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Problem 78 Easy Difficulty

For a planet in our solar system, assume that the axis of orbit is at the sun and is circular. Then the angular momentum about that axis due to the planet's orbital motion is $L = M$$\upsilon$$R$.
(a) Derive an expression for $L$ in terms of the planet's mass $M$, orbital radius $R$, and period $T$ of the orbit. (b) Using Appendix F, calculate the magnitude of the orbital angular momentum for each
of the eight major planets. (Assume a circular orbit.) Add these values to obtain the total angular momentum of the major planets due to their orbital motion. (All the major planets orbit in the same
direction in close to the same plane, so adding the magnitudes to get the total is a reasonable approximation.) (c) The rotational period of the sun is 24.6 days. Using Appendix F, calculate the
angular momentum the sun has due to the rotation about its axis. (Assume that the sun is a uniform sphere.) (d) How does the rotational angular momentum of the sun compare with the total orbital
angular momentum of the planets? How does the mass of the sun compare with the total mass of the planets? The fact that the sun has most of the mass of the solar system but only a small fraction of its total angular momentum must be accounted for in models of how the solar system formed. (e) The sun has a density that decreases with distance from its center. Does this mean that your calculation
in part (c) overestimates or underestimates the rotational angular momentum of the sun? Or doesn't the nonuniform density have any effect?

Answer

a) $$2 \pi m R^{2} / T$$
b) $$3.13 \times 10^{43} \mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s}$$
c) $$1.14 \times 10^{42} \mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s}$$
d) $$746$$

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Video Transcript

This is a question about angular momentum in the solar system. In this question, we should assume that all planets orbit the sun in perfectly circular orbits then, because off that the angular momentum off each of those planets in orbit is given by this expression. So the angular momentum is equals to the mass of the planet times the velocity off that planet times the radios off the orbit. In the first item off this question, we have to derive an expression for the orbital angular momentum as a function off the mass, the radios off the orbit and the period off that orbit. We begin by rewriting the regional expression. So the angular momentum is given by the mass times, the velocity times, the radios. We want to eliminate the velocity. For that. We can use this expression, which tells us that the velocity is equal to the angular frequency times the radius off that circular movement, the reform we can say the following the angular momentum is equal to the mass times. The angular frequency times the radius squared squared because we already have a factor off are here, and there is an extra factor off our when it's substitute V by Omega Times are now remember about the relation between the angular frequency on the act off frequency in units off one over time that relation is the following. The angular frequency is equal to two pi times the act off frequency in units off one over time. Using that relation, we can see the following. The angular momentum is given by the mass times two times spike times the frequency times their radios off the orbit squared. Now there is a relation in between the frequency and the period. That relation is the following the frequency is the inverse off the period. Then, using this equation, we get that the angular momentum is equals to two times spy times, the mass times the radius squared, divided by the period. And this is the expression that the first item off this problem was asking us about in the second item off this question, we have to apply this expression toe all the eight planets off the solar system using the data there is available in appendix F, making it easy for both of us. I had already separated all that data right here, so let me put it here so that we can see. Now all you have to do is feel this column using this data and the equation for L. As an example, let me show you how to perform the calculation for mercury. So it goes as follows the angular momentum l is equal to two times spy times the mass off mercury, which is already in kilograms. So we have three 0.30 time stand to 23 times the radius off the orbit squared the radios off the orbit is already in meters, so we just have to plug it in here than 2.44 times 10 to the sixth squared, divided by the period off that orbit, the period is in days. It's not in seconds, so we have to convert when plugging in. The conversion is fairly easy to perform, so we have 88 days, but each day has 24 hours, but each hour has 60 minutes on. Each minute has 60 seconds, then it's already converted. By serving that calculation, we get a value for the angular momentum that is approximately 9.14 times. Stand to 38 kilograms, meters squared per second Now all we have to do is repeat this process to feel this column. By doing that, we get the following, and this is the answer to the second item off this question. Just a small advice. Here we have several orbital periods, which are in ears and not in days, so the conversion process outdoors similar is not exactly the same. Converting from years. Two seconds goes as follows. One year has 365 days. Each day has 24 hours. Each hour has 60 minutes, and each minute has 60 seconds. The only difference, as you can see, is an extra factor or 365. As long as you will be careful with this conversion, you should have no problem in arriving at these results. In the third item off this question, we have to evaluate what angular momentum the sun has due to the rotation about it's axis again. Using the data from Appendix F in Appendix F, you can find the mass of the sun and the radios off the sun. On top of that, the problem gives us the rotational period off the sun to so desired, and you have to remember that the rotational, angular momentum is given by the product in between the moment off inertia and the angular frequency. And this is our starting point. So the rotational, angular momentum off the sun about its own axis is given by the moment off inertia off the sun times the angular frequency off rotation. Now, for the moment off inertia, we assume that the sun is a uniforms fear in that setting the moment off inertia off the sun will be given by this expression, which is just the moment off. Inertia off a uniforms fear it is true times the mass times the radio squared, divided by five Then here we have chill times the mass off the sun Let me call this M s times the radios off the sun squared divided by five Now we multiply by the angular frequency off rotation The angular frequency off rotation is nothing else than the interval Delta, Tita that the sun rotates and the time it takes for the sun to rotate by Delta Tita Now we want a complete rotation. Then Delta Tita is equals to chew pie Radiance. So here we have two pi and How much time does the sun takes to make a complete rotation? Well, we call that interval of time exactly the period. Then we reached the conclusion that the angular frequency of rotation is equals to two pi divided by the period. Now all we have to do is plug in the values that we have and get our answer. So the orbital angular momentum is given by two times the mass off the sun. 1.99 times. Stand to the 30 times the radios off the sun squared 6.96 times stan to the eight squared, divided by five, multiply it by two pi and then divided by the period off rotation. That period is off 24.6 days, but we have to convert it into seconds in order to use in this equation. Then converting from these two seconds goes as follows. One day has 24 hours, one hour has 60 minutes and each minute has 60 seconds. Then we multiply the number off days by this conversion factor to have it in seconds. By solving that calculation, we get approximately 1.14 times stand to 42 kilograms meters squared per second, and this is the answer to the third item off this question before proceeding to the next item. How you have to organize my board a little bit In the third item off this question, we have to compare the total angular momentum off all planets. Together with the angular momentum off the sun, we begin by calculating the total angular momentum off all planets together. For that, all we have to do is add all these values by doing that to get following 9.14 times 10 2 38 plus ah, lot of values plus 2.50 times 10 to 42. By solving that calculation, you'll get the following total angular momentum. 3.13 times tend to 43 kg meters squared per second. Now we compare the angler momentum's by taking the ratio in between the angular momentum off the sun on the angular momentum off the planet's. These results in 1.14 times 10 to 42 divided by 3.13 times stand to 43 which is approximately 0.3 63. Therefore, the angular momentum off the sun is on. Lee 3.63% off the angular momentum off other planets together. On the other hand, if we add all the masses off other planets, we get the following. The total maths off all planets together is equals to 2.669 times 10 to 27 kg. But the mass off the sun is actually 1.99 times 10 to the 30. Therefore, the ratio in between the mass off the sun and the mass off the planets is given by 1.99 times 10 to the 30 divided by 2.669 times 10 to 27 which is approximately 746 that is. The sun has 746 times more mass than all the planets together, but at the same time has only 3.63% off the angular momentum off all planets together. Finally, the answer to this item are the ratios in between the angler momentum's on the masses. Before moving to the next item, I you have to organize my board again. Now we go to the final item off this question In this item, we have to tell if our approximation for the angular momentum off the sun is overestimated or underestimated, provided that the density off the sun increases as we get closer to the center, this one is not difficult to answer. The idea is the following. Suppose that this is the sun and there is nothing inside it. Now let me put a small point mass inside the sun. Suppose that this point mass is right here at a distance, say small are to the center. Then the moment off inertia off That small point particle will be given by the mass off that particle times the radios squared and that's it. Now suppose that we have another point particle that is closer to the center at a radius. Say our prime. The moment off inertia off that second point particle is M. Times are prime squared. Now compare them. As you can see, our is bigger than our prime. Therefore, M Times R squared is bigger than M. Times are prime squared. What I'm trying to say is that a particle that is further away from the access off rotation has a bigger moment off inertia under a form, a bigger angler momentum than a particle that it's closer to the axis of rotation. The Sun actually has more particles that are closer to the access off rotation than particles that are further away from the access off rotation. It means that our approximation off the sun as a uniforms fear is an overestimation because we consider that there were as many particles away from the center as there are particles close to the center. And this is not the case. There are more particles close to the center and the reform. There are more particle with less angular momentum than our approximation overestimates the angular momentum off the sun, and this is the answer to the final item off this question.

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