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Planets beyond the solar system. On October $15,2001,$ a planet was discovered orbiting around the star HD68988. Its orbital distance was measured to be 10.5 million kilometers from the center of the star, and its orbital period was estimated at 6.3 days. What is the mass of $\mathrm{HDG} 8988$ ? Express your answer in kilograms and in terms of our sun's mass. (Consult Appendix E.)

a) $8.64 \times 10^{4} \mathrm{s}$b) $\mathrm{F}_{\mathrm{g}}$c) $3.58 \times 10^{7} \mathrm{m}$

Physics 101 Mechanics

Chapter 6

Circular Motion and Gravitatio

Physics Basics

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Applying Newton's Laws

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we know the radius and the time period. Um, we can figure out the mass by falling in question 6.10 which says that time period is Dubai are to the part three by two divided my square Rudolph G times must off hd. So this is the a Mars that we need to figure out from this equation. Now, the given set of parameters are the radius, which is 10.5 times in the par nine meter, then time period, which is 6.5 days. We can weather two seconds, too, just to be consistent with the unit. So that's if I find 44 three times 10 to the bar. Fine, second and ah from their weekend. Um, actually right this mass hd in terms of the other parameters. So that's for by squared R cubed, divided by cease fire G. So what we did here is we took square on both sides, and if we if we take square on both sides, we can get rid of this one over two part right here and this square root turn, and then he becomes t squared and we are All we have is our Cuba and then we have a square over here. So what we're doing here is we're taking square on both sides. And by doing so, we're getting it off this part and this part. And then, of course, because of this square turn, this guy becomes four. Pi squared. So we have four pi squared R Q. And then since we're taking him here, he comes here with his t squared and then we have tea. So yeah, let's put a plug in the numbers and let's see how much we get as the mass. So that's four by squared, then for our it's 10.5 times into the lower nine meters, divided by 5.4 for three times, 10 to par five seconds and then for G, it's six point 673 times and far minus 11. Newton Meena spread Sankaty squared. And from here we get three mass as 2.3 time stands The bar 30 Katie. Now, if we want to express this mass in terms of solar mass Ah, oui. She'd do the following so we know that the solar masses 1.99 times 10 to the power 30 kg so That means we have 2.3 times into a 30. So that means em off. H d is approximately 1.2 times this older muss. And the way we do it is we divide HD by s. And then we write the quantities. So for here, for this case, we have 2.3 time sent over 30 Clinton. And then we have 1.99 times into a 30 Katie and we get off this part and that part, and if we calculate this, we get one point toe and then we multiply a Emma's on both sides. So get rid of this part on DH that way. See that I am myself. Am HD is 1.2 times a solemn us. Thank you.

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