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Problem 77 Hard Difficulty

Most of our Solar System's mass is contained in the Sun, and the planets possess almost all of the Solar System's angular momentum. This observation plays a key role in theories attempting to explain the formation of our Solar System. Estimate the fraction of the Solar System's total angular momentum that is possessed by planets using a simplified model which includes only the large outer planets with the most angular momentum. The central Sun (mass $1.99 \times 10^{30} \mathrm{kg}$ ,
radius $6.96 \times 10^{8} \mathrm{m}$ ) spins about its axis once every 25 days and the planets Jupiter, Saturn, Uranus, and Neptune move in nearly circular orbits around the Sun with orbital data given in the Table below. Ignore each planet's spin about its own axis.

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Physics 101 Mechanics

Physics for Scientists and Engineers with Modern Physics

Chapter 11

Angular Momentum; General Rotation

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Video Transcript

so we will calculate the angular velocities omega equaling two pi over t the period and so we can calculate the angular momentum of the sun. Uh, this would be equally to the moment of inertia of the sun multiplied by the angular velocity of the sun. This is equaling 2/5 times the mass of the sun Modeling is a sphere times the radius of the sun squared and then multiplied by two pi times the period of the sun, and so we can actually solve this. And the angular momentum of the sun is equaling 2/5 times the mass of the sun, 1.99 times 10 to the 30th kilograms multiplied by 6.96 times 10 to the eighth meters quantity squared. We're going to multiply this by two pi divided by 25 days because it takes 25 days for the sun to rotate about about its own axis. And then we're gonna multiply this by one day for every 86,400 seconds, and we find that the angular momentum of the sun is equaling one point 1217 times 10 to the 30th. Rather tend to the 42nd. My apologies, kilograms meters for a second, and so we can then calculate the angular momentum of all of the planets. So we're going to be doing the exact same thing. However, we're gonna be treating the planets as a ah appoint mass orbiting around the sun. And then we're gonna do this for Jupiter, Saturn, Uranus and Neptune. So we can say that first, we can say that the, uh, uh, angular momentum of Jupiter would be equaling the mass of Jupiter with the same mass sub j radius of Jupiter Square. And now this is not the radius of Jupiter itself, but the radius of Jupiter's orbit. Because if retreating Jupiter as a point mass, we're not going to be taking the moment of inertia of Jupiter itself with other. Yeah, we're not taking a moment of Jupiter itself for taking the moment of inertia of Jupiter rotating about the sun. So we're gonna have to say the radius or not really the radius, but rather more appropriately, the distance from the center of Jupiter to the center of the sun squared and then multiplied by to pi over again following the same lines this would be the the ah period not of Jupiter's rotation about its own axis. But Jupiter's rotation. Jupiter's period are about the about the sun, so when it makes a full revolution around the sun, and so this would be the period of Jupiter's orbit. Uh, and so we can say that this is gonna be equaling 190 times 10 to the 25th kilograms multiplied by 778 times 10 to the ninth meters again, this is obviously much larger than the than the radius of Jupiter, because this is again the rate of the distance between the center of Jupiter and the center of the sun, and then it takes to pi over 11.9 years. So it takes 11.9 years for the juke for Jupiter to create one revolution around the sun. And then we're gonna multiply this by for every one year. There are 3.156 times, 10 to the seventh seconds in one year. And so the angular a momentum of Jupiter would be 1.9 240 times 10 to the 43rd kilograms meters per second and so keep check this number, keep track of this value, and then we're gonna do it for Saturn, Uranus and Neptune. So let's do the angular momentum of Saturn. This would be the massive Saturn Times. The radius of Saturn's orbit squared, multiplied by to pi times, rather to pi divided by the period of Saturn's orbit. And so this would be equaling. You could just plug in all of the values there, they're in the appendices or they're all tabulated online. And this would be equaling 7.806 times, 10 to the 42nd kilograms meters per second. Similarly, um, the angular momentum of Uranus equals the massive year ness times The distance between the center of Uranus and the sun squared multiple multiplied by two pi over the period of Uranus is over orbit, and this is equaling 1.695 times, 10 to the 42nd and kilograms meters per second. As you can see, this is a little bit less than Saturn, and that smoke from that's most likely because, um, the increased distance from the centre of Uranus to the sun is not great enough. In order to account for the much, much larger period of Uranus is orbit. That's why that's probably why this is, um, much less. And then the angular momentum of Neptune would be the massive Neptune times the distance between the center of Neptune and the center of the sun. Or we can say the radius is Neptune's orbit squared, multiplied by two pi over the period of Neptune's orbit. And this is gonna equal 2.492 times, 10 to the 42nd kilograms meters per second. And then what? They want us to find a CE the angular momentum we'll call it F the fraction of the angular momentum of the total of all planets. And then this would be divided by the angular momentum of the planets, plus the angular momentum of the sun. So, essentially, does the sun's own angular momentum contribute to the total angular momentum of our solar system? And so we could add, this would be 19.240 plus 7.806 plus 1.695 plus 2.492 times 10 to the 42nd. We'll lose the units because, of course, we're dividing by the same variable so that units were going to cancel out anyways. And then we're going to divide this by again. 19.240 plus 7.806 plus 1.695 plus 2.492 and then plus that of the sun. So one point around 1.122 times tend to the 42nd and we see that this fraction will put it here. This fraction is equaling 0.965 So essentially the anger momentum of the planets account for much more. Then the angular, um, momentum of the of the sun about its own axis in relation to the total angular momentum of our solar system. So again, after his equal in 0.965 this would be our final answer. That is the end of the solution. Thank you for a while.

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Video Transcript

Douglas C. Giancoli

Physics for Scientists and Engineers with Modern Physics

Liev Birman

Aspen Fenzl

Jared Enns

Meghan Miholics

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Physics for Scientists and Engineers with Modern Physics

Liev Birman

Aspen Fenzl

Jared Enns

Meghan Miholics

Douglas C. GiancoliPhysics for Scientists and Engineers with Modern Physics

Liev Birman

Aspen Fenzl

Jared Enns

Meghan Miholics

Douglas C. Giancoli

Physics for Scientists and Engineers with Modern Physics