Bohr-Sommerfeld model question "Old" Quantum Theory

[RJLiberator]

Homework Statement

In analogy to the Bohr theory of the hydrogen atom, develop a quantum theory of Earth satellites, obtaining expressions for the orbit radius, r_n, and the energy, E_n in terms of the quantum number n and other relevant parameters. Use the "Old" Quantum Theory. A satellite of mass 1000kg is in a circular orbit of radius 7000 km. To what value of n does this correspond? What is the satellite energy? Determine the differences in the radius, Δr_n, and in the energy ΔE_n, for successive orbits at this radius. (Hint: Differentiate with respect to n and set Δn = 1. Why is this legitimate since n is not a continuous variable?)

Homework Equations

$$V_{Sat} = \sqrt{\frac{GM_e}{r}}$$

L=angular momentum of satelite:
$$L_{sat} = rm_{sat}v_{sat}$$

T = period
$$T = 2*\pi\sqrt{\frac{r^3}{GM_e}}$$

Bohr–Sommerfeld model,

$$\int_0^T p_rdq_r\ =nh$$
h = Planck's constant = 6.626*10^(-34) J*s

The Attempt at a Solution

So, based on the above equations and m = 1000kg, r = 7000km, I can find V_s = 7548.6 m/s and the period, T = 5826.6 s.

If I have angular momentum correctly, then that is equal to 5.284*10^(13) kg*m^2/s .

I can easily set up the bounds of integration from 0 to the period, T.
But I've never used the Bohr-Sommerfeld model. I don't know what is meant by the wiki's description of "where p_r is the radial momentum canonically conjugate to the coordinate q which is the radial position and T is one full orbital period."

Once I am able to calculate the integral, it will be easy to solve for n, and thus should be easy to find the Energy.

I believe that we are allowed to treat n like a continuous variable due to how minute it will be compared to the other numbers. n will likely be so small, that it will act like a continuous variable, thus we can differentiate it.

Any help on how to get started with the Bohr-Sommerfeld model in relation to this problem? What is "dq" ?

 

[Charles Link]

A simpler form of the Bohr model might be helpful: The results are the same: $\\$ 1) $L=mvr=n \frac{h}{2 \pi}$ (Angular momentum is an integer number times $h/(2 \pi)$). $\\$ 2) $\frac{mv^2}{r}=\frac{GMm}{r^2}$ (centripetal force=gravitational force) $\\$ 3) $E_n=\frac{1}{2}mv^2-\frac{GMm}{r}$ (Energy is the sum of the kinetic energy plus the potential energy). $\\$ These are 3 equations, and 3 unknowns ($v$, $r$, and $E_n$). $\\$ You can solve for the unknowns and then compute the required quantities. $\\$ Note: The integral $\int\limits_{0}^{T} p \, dq=mv(2 \pi r)=n h$, which is the same as $mvr=n \frac{h}{2 \pi}$. And you also have $v=2 \pi r/T$, but I don't think you need to solve for $T$.

 

[RJLiberator]

Hi Charles, thank you kindly for helping me.

I have some follow up questions.

1. Are you sure r is an unknown? The question states that radius = 7000km. I am finding "n", E_n, and v as my unknowns.

2. So, the bohr-sommerfeld statement (that integral) comes down to be simply nh = mv(2*pi*r)?

3. How do you type latex in the middle of the sentence on PhysicsForums like you did? :)

Thank you.

 

[Charles Link]

1) $r$ will be a function of $n$ and the other constants. They then give you $r$ so that will tell you what $n$ is. In these 3 equations, you might not need $v$, but you need to solve for $r$ and $E_n$ (I could have called it $r_n$, and the $v$ I could have called $v_n$). $\\$ 2) The answer is yes. $\\$ 3) just type ## on both sides of the Latex expression.

 

[RJLiberator]

Ah, I see. That makes sense, so using the first equation:

$$mvr = n\frac{h}{2*\pi}$$
We get a function r of n, namely $r(n)=\frac{n}{mv}\frac{h}{2*\pi}$.

Couldn't I use r = 7,000,000 m to find what v is in the second equation? $v=\sqrt{\frac{GM}{r}}$ and we have r, G (grav. constant), M (mass of earth). This nets us a v of 7548.6 m/s.

With that v and using an r value again of 7,000,000 m, we can find what n is in the first equation which comes out to be a huge 5.0106*10^(47).

 

[Charles Link]

Ah, I see. That makes sense, so using the first equation:

$$mvr = n\frac{h}{2*\pi}$$
We get a function r of n, namely $r(n)=\frac{n}{mv}\frac{h}{2*\pi}$.

Couldn't I use r = 7,000,000 m to find what v is in the second equation? $v=\sqrt{\frac{GM}{r}}$ and we have r, G (grav. constant), M (mass of earth). This nets us a v of 7548.6 m/s.

With that v and using an r value again of 7,000,000 m, we can find what n is in the first equation which comes out to be a huge 5.0106*10^(47).

The second equation gives you $v=\sqrt{\frac{GM}{r}}$, so for this case, yes, you can take that shortcut to the answer. $\\$ Meanwhile the huge $n$ does not surprise me. I might try to compute it, and see if I concur.

 

[RJLiberator]

Yes, my thoughts when first reading this problem was that n would be very extreme and the changes from the change in n should result in very minute differences. This is what explains why we can differentiate r(n) without n being a continuous variable.

 

[Charles Link]

Yes, my thoughts when first reading this problem was that n would be very extreme and the changes from the change in n should result in very minute differences. This is what explains why we can differentiate r(n) without n being a continuous variable.

The arithmetic checked out=I agree with your answers. $\\$ Meanwhile $r_n$ can be written as a Taylor series: $r(n)=r(n_o)+r'(n_o)(n-n_o)+\frac{1}{2}r''(n_o)(n-n_o)^2+...$. The derivatives get successively smaller. Your explanation is ok too, but I think justifying it with the Taylor series is more mathematical. (Even though $n$ is integer, you can assume it to be a continuous variable in the formula for $r(n)$, and when you put in integers $n$ and $n_o$, the Taylor Series would get you the exact answer for $r(n)$ if you included all of the terms in the series. In this case you find $\Delta r=r(n)-r(n_o)$ with just the first derivative term).

 

[RJLiberator]

Charles, thank you for helping me with my problem and even cross-checking your solution with mine.