Modeling the Earth-Sun system in Quantum Mechanics

[eep]

Hi,
We recently solved the hydrogen atom and one of our homework problems asks us to replace the coloumb potential with a gravitational potential. I have the potential as being

$$V(r) = -\frac{GMm}{r}$$

Where M is the mass of the sun, m the mass of the earth. I have calculated the Bohr radius to be:

$$a_g = \frac{\hbar^2}{GMm^2}$$

However my calculator chokes on this value. I managed to get an approximation for it by first multiplying and dividing all decimal values, then doing the needed multiplication, addition and subtraction for the powers of 10. Needless to say I get something ridiculously small.

I have an expression for the energy of state n as

$$E_n = \frac{-G^2M^2m^3}{2n^2\hbar^2}$$

I am then supposed to find out what Earth's quantum number is by assuming a classical, circular orbit, so I do:

$$E_n = \frac{-GMm}{2r}$$

where r is the distance from the Earth to the sun. Substitute in the above expression and calculate n. This gives me that

$$n_e = \sqrt{\frac{r}{a_g}}$$

Which is some insanely large number (~$2.5 * 10^{71}$). I'm then supposed to calculate how much energy would be released if the Earth were to jump down to the next quantum number, and what the wavelength of the photon (or graviton, as the book suggests) would be in light years. I'm having some serious problems computing this quality as my TI-83 chokes, and when I make some approximations I get something like $\deltaE = 7.68 * 10^-6$ which seems much much too small.

First, are my calculations correct? Second, how can I compute these numbers?

 

[marlon]

Err, do you know at what distance scale QM becomes relevant ?

What's the criterium we use to decide at what distance scale quantummechanical effects will arise ?

Think about that for a while, please, it'll save you a lot of trouble.

marlon

 

[quantumdude]

my TI-83 chokes

Try the Microsoft calculator on your computer. It can handle very large numbers.

Err, do you know at what distance scale QM becomes relevant ?

Something tells me that this exercise is supposed to give him an idea of the answer to that question.

 

[nrqed]

Hi,
We recently solved the hydrogen atom and one of our homework problems asks us to replace the coloumb potential with a gravitational potential. I have the potential as being

$$V(r) = -\frac{GMm}{r}$$

Where M is the mass of the sun, m the mass of the earth. I have calculated the Bohr radius to be:

$$a_g = \frac{\hbar^2}{GMm^2}$$

However my calculator chokes on this value. I managed to get an approximation for it by first multiplying and dividing all decimal values, then doing the needed multiplication, addition and subtraction for the powers of 10. Needless to say I get something ridiculously small.

I have an expression for the energy of state n as

$$E_n = \frac{-G^2M^2m^3}{2n^2\hbar^2}$$

I am then supposed to find out what Earth's quantum number is by assuming a classical, circular orbit, so I do:

$$E_n = \frac{-GMm}{2r}$$

where r is the distance from the Earth to the sun. Substitute in the above expression and calculate n. This gives me that

$$n_e = \sqrt{\frac{r}{a_g}}$$

Which is some insanely large number (~$2.5 * 10^{71}$). I'm then supposed to calculate how much energy would be released if the Earth were to jump down to the next quantum number, and what the wavelength of the photon (or graviton, as the book suggests) would be in light years. I'm having some serious problems computing this quality as my TI-83 chokes, and when I make some approximations I get something like $\deltaE = 7.68 * 10^-6$ which seems much much too small.

First, are my calculations correct? Second, how can I compute these numbers?

I agree with all your expressions.
And the fact that you get "ridiculously" large numbers is to be expected, it is part of the goal of the exercise, to make you realize that it would be ridiculous to use quantum mechanics on the scale of planets or even moons, space shuttles, etc.


For the problem with the choking part, I haev a suggestion. You iwl have to calculate $-{C \over n^2} + {C \over (n-1)^2 }$ where C is the bunch of constants appearing in the energy equation. The problem is due to the n being so large. So what you should do is to use a Taylor expansion.

Write

$${1 \over n^2-2n+1} \approx {1 \over n^2 -2n} = {1 \over n^2(1-2/n)} = {1\over n^2} {1 \over (1-2/n)} \approx {1\over n^2} (1+ {2 \over n}) = {1\over n^2} + {2 \over n^3}$$

Now, when you take a difference of energy, the 1/n^2 pieces will cancel out leaving only 2C/n^3 and that's your difference of energy.

Hope this helps

Patrick