approximately electrostatically neutral.

 

[Steve4Physics]

Homework Statement:: Let the Sun and Earth are put as part Hydrogen atom. Find the Bohr radius in this case.

The question could be worded better. It looks like you are being asked to find the gravitational equivalent of the Bohr radius for the Earth orbitting the sun.

Make sure you can follow the (quite simple) derivation of the usual Bohr radius formula for an electron in a hydrogen atom. Look it up if needed.

Then repeat the derivation, but - as already hinted by @PeroK - using the gravitational (rather than electrostatic) force between the sun and earth.

Edit - typo's corrected.

 

[songoku]

The question could be worded better. It looks like you are being asked to find the gravitational equivalent of the Bohr radius for the Earth orbitting the sun.

Make sure you can follow the (quite simple) derivation of the usual Bohr radius formula for an electron in a hydrogen atom. Look it up if needed.

Then repeat the derivation, but - as already hinted by @PeroK - using the gravitational (rather than electrostatic) force between the sun and earth.

Edit - typo's corrected.

$$G\frac{Mm}{r^2}=m\frac{v^2}{r}$$

Is this what you mean?

Thanks

 

[hutchphd]

Yes then write this in terms of angular momentum and then assume it is "quantized". Solve for (smallest) r.

 

[songoku]

Using quantization of angular momentum:
$$mvr=\frac{nh}{2\pi}$$
$$v=\frac{nh}{2\pi mr}$$

Substitute to equation of force:
$$G\frac{Mm}{r^2}=m\frac{v^2}{r}$$
$$G\frac{M}{r}=\frac{n^2h^2}{4\pi^{2}m^2r^2}$$

To find the smallest radius, I just need to use $n=1$ and solve for $r$

Thank you very much PeroK, Steve4Physics, hutchphd