For each of the eight planets Mercury to Neptune, the...

For each of the eight planets Mercury to Neptune, the semi-major axis $a$ of their orbit and their orbital period $T$ are as follows: (a) Explain why these values, when plotted as ${T^2}$ versus ${a^3}$, fall close to a straight line. Which of Kepler's laws is being tested? However, the values of ${T^2}$ and ${a^3}$ cover such a wide range that this plot is not a very practical way to graph the data. (Try it.) Instead, plot log$(T)$ (with $T$ in seconds) versus log($a$) (with $a$ in meters). Explain why the data should also fall close to a straight line in such a plot. (b) According to Kepler's laws, what should be the slope of your log$(T)$ versus log$(a)$ graph in part (a)? Does your graph have this slope? (c) Using $G =$ 6.674 $\times$ 10$^{-11}$ N $\cdot$ m$^2$/kg$^2$, calculate the mass of the sun from the $y$-intercept of your graph. How does your calculated value compare with the value given in Appendix F? (d) The only asteroid visible to the naked eye (and then only under ideal viewing conditions) is Vesta, which has an orbital period of 1325.4 days. What is the length of the semi-major axis of Vesta's orbit? Where does this place Vesta's orbit relative to the orbits of the eight major planets? Some scientists argue that Vesta should be called a minor planet rather than an asteroid.

Transcript

00:01 This question gives us data for eight planets of the solar system.

00:04 Precisely, it tells us what is the same major axis and the orbital period for each of those planets.

00:11 Then it asks us a bunch of questions about that data.

00:14 Let's begin by remembering about what is the same major axis and what is the orbital period.

00:20 So suppose that a big star, maybe our sun, is this one.

00:25 Now there is some planets orbiting around that big star.

00:28 Let me draw the orbit of one of one.

00:30 Of those planets.

00:30 We know that the orbit is like an ellipse, so something like that.

00:35 Then our planet goes around the sun on top of that ellipse, like this.

00:41 The semi -major axis is this distance in between the center of the ellipse and the outermost part of that ellipse.

00:50 This is the same major axis.

00:52 The orbital period, on the other hand, is the time that planet takes to complete a revolution around the sun on top of that ellipse.

00:59 Now we can begin the question.

01:01 So let me erase that drawing so that we can begin the first item.

01:05 We begin the first item by making a plot of t squared versus a to the third, and that plot is as follows.

01:13 In order to reproduce that plot, you have to convert the orbital period from days to seconds and the semi -major axis from 10 to the 6 kilometers into meters.

01:24 To convert from days into seconds is not that difficult.

01:29 All you have to do is multiply the number of days by 24 because we have 24 hours in each day.

01:36 Then you multiply the result by 60 because we have 60 minutes in each hour.

01:42 And finally, you multiply this result by 60 again because we have 60 seconds in each minute.

01:48 This will give you the number of seconds in each day.

01:52 And to convert from 10 to the 6 kilometers into meters is kind of straightforward.

01:57 Forward.

01:58 You just keep that 10 to the 6th and you multiply that by 10 to the 3rd because each kilo is a 10 to the 3rd.

02:06 Then one of this unit, 10 to the 6th kilometer, is the same as 10 to the 9 meters.

02:13 By doing that and making a plot of t squared by 8 to the 3rd, you will get a plot that is just like this one.

02:20 And this plot has an interesting feature which is the following.

02:23 If you connect all of those points, you will get a straight line.

02:27 Now we have to explain why is this the case.

02:30 For that, all you have to do is remember about kepler's third law.

02:35 Kepler's third law tells us that the square of the orbital period, so t squared, is proportional to the semi -major axis a cubed.

02:48 So this is what kepler's third law tells us.

02:52 In other words, we can say that t squared is equal to some unknown, coefficient, let me call it gamma times a to the third.

03:02 Then you can see that plotting t squared versus a to the third will be a straight line.

03:08 Now just in case you are interested and we will need that, that coefficient gamma is actually well known.

03:16 So using the known value for the gamma coefficient, we have t squared being equals to 4 pi squared divided by the newton constant times the mass of the same.

03:27 Sun times a to the third.

03:30 This is the complete form of kepler's third law.

03:34 Now you might notice something in this plot.

03:37 We plotted eight points but we can only see one, two, three, four points.

03:44 Where are the other four points? the answer is the following.

03:47 They are right here, one on top of the other.

03:50 If you zoom in that plot you'll be able to see them.

03:53 This happened because our data have a very big amplitude.

03:56 In this case, it's better to do a log plot instead of a linear plot like this one.

04:01 This is what the question asks us to do next, and by doing that we get the following.

04:06 Now we have a plot of the logarithm of the orbital period versus the logarithm of the semi -major axis.

04:14 Now, as you can see, this plot is again a straight line.

04:18 If you connect all of those points, we still get a straight line.

04:21 Why is this the case? to see that we can start begin from this equation for capular's third law.

04:28 Taking the logarithm of this equation, we get the following.

04:32 So the logarithm of t squared is equal to the logarithm of that complicated coefficient, let me call this gamma, times a to the third.

04:43 Then you use the property that the logarithm of t squared is equal to two times the logarithm of t.

04:50 Also the logarithm of gamma times a to the third is equals to the logarithm of gamma plus the logarithm of a to the third.

05:01 Now, you can say that two times the logarithm of t is equal to the logarithm of gamma plus three times the logarithm of a.

05:12 And finally, by dividing everything by two, you get that the logarithm of t is equal to 1 over 2 times the logarithm of that complicated coefficient plus 3 over 2 times the logarithm of a.

05:28 So, as you can see, the logarithm of t as a function of the logarithm of a is a first -degree function.

05:36 That is, this is of the form y is equal to some constant coefficient b plus some coefficient a times x.

05:46 Then we expect this plot to be a straight line, and this is why the plot still a straight line.

05:53 Now we proceed to the second item.

05:55 In the second item, we still use this plot, and we still use all of those equations.

06:01 We have to tell what would be the coefficient of that straight line according to kepler's third law.

06:09 According to kepler's third law, we know that t squared is equal to this complicated coefficient times a to the third.

06:15 Then, by doing what we have just done, taking the logarithm of both sides, we get the following.

06:22 The logarithm of t is equal to one -half times the logarithm of that complicated coefficient plus 3 over 2 times the logarithm of a.

06:31 Then, as you can see, when we look at this equation, like a first -degree equation, we see that the linear coefficient of that straight line should be 3 over 2.

06:42 So this is the expected value.

06:45 Now we have to tell if our plot has the correct slope.

06:49 For that, we use the computer to perform a linear fit of this data on a straight line...

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