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Extension Tke planet: in our Solar System orbit tre Sun ir almost circular orbit:. Skowr that the orbital speed cfa plane: 2t dstance ffom tze cente of the Sur 1 g1...

Question

Extension Tke planet: in our Solar System orbit tre Sun ir almost circular orbit:. Skowr that the orbital speed cfa plane: 2t dstance ffom tze cente of the Sur 1 g1tren %}" GMThe mear distarce betiveen tke Su and tze Earth i: 1* 10" and tke mas? ~ t.e 120 x 10"k: Calcula e the orbita speed of-he Earth 28 it Tzvel: Icudte Sun: Tzere point betireen tke Earth ard the Mccn where tke ret ravitational field sTerzth zer? Atthi: point the Earth : gratitaticnal field strergth equa- magit

Extension Tke planet: in our Solar System orbit tre Sun ir almost circular orbit:. Skowr that the orbital speed cfa plane: 2t dstance ffom tze cente of the Sur 1 g1tren %}" GM The mear distarce betiveen tke Su and tze Earth i: 1* 10" and tke mas? ~ t.e 120 x 10"k: Calcula e the orbita speed of-he Earth 28 it Tzvel: Icudte Sun: Tzere point betireen tke Earth ard the Mccn where tke ret ravitational field sTerzth zer? Atthi: point the Earth : gratitaticnal field strergth equa- magitude but opposite direction the gravitational feld strength of the Moon Given tha:: 2283of Eartk mass 0f Moor calculate bow far this point is fom the certre of tbe Moon in teras of R; Where R i: tbe separation betireen tre centres ofthe Earth andtre Moon



Answers

In Exercises 16 and $17,$ two planets, planet $A$ and planet $B$ , are orbiting their sun in circular orbits with $A$ being the inner planet and $B$ being farther away from the sun. Suppose the positions of $A$ and $B$ at time $t$ are

$$
\mathbf{r}_{A}(t)=2 \cos (2 \pi t) \mathbf{i}+2 \sin (2 \pi t) \mathbf{j}
$$

and
$$
\mathbf{r}_{B}(t)=3 \cos (\pi t) \mathbf{i}+3 \sin (\pi t) \mathbf{j}
$$
respectively, where the sun is assumed to be located at the origin and distance is measured in astronomical units. (Notice that planet $A$ moves faster than planet $B . )$ The people on planet $A$ regard their planet, not the sun, as the center of their planetary system (their solar system).

Using planet $A$ as the origin, graph the path of planet $B .$ This exercise illustrates the difficulty that people before Kepler's time, with an earth-centered (planet $A$ ) view of our solar system, had in understanding the motions of the planets (i.e., planet $B=$ Mars). See D. G. Saari's article in the American Mathematical Monthly, Vol. 97 (Feb. $1990 ),$ pp. $105-119$ .

Well let's R A B T denotes the vector from planets A two planets be at time. T. So then it implies that R A B C is equal to R B minus are. So then this is going to be equal to what is mm This is going to be equal to you have three codes by t minus two. Because Hi see I plus you have three saying pity minus to sign pride. See G. This is equal. Soon we have three cause by t minus two. You have caused squared. Alrighty my net sign squared. I thi I blood so lives. I have three. Same by t minus full saying by T cause piety G. This is going to be equal to you have three. Because do you call spicy minus four? Course. Quick. I see. Plus two. I then Nigeria components. We have three minus four. Because up I see. Hi eight C. Saying Fizi G. So this implies that our parametric. I work for a metric equations, equations for the path for the past. Ah ah You have the X and Y companies so you have X to be equal to two bloods three minus four. Cause by T because Hi. See then that's of the why component is going to be equal to do we minus four course high tea saying hi, etc. As our finer results.

Hello. In this question here, our task is to calculate the mutual gravitational force between the moon and our planet earth. Were given the following info the mass of the moon, Which is equal to 7.35 Times 10 to the power of 22 measured in kilograms. And also were given the mass the mass of planet earth here Being 5.98 Times 10 to the power of 24 kg. We're also giving the main distance or between the center of both objects here. All right. And it's around 3.85 Times 10 to the power fate measured in meters. Also, we need to mention that here we have G. Which is the constant of the University of constant of gravitation here. And this is equal to 6.67, 3 Times 10 to the power of -11. And here the only thing left is to substitute with these numbers here. In the force or in the law of mutual gravity that Newton stated hundreds of years ago. So here we have the force of mutual gravitation is equal to G. Multiplied by the first mass multiplied by the second mass. So here this is a massive moon and this is the mass of Earth and divided by that square of the distance between them. So here, just substituting numbers G is equal 6.67, 3 times 10 to the power of negative 11 times the math of the moon. 735 Times 10 to the power of 22 multiplied by the mass of Earth. All of this is divided by the square of the distance between them. So here we have 3.85 times 10 to the power of eight. All of this is To the power of two, and we have the answer Of 1.9, 8 Times 10 to the power of 20 Noten's. So that's here, the mutual gravitational force between the moon on Earth. Or in other words, this is the force that the moon gravitates the earth towards it, and also the force that the earth gravitate the moon towards them.

Hi. So this question we have Especially orbits the moon at the height of 20,000 m. I want to find The speed and the time it takes one orbit. So first of all we can to quit the gravitational force to the centripetal force. Just the friendship is in stable over its should gravitational force physicals or G. Um was a satellite maps of the moon divided by the radius of the orbit squared. And this would be cool to. Mhm. The mass of the satellites times velocity squared divided by the radius. This council is out. We get go to you. Did you call soon? You scared the G. Um of the moon Ready by the ridges substitute. And we have The height is 20,000 m. Mhm and we have the security so the moon is 1 7. It's 1.738. I'm still into politics. So we need to add those two heights to find the riches of the orbit. The skirt of NGS 6.67 Time stands about 11 times. Marshall Dimon has given us 7.34. I'm standing about of 22 kg wanted by 1.738 Time standard power of six class 2000 m. Good. And this would be a culture And is approximately equal to 1,617. It was my second. Yeah. So now I can find the time for one orbit and the time for one or but you know that we lost easy calls. It just starts divided by time and so time we called to the stance whereby velocity and the distance here with the circumference with the orbit so time will be called for to buy are provided by feet. This is 25 times Radius of Europeans, which is 1.7 38 times 10. About six plus 20,000 created by 167 year. It should be called him. There's approximately called to 6.62 Time stands about three seconds, Mom.

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.


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