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Problem 24 Hard Difficulty

(a) Calculate the angular momentum of the Moon due to its orbital motion about Earth. In your calculation use $3.84 \times$ $10^{8} \mathrm{m}$ as the average Earth-Moon distance and $2.36 \times 10^{6} \mathrm{s}$ as the period of the Moon in its orbit. (b) If the angular
momentum of the Moon obeys Bohr's quantization rule $(L=n \hbar),$ determine the value of the quantum number $n .$ (c) By what fraction would the Earth-Moon radius have to be increased to increase the quantum number by 1?

Answer

a) $2.89 \times 10^{34} \mathrm{kg} . \mathrm{m^2} / \mathrm{s}$
b) $2.75 \times 10^{68}$
c) $7.30 \times 10^{-69}$

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Top Physics 103 Educators
Andy C.

University of Michigan - Ann Arbor

LB
Liev B.

Numerade Educator

Farnaz M.

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Zachary M.

Hope College

Video Transcript

question A. We have to calculate the angular momentum off the moon. L given that the radius of the orbit of the moon is approximately 3.84 times 10 to the eight meters and that the period of rotation of the moon is 2.36 times 10 to the sixth seconds. I'm just right. Seeks a little better here. Okay, well, we know that the angular momentum is given by M V. Times are where m is the mass of the moon boot and V's the speed. So we're gonna need to look up the mass of the moon somewhere. I'm going to just write it here. It's seven point 35 times 10 to the 22 kilograms. Okay? And remember that the speed for the speed V for a circular motion is given by two pi r Over the period of rotation T, this is M actually two pi and R squared over tea. Okay, So have two pi times and which is seven point thirties five times sent to the 22 times the radius, which is 3.80 four times into the eight. All this is in the international system of units. So I'm just leaving it implicit. Okay. But ah, the masses and kilograms And the radius is in meters divided by the period, which is 2.36 times 10 to the sixth seconds. So this here is equal to now it will stoop 2.89 times 10 to the 34th kilograms Meters squared for second. Okay, this is the angular momentum of the moon in question be we have to assume that the angular momentum is going ties social, that El is able to end times a bar. And we have to find what is the in what Quintal number. What is the consul number? End for the moon in this case. Okay, we have a right accommodate. Complicated l. So all I have to do is to isolate and in the formula, so have n equals. Fish bar over l H bar is h over two pi. So this is Asia, where two pi l age is 61 63 Thompson to the minus 34. Your second divided by two pi times. L l is 2.89 time. Stand to the 34th kilograms meters square her second. So and is equal to 2.74 times 10 to the 68. Ah, just Well, no units, because this is just a number. So this is a control number and and in C, we have to We have to find what is the fractional difference in the radius? That is the difference in the rate is divided by the original radius if in the quantum number associated with the the angular momentum increases by one. So initially, the who suppose that the angry momentum of the moon is at the end quantum number and he jumps to the and plus one. And we have to calculate the fractional difference in the radius of the orbit. So in order to do that well, the first thing I'm gonna notice is that l is able to m v are as a right before, but also that the gravitational force requires that the force F G gravitational force is equal to g times the mass of the the Earth which I'm writing as this capital. I am here times The mass of the moon, divided by the radius square is equal to since the moon is moving. In a circular circular motion. This is just gonna be this centripetal force. So the mass of the moon times three square over our and my goal here is to isolate V. Okay, so I'm gonna actually be cancel the ends one of the ours. So you have a V is equal to the square root of G capital him over our and I'm gonna take this formula for Vienne substituted back in the former for the under momentum. So l is equal to m times V, which is the square gm over r times are So this is m times G capital M over our This is the anger more mental. And we're assuming that this is equal to n times H bar on. Now I'm gonna isolate Ah, and I'm gonna index the are by and then here. Okay, this is the radius of the orbit when the when the the angular momentum has the total number. N I'm just putting a little index here and I'm gonna isolate the radius are in so that we have and squared h bar square over AM Square. I'm sorry or m spared gm. Okay. This is an important formula whom that I highlighted when we want to calculate is though they are, which is our n plus one minus one. That's the difference off radius of the radius When we when we change the quintal number by only one And we want to divided by R N. Okay, so let's talk late. Don't our don't our my definition that's our endless one minus r and r n plus one from our formula is and plus one square H Maher squared, divided by m squared G m and are in is just and squared each bar squared, divided by m square G m So I can write this as H bar square over in squared GM, I can pull out the end square and right, whatever comes next as one bless one over n squared minus What? Okay, No notice. I'm gonna, uh, right and rocks approximation down here. Blue, That one plus one over in, um, square is approximately one bus chu over in if one over and is much smaller than what in our case, remember in question be we found that end. It's very large. So one over and is much, much smarter than one. So we're ah, we are. We have Ah, reason to to consider that one over And is my swatter than one. So it's this approximation Here is a good approximation. Okay, so there are is able to h bar, square and square over M square GM times one plus two over and minus one. So this is to h bar, square and square, actually without the square divided by M square gm. Okay, this is Delta R, and I can divide it. Don't our by our end. So this is just too h bar squared in divided by AM square G I m. Times one over r n, which is m squared GM divided by and square H bar square. So age bar cancels out. So there's the G V m and the M squared, and we're left with two over. Okay, well, in our case and is equal to 2.22 point 74. So I'm staying to the 68 so that our over our end is equal to 7.3 times 10 the minus 69. Okay, eso the radius increases by a very small amount when the quantum number only increases by one

Universidade de Sao Paulo
Top Physics 103 Educators
Andy C.

University of Michigan - Ann Arbor

LB
Liev B.

Numerade Educator

Farnaz M.

Other Schools

Zachary M.

Hope College