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# As the Earth moves around the Sun, its orbits are quantized. (a) Follow the steps of Bohr’s analysis of the hydrogen atom to show that the allowed radii of the Earth’s orbit are given by $$r_{n}=\frac{n^{2} \hbar^{2}}{G M_{S} M_{E}^{2}}$$where $n$ is an integer quantum number, $M_{S}$ is the mass of the Sun, and $M_{E}$ is the mass of the Earth. (b) Calculate the numerical value of $n$ for the Sun-Earth system. (c) Find the distance between the orbit for quantum number $n$ and the next orbit out from the Sun corresponding to the quantum number $n+1 .$ (d) Discuss the significance of your results from parts (b) and (c).

## a) $r_{n}=\frac{n^{2} \hbar^{2}}{G M_{S} M_{E}^{2}}$b) $2.54 \times 10^{74}$c) $1.18 \times 10^{-6.3} \mathrm{m}$d) answer not available

Atomic Physics

Nuclear Physics

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##### Top Physics 103 Educators  ##### Christina K.

Rutgers, The State University of New Jersey

LB ##### Jared E.

University of Winnipeg

### Video Transcript

question A. We have to con ties the earth's orbit around the sun and show that the radius of the orbit is equal to and squared H m r squared, divided by g times the mass of the earth Square times the mass of the sun. Ah, so in order to do that, first of all, I'm gonna remind you that the gravitational force between the earth and the sun, even by the gravitational constant g times the mass of the earth times the mass of the sun divided by the radius of the orbit square. And if the earth is moving in a circular motion than this force is equal to the mass of the Earth Times dispute square divided by r And what I'm gonna do here is to isolate TV so one of the art cancels out. So there's an E. So have a V is able to the square root of G. M s divided by r. Okay. Ah, that's the speed of the earth throughout the trajectory as a function of our and what we can do now is to Kwan ties the angular momentum No of the earth, just like boarded for the hydrogen atom. So the angular momentum can be redness, the mass of the earth times the speed V times the radius of the orbit are. And if we want to Colin ties it. This is able to end h book, Okay? And now I'm gonna substitute V the V that we found before so V is equal to G and s over our Sam's r equals and H bar. So here we have the mass of the earth times you times the mass of the sun Times are it goes in each bar. You know I'm gonna divide both sides by EMI So you have g a mess are it was an age over and he and so have G m s r is able to end square ish bar square or an e square. And finally we have it. Our is equal to end squared H squared, divided by G m E Squared m s. Okay, this is the answer to the first question. Now we move on suppression be where we have to apply this one of it. We just found and find the principal quantum number and for the rial orbit of the earth around the sun. So we want to find in and we're gonna need the information is that the mass of the earth is given by 5.972 times 10 to the 24 kilograms. The mass of the sun is given by 1.99 time still to the 30 kilograms, and the radius of the orbit is given by 1.496 time. Stand to the 11th meters, okay? And I'm gonna take end in the equation that we found in question. They that's highlighted in, uh, rad here and I'm I'm gonna isolate it. So you have that in is equal to the square root of G times in E square a mess are divided by H bar square and this is just an e divided by h bar times. The square root of G. M s times are and I can answer the numbers So an is equal to the mass of the earth. That's nine points out of 465.9 72 times Center. The 24 kilograms divided by H Bar and they spar is one point 055 times 10 to the minus 34 usual second times G, which is 6.16 seven times sent to them in his 11th Newtons. Meters squared kilograms square times the mass of the sun that's 1.99 times tend to the 30th kilograms times the radius of the orbit. And that's one point or 96 times center, the 11th meters. So in the end, and is equal to 2.5 times 10 to the 74. Okay, this is the principal quantum number or the orbit off the the earth around the sun. Questions See, we have to calculate difference Don't are between the radius of the orbit for n plus one or the principal quantum number and plus plus one and the Raiders of the orbit for de Quantum number in. So let's do it. Dr. R is equal to r N plus one. So I'm just gonna substitute and plus one in the our formula. So have endless one square H bar square over G and E square and mess minus and square h bar, square Z every square a mess. So I'm just gonna pull out the end square. So having squared each more square divided by G and E square, him s. And we have one plus one over n squared minus one. And notice that our end, the end that we found in question B is very large. Okay, so we can make this approximation that I'm gonna make him make it here in blue. We have that one every one plus X to the Actually it one plus x two. The M is approximately one plus and acts if X is much smaller than what? For that reason, one plus one over in square is approximately one plus two over end. Because we have the one over N is more smaller than what. Okay, so this means that don't are is able to end square H bars. Where ever g m e square in mass. Times one plus two over end minus one. So this is just to in h bars. Where over g m e square in s okay. And I can simply substitute the numbers that we have. So, uh, those are is two times and it is 2.5 times 10 to the 74th. As we obtained in question be time daesh bar swear That's one point 0 55 temps. 10 to the minus 34 square divided by G, which is 6.67 times 10 to the minus minus 11th. Everything here is in the international system of units, so I'm not gonna bother to put the units here. But just suppose that everything is in the international system. Times the mass of the earth Square toe this five Wait 972 times 10 Lemonis, 24th kilograms squared times the mass of the sun, which is 1.99 time soon. So I started here. The mess of the earth. It's plus 24. Of course. Ah, to the 30th kilograms. Okay, Uh, so what we have is that there are is equal to 1.18 time Step to them on a 63 meters. Okay, this is the change in radius. If we just increase the principal mental number by one. And in question D, we have to, uh, discuss how this number is significant. We have to to make a little analysis here. So well, notice that Delta Har is much smaller than the radius. The nucleus. Ah, nuclear radius. It's much smaller than the proton radius. Its smaller looked through the ducts are is smaller than any length scale we know. Okay, So this means that, uh, quantum mechanics shouldn't but least Or at least it won't help if quantum mechanics is applied to the Earth's orbit around the sun. It's a capitalist, characteristically classic, uh, orbit. So there is no use in applying quantum mechanics to to the orbit off the earth around the sun because we are not able to see quantum effects. That is the answer to our problem. Universidade de Sao Paulo

#### Topics

Atomic Physics

Nuclear Physics

##### Top Physics 103 Educators  ##### Christina K.

Rutgers, The State University of New Jersey

LB ##### Jared E.

University of Winnipeg