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The table shows the mean (average) distances $d$ of the planets from the sun (taking the unit of measurement to be the distance from planet Earth to the sun) and th...

Question

The table shows the mean (average) distances $d$ of the planets from the sun (taking the unit of measurement to be the distance from planet Earth to the sun) and their periods $T$ (time of revolution in years).(a) Fit a power model to the data.(b) Kepler's Third Law of Planetary Motion states that "The square of the period of revolution of a planet is proportional to the cube of its mean distance from the sun." Does your model corroborate Kepler's Third Law?

The table shows the mean (average) distances $d$ of the planets from the sun (taking the unit of measurement to be the distance from planet Earth to the sun) and their periods $T$ (time of revolution in years). (a) Fit a power model to the data. (b) Kepler's Third Law of Planetary Motion states that "The square of the period of revolution of a planet is proportional to the cube of its mean distance from the sun." Does your model corroborate Kepler's Third Law?



Answers

The table shows the mean (average) distances $ d $ of the planets from the sun (taking the unit of measurement to be the distance from the earth to the sun) and their periods $ T $ (time of revolution in years).

(a) Fit a power model to the data.
(b) Kepler's Third Law of Planetary Motion states that " The square of the period
of revolution of a planet
is proportional to the cube of its mean distance from the sun."
Does your model corroborate Kepler's Third Law?

All right. Here's another great problem to do on a graphing calculator. And so we're going to go into the statistics menu and go into edit, and then we're going to type our numbers into list one endless, too, Where List one stands for the distance. The average distance of the planet from the Sun analyst to stands for the period. The time of revolution around the sun. Okay, once we have those numbers typed in, we want to find the power model. So we go into the stat menu, and then we go over to calculate, and we go down until we find power. It's a little ways down in the list. They're ago. Power regression, repress. Enter and we are Lee using List one and list, too. We don't need to worry about frequency list. We don't need to store the equation anywhere we could just calculate. So here's our power regression model. It's approximately y equals one times X to the 1.5 power. Okay, let's move on to Part B. So Part B talks about Kepler's third Law of planetary motion, saying that the square of the period of the revolution of a planet is proportional to the cube of its main distance from the sun. Well, that can be translated into this equation. The square of the period would be t squared is proportional to means we have some constant of proportionality K. And then the cube of the mean distance would be the Distance Cube. We want to know if that equation is similar to this equation that we got well in the equation we got from the calculator. RT is actually why, and it's not squared. It's just tea. Just why? So what we're going to do is square root both sides of the equation to get t equals instead of t squared equals. And when we do that, when we square root both sides of the equation, we end up with some other K. This K may not be the same as the other K because we've square rooted it, but it's still just some constant K times D to the three halves power. Now, the three halves power is equivalent to the 1.5 power. So I would say that the model we found is consistent with Kepler's third Law

6.43. So we have this. This is part of a series of problems involving this table to do with periods, uh, to do with the orbits of Jupiter's moons or the four largest moons of Jupiter. And so, for this part problem in this series, we want to use the mean distance from Jupiter value, uh, for IO to find the other, uh, mean distances using Kepler's third Law. And then you want to compare our results with what we get in the table so we'll use the period and mean distance from Jupiter of Io. So we'll call these tea too, and are too. And so we're going to want to find our one true Europa, Ganymede and Callisto. So basically, we, uh, have one problem to solve here. So let's go ahead and solve that. So we have Kepler's third Law. But the period and average distance the period squared being proportional to the average distance cube, and this means that are one which is what we're calling these other things for her. The distance for Europa, Ganymede and Callisto is equal to our two times the ratio of the two periods. Two of the 2/3. So for Europa, we put in its period is 3.55 and to this and we get 671 times 10 to the third kilometers. Bitch agrees with the table. So far, so good. Agana me, We have that its period of 7.16 days. So we put that in here and we find that it's average distance from Jupiter must be 1070 times 10 to the third kilometers. And this also agrees with what the table says. So the table so far at least, is consistent with itself and Kepler's third law and presumably being compiled from actual data from observing these things. Uh, this means that, you know, kept listed Law actually works as you'd hope so for Callisto, then its period of 16.7 days. And so we end up with an average distance of 1880 times 10 to the third kilometers. And you know, this is this is a wrong in the fourth, you know, digit here on that probably has more to do with the number of digits of precision we have for the the orbital radius and period of hyo than it Callisto for the period than it does to do Is with Kepler's Law Kepler's third law, not working. So this is, you know, this is still quite good, and probably if you had better numbers for these, you would get an exact agree.

6.79. So we want to take this table and first we want. First. We want the plot squared periods versus the average average distance is huge, okay? And then find a best fit straight line. And then we want to finds the average distance for Ludo. Given that it's period is 247.7 years. So now for the flooding part and then finding this best fit straight line, you just put these into a spreadsheet or whatever sort of thing you like to do for for plotting and you end up with a graph that looks like this. And so we see we have our best fit line here, which is the T squared is equal 0.999 nine our cue plus 0.3412 with units where the periods and years and the distance is an astronomical units. So then, just rearranging this, we get that the average distance is the periods where minus this constant must have units of years squared. Invited by 0.999 nine, then the cube root of this. So putting in our 247.7 years in here, we get that Pluto's average distance from the sun is 39.44 astronomical units, and this is very close to the sort of accepted value, which is 39.47 So we don't have sort of a perfect fit. But it's white gloves. And also there are probably other factors influencing the orbit of Pluto's because it's very sort of eccentric and tilted. There's other things in the solar system that could be could be affecting that, Uh, so in any event, I mean, this is our fourth digit. Here is where it disagrees, so this this works pretty well.

So we're talking about Kepler's third Law, and it talks about the period of time it takes to travel around the sun. The square of that is directly proportional to the distance that the average distance of that planet from the sun So there's our relationship, and it looks like they're talking about, uh, looks like they're talking about Earth in this one. So if the period is 365 days, yeah, and if the distance to the sun is 93 million miles, I remember that being a question on Who Wants to Be a Millionaire? That was the first guy that won the Millionaire. That was his question. And so if we take this 3 65 squared and divided by that 93 times 10 to the sixth Power Cube, that will give us our K constant, Let's do that. So from a calculator on and, um, clear out some stuff 3. 65 squared, divided by and I'll do a left parenthesis E 93 times, 10 to the sixth Power and then we need to cube that to the third power. And when I do that, I get that that K constant is 1.656 times 10 to the negative 19th power. And make sure you don't write that as e to the negative 19. That's that's That's not scientific notation. You want to write it in scientific notation, that calculator notation. Now we're dealing with Neptune, and we want to find out what the period is of Neptune, and we're going to utilize that same K constant. I'm just going to write K there. I have it in my calculator, so that's going to be convenient. And we know the distance of Neptune is 2.79 times 10 to the ninth power, and so we'll be able to find our our value just by multiplying. So if I take that k constant that I have in my calculator times, this, uh, left parentheses to 2.79 times 10 to the on, and that's a nine not very clear night power. And then we need to cube that, and that's what our T squared is going to end up being. So our T squared is and 3597075 and then three zeros helps, So that was like, That's a 3.5 billion, and then now we need to square root it. I actually take it to the 0.5 rather than square boat myself, and I find that the period is pretty darn large. Yeah, mhm. And that's how many days. So Neptune takes a long time for it to travel around.


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