So this problem is on ly tedious. It's not very difficult, however we need to. We do need to do a lot of calculations and there are a lot of room. There's a lot of room for computational error, so definitely just be careful. Let's start the angular momentum. The formula is going to be Iomega, and we can say that the, um, angular momentum of the sun is going to be a model is gonna be Iomega, but I of the sphere Omega So this is going to be equal to two times the mass of the sun radius of the sun squared, divided by five times two pi over the period of the sun in its orbit. So we can say that out of the sun equals two times one point nine nine times ten to the thirtieth, and these were all tabulated values. So you simply need to look in any physics textbook in order to find these values. Times two pi I apologized to five clean this up and this will be divided by five times twenty five days times twenty four hours per day, times thirty six hundred seconds per hour. And we have the angular momentum of the sun being one point one two times ten to the forty second kilogramme meters squared per second. So, in order to find the formula for the the moment of inertia for all the planets, well, the planets are in orbit with sun. So the center of the axis going going through the center of the sun, that would be the axis of rotation here. So the distance between a planet to the to the axis of rotation which simply be the distance from the planet to the sun. And ah, these are tabulated. So we can say that l of Jupiter is going to be equal to the mass of Jupiter Times the radius of Jupiter squared times to pie over the period of Jupiter and again, R, this term right here are subject simply means the distance from Jupiter to the sun. And this is going to be equal two. Brother, you don't need the pregnancies, so we can just say a to pi. One hundred and ninety points times ten to the twenty fifth kilograms seven hundred and seventy eight times ten to the ninth meters to the squared, uh, squared. Divided by eleven point nine years times three hundred and sixty five days per year times twenty four hours per day, times thirty six hundred seconds per hour. And this is going to give us one point nine two times ten to the forty third kilogramme Meters squared per second. The same thing is gonna happen for Saturn, so we'LL say Allah vests. The angular momentum of Saturn equals the massive Saturn times The distance from Saturn to the sun squared times two pi over the period of Saturn and this is going to be equal to two pi a fifty six point eight times ten to the twenty fifth, one thousand four hundred twenty seven times ten to the ninth squared, all divided by the period of Saturn is twenty nine point five years. So we're going to have to do twenty point five years times three hundred sixty five days in one year times twenty four hours in one day times thirty six hundred seconds in one hour and we're approaching seven point eight one times ten to the forty second, um, rather over anything. My apologies, simply the units kilograms meters squared over seconds. So at this point, we are going to do the same exact thing for Uranus and Neptune. So this will be the mass of Uranus. Times the radius of Uranus squared over to pie time developer the radius of Uranus, rather to pi divided by my apologies the period of fairness. This is again the angular velocity of Uranus. So this would be to pie. Okay, eight point six eight times ten to the twenty fifth time's two thousand eight hundred seventy times ten to the ninth squared. Invited by This will be eighty four years times three, sixty five times twenty four times thirty six hundred. And we're approaching one point six nine times ten to the forty second kilogramme meters squared per second. And then finally, angular momentum of Neptune Massive Neptune rate distance from no tune to the sun squared times two pi Over the period of Neptune to pie. This will be ten point two times ten to the twenty fifth, forty five hundred times ten to the ninth squared. Divided by here, it's going to be one hundred sixty five years times three, sixty five times twenty four times thirty six hundred and this is going to be two point four and nine times ten to the forty seconds kilogram meters squared over second. And so, if they're asking us for the angular moment, the kinetic and the rotational kinetic energy rather the total angular momentum of the solar system and the percentage of the total angular the momentum of the solar system that is accounted for by the rotation by the planets rotating the sun, we're going to simply take the angular momentum of the planets and divided by the angular momentum of the solar system. And this will be equal to all of the sun close out of Jupiter plus, ah, annual rooms of Saturn, plus angular momentum of Neptune, plus angular momentum of Uranus and then angular momentum of Jupiter plus Saturn, plus not tuned. Plus, Eunice. And I'm not gonna write down all of these numbers because they're already down here so you can simply add them up and divided by the some of the total angular momentum of the solar system, and you find that this is equal two point nine six six. So ninety six point six percent of the of al solar system equals L planets. Essentially ninety six point six percent of the angular momentum of the solar system is accounted for by the rotational energy of the planets orbiting the sun. So again, it's tedious. However, it's not hard. You simply have to make sure that you don't make any computational airs. And that's the end of the solution. Thank you for watching.