## a) $$\frac{3}{2} \log a+\log \left(\frac{2 \pi}{\sqrt{G m_{\mathrm{s}}}}\right)$$b) $$3 / 2$$c) $$10^{-9.2476}$$d) $$353 \times 10^{6} \mathrm{km}$$

Gravitation

### Discussion

You must be signed in to discuss.
##### Top Physics 101 Mechanics Educators ##### Christina K.

Rutgers, The State University of New Jersey ##### Andy C.

University of Michigan - Ann Arbor

LB Lectures

Join Bootcamp

### Video Transcript

this question gives us data for eight planets off the solar system. Precisely. It tells us what is the same major access on the orbital period for each of those planets. Then it asks us a bunch of questions about that data. Let's begin by remembering about what is the same major access and what is the orbital period. So I suppose that a big star, maybe our son, is this one. Now there is some planets orbiting around that big a star. Let me draw the orbit off one off this planet. We know that the orbit is like an ellipse. So something like that, then our planet goes around the sun on top off that lips like this, the same major access is this distance in between the center off the ellipse on the altar, most part off their lips. This is the same major access. The orbital period, on the other hand, is the time that plan it takes to complete a revolution around the sun. On top of that ellipse. Now we can begin the question. So let me raise that drawing so that we can begin the first item. We begin the first item by making a plot off T squared versus a to the third on that plot is as follows. In order to reproduce that plot, you have to convert the orbital period from these two seconds on the same major access from 10 to the six kilometers into meters. To convert from these in two seconds is not that difficult. All you have to do is multiply the number off days by 24 because we have 24 hours in each day. Then you multiply the result by 60 because we have 60 minutes in each hour and finally you multiply this result by 60 again because we have 60 seconds in each minute. This will give you the number off seconds in each day and to convert from 10 to the six kilometers into meters is kind off straightforward. You just keep that 10 to the sixth and you multiply that by 10 to the third because each kilo is a 10 to the third, then one off this unit 10 to the sixth kilometer is the same as 10 to the nine meters. By doing that and making a plot off T squared by eight of the third, you will get a plot that it's just like this one. And this plot has an interesting feature, which is the following. If you connect all of those points, you will get a straight line. Now we have to explain why is this the case for that? All you have to do is remember about Kepler's third law. Capital stirred Law tells us that the square off the orbital period so T squared is proportional to the same major axis. A cube. So this is what Kepler Store Law tells us. In other words, we can say that the squared is equals to some unknown coefficient. Let me call it Gamma Times a to the third. Then you can see that plotting T's cornered versus a to the third will be a straight line. Now, just in case you're interested and we will need that that coefficient gamma is actually well known. So using the known value for the gamma coefficient, we have d squared. Being equals to four pi squared, divided by the Newton constant times The mass off the Sun Times a to the third. This is a complete form off Kepler's third law. Now you might notice something in this plot, we plotted eight points, but we can only see 1234 points. Where are the order? Four points. The answer is the following. They are right here, one on top of the other. If you zoom in that plot, you will be able to see them. This happened because our data have a very big amplitude. In this case, it's better to do a log plot instead off a linear plot like this one. This is what the question asked what to do next. And by doing that, we get the following. Now we have a plot off the logarithms off the orbital period versus the logarithms off the same imager. Access. Now, as you can see, this plot is again a straight line. If you connect all of those points, we still get a straight line. Why is this the case? To see that we can begin from this equation for Kapler stored law taking the longer written off this equation, we get the following so the longer it, um off the squared is equal to the longer it, um, off that complicated coefficient. Let me call this Gamma Times a to the third then you use the property that the logarithms off T Square is equal to two times the logarithms off t. Also, the longer it, um off gamma times eight of the third is equals to the longer item off gamma. Plus, the longer it, um, off a to the third. Now, you can say that two times. The longer it, um, off t is equals to the longer it, um, off gamma plus three times, the longer it, um off a And finally, by dividing everything by two, you get that the longer it, um, off t is equals to 1/2 times the logarithms off that complicated coefficient plus 3/2 times the logarithms off A. So as you can see the longer rhythm off T as a function off the logarithms off A is the first degree function that is, this is off the form. Why is equals to some constant coefficient? B plus some coefficient a Times X. Then we expect this plot, Toby a straight line. And this is why the plot still a straight line. Now we proceed to the second item in the second item. We still use this plot, and we still use all of these equations. We have to tell what would be the coefficient off that straight line? According to Kepler's Third Law, According to Kepler's third Law, we know that he squared is equal to this complicated coefficient times a to the third. Then by doing what we have just done taking the logarithms off both sides, we get the following. The logarithms off T is equal to one half times the longer it, um, off that complicated coefficient plus 3/2 times the logarithms off a. Then, as you can see, when we look at this equation like a first degree equation, we see that the linear coefficient off that straight line should be tree over two. So this is the expected value. Now we have to tell if our plot has the correct slope for that, we use the computer to perform a linear feet off this data on a straight line. By doing that, we got a slope that is 1.49 86 which is approximately 1.5, which would be equivalent to three over to meaning that our data has a very good agreement with the theory in the turd item. We're giving the value off Newton's constant. And we are asking to calculate what is the mass off the sun m s using our feet. So according toa our feet, the slope is this one that I had just said on the linear coefficient that is one half times the local rhythm off. That complicated gamma is approximately minus 9.24 7 to 6. That ISS B is equal to minus 9.24 76. Now, how can we use this data to evaluate the mass of the sun? We begin by equating one of the true longer rhythm off gamma. With this. By doing that to get the following one of virtue times, the local rhythm off gamma is equals to minus 9.24 76. Now we can send one over to inside the logarithms at the square root. So the longer it, um off the square root off gamma or gamma to the power off. One half is equal to minus 9.24 76. Now, ISP, initiate both sides off this equation. So 10 to the longer it, um off the square root off gamma is equals to 10 to minus 9.24. 76. Now, since we were using longer written in base 10, we can simplify 10 with the longer it, um like that And we're left with the following equation. The square root off gamma is equal to 10 to minus 9.24. 76. Now we have to use the equation for the complicated coefficient gamma, which is this one then what we know is that the square root off for pi squared, divided by G times, the mass off the sun is equals to 10 to minus 9.24. 76. Now it's square both sides to get the following four pi squared divided by G mass off the sun is equal to 10 to minus 9.24. 76 on disease squared Now we saw for the mass off the sun. To do that, we sent this number to the left hand side dividing and the mass of the sun to the right hand side. Multiplying by doing that to get the following the mass off the sun is equals True g dividing the Newton constant times 10 tu minus 9.24 76 squared on this wall thing is dividing four times by squared. Then, by plugging in the value off Newton's constant, we got the following. The mass of the sun is equals to 6.674 times 10 to minus 11 times 10 tu minus 9.24 76 squared, dividing four times by squared. And by solving that calculation, we got a value. There is approximately 1.85 times stand to the turd kilograms by looking at the appendix. So let me call this the mass of the sun. According to the appendix F, we have a value off 1.99 times 10 to the third kilograms. So as you can see, we have a value that has the same order off magnitude as the value in the appendix, meaning that our data is good and our calculations were good too. Now, for the margin off error, we can do the following the error. Let me call this Absalon is equals to one minus 1.85 times stand to the 30 divided by 1.99 times stand to the 30. This results in an error off approximately 0.7 or 7%. So our results has a good agreement with the reference value. Now, before beginning the last item, let me organize my board to keep on Lee things that we're going to use. Okay, in the last item, we're given the value off the orbital period for the asteroid Vesta, and we have to evaluate what is the same major access off its orbit. For that, we can use Kepler's third law, which is this one. All we have to do is the following. We have to solve this equation for a plaguing the values that were given and then get the result. So we begin by solving this equation for a for that, we sent this term to the other side. Dividing by doing that to get the following eight of the third is equals to t squared, divided by four pi squared, divided by G times the mass off the sun. Now we take the cube root on. By doing that, we got that a is equals to the cube root off t squared, divided by four pi squared, divided by G times the mass off the sun. Then By simplifying this equation, we get the following a is equals to the cube root off T squared times the Newton constant G times the mass off the sun and all this is divided by four times spy squared. Now all we have to do is plug in the values that were given by the problem. So we have the cube root off t squared T is the orbital period, but the orbital period was given in days before plugging into this equation. We have to convert it into seconds. This is not difficult to do. Let's do it here. We have that period in days. Now we have to convert it to seconds. What can we do? We just have to multiply it by 24 because each day has 24 hours, then by 60 because each hour has 60 minutes and then by 60 again because each minute has 60 seconds and Walla it's already converted. Now it's square it and then multiplied by the Newton Constant, which is 6.674 times 10 to minus 11. And finally we multiplied by the mass off the sun. For this item, I'm using the mass of the sun that is given in the appendix. So one 0.99 times 10 to the third and to finish, we divide this whole equation by four times by squared. Then, by solving this calculation, we get a value for the same major axis off the orbit off the asteroid Vesta. That is approximately three 0.53 times 10 to the 11 m. Now, to compare this value with the values that were given by the problem in the table, we have to convert it into kilometers. And to do that, we divide by 10 to the third to get the following. This is the same as 3.53 times, 10 to the 8 kg meters. But now we have to write it like 10 to the six kilometers. Then we can split this value as follows. This is 3.53 times Stand to the second times, 10 to the sixth kilometers. Now we can compare 3.53 with the values that were given by the problem in that table. To make that comparison, it's easier to write 3.53 as 353 then, comparing with the values. We conclude that the Vesta asteroid orbits in between Mars and Jupiter. With that, we finished this long question. The numerical answers are circled in green. Brazilian Center for Research in Physics

#### Topics

Gravitation

##### Top Physics 101 Mechanics Educators ##### Christina K.

Rutgers, The State University of New Jersey ##### Andy C.

University of Michigan - Ann Arbor

LB Lectures

Join Bootcamp