kingwinner Posted June 6, 2006 Report Share Posted June 6, 2006 I really need help with this question: 1) A certain "double star" consists of two identical stars, each of mass 3.0x10^30 kg, separated by a distance of 2.0x10^11 m between their centres. They orbit around a common centre of gravity. How long does it take to complete one cycle? (in seconds) [answer: 7.9x10^7 s] Method 1: (my method)This is the method I did. I assumed one of the stars to be stationary, i.e. viewing from the frame of reference on the star, and let the other star orbit aorund it in circular orbit. This becomes like a problem of a satellite orbitting around the earth. Let Fc = centripetal force Fc=Fg......T=2pi * square root of (r^3/G m) where m=mass of one of the stars, and r=2x10^11 m=39719490.78 s=4.0x10^7 s Method 2: (my instructor's method)This is what my instructor did.D=Diameter of the circle=2x10^11 mr=raiuds of the circle=1x10^11 mm=mass of one of the stars Fc=Fg...v = square root of (G m r / D^2)v=22371.299m/sv = 2 pi r / TT=2.8x10^7 s But, both answers are wrong. The correct answer is 7.9x10^7 s. This is a tough tough question for me. Can someone please explain? My method is wrong. I doubt that even my instrctor's method is wrong. Quote Link to comment Share on other sites More sharing options...

ronthepon Posted June 6, 2006 Report Share Posted June 6, 2006 Make yourself sure if the stars are separated from the centre of gravity at the farthest position or is it some other case. Quote Link to comment Share on other sites More sharing options...

Roadam Posted June 6, 2006 Report Share Posted June 6, 2006 Are you sure that the first answer is right? I myself come to some wierd solutions. Maybe you miscalculated. Check for some 2s, your answer is exactly half of the supposedly right one. Or you could post how did you pull that equation out, becouse in mine there is Gm/r^3 in the root but that is obviously wrong becouse I get resoults of 10^-(something). Check it out. Quote Link to comment Share on other sites More sharing options...

Tim_Lou Posted June 6, 2006 Report Share Posted June 6, 2006 firstly, you must find out where the center of mass is.let r1 be the distance between the center of mass and the first body, r2 for the due to symmetry, r1=r2*i'm not sure what it is meant by "between their centres", anyhow...find r1 and total distance. Now if it is assumed that the orbit is circular, it would be quite simple.if not, we gotta use elliptic integral. equate[math]{4\pi^2r_1}/T^2={Gm}/r^2[/math]and solve for the answer, Tr is the total distance, r1 is... r1, half of total distance, m is the mass. edit: wait this is the method your instructor told you, it should be correct. Quote Link to comment Share on other sites More sharing options...

Tim_Lou Posted June 6, 2006 Report Share Posted June 6, 2006 oh, i see why, this is just a bit of misunderstanding. your instructor's method is definitely correct, however he interpreted the question wrongly. see, "2.0x10^11 m" is not the "diameter", it is the radius. if you use this number and apply it to the second method, you'll arrive at:7.94449*10^7 seconds (ignoring sig. fig.) Quote Link to comment Share on other sites More sharing options...

kingwinner Posted June 6, 2006 Author Report Share Posted June 6, 2006 oh, i see why, this is just a bit of misunderstanding. your instructor's method is definitely correct, however he interpreted the question wrongly. see, "2.0x10^11 m" is not the "diameter", it is the radius. if you use this number and apply it to the second method, you'll arrive at:7.94449*10^7 seconds (ignoring sig. fig.)Can you show me what values you substitute into v = square root of (G m r / D^2) ? I can't get the answer. Besides, using the frame of reference of one of the stars, wouldn't you see the other star as orbitting around you in a circular orbit? (the logic of method 1) Quote Link to comment Share on other sites More sharing options...

Tim_Lou Posted June 6, 2006 Report Share Posted June 6, 2006 you would not want to mess with an accelerated frame. when you do, all the forces formula are altered (ficitious forces emerge). you would need to apply some nasty things like Hamiltonian mechanics which i do not even know.so, always pick the non-accelerated frame if you can. to find the solution, use:[math]{4\pi^2r_1}/T^2={Gm}/r^2[/math]r1 would equal 2.0*10^11, r would equal 4.0*10^11 the answer should be around 7.943188*10^7 seconds Quote Link to comment Share on other sites More sharing options...

Jay-qu Posted June 7, 2006 Report Share Posted June 7, 2006 It appears to be a badly worded question, they give you the distance between the stars centers and when they calculated the answer used that distance as the radius. At any rate the method tim and your instuctor used is correct. The problem with yours is that you cant just call one stationary and in the middle, because the center of mass will always be halfway between two equal masses. Quote Link to comment Share on other sites More sharing options...

kingwinner Posted June 7, 2006 Author Report Share Posted June 7, 2006 Method 2:Fc=Fgmv^2/r = Gmm/D^2 <=====I don't understand this line! Why is r (radius) used for Fc and D (diameter) used for Fg? Shouldn't the radius be used in both cases? Why is the centripetal force on one star equal to the force of gravity supplied by the other star, located in a distance away that equals the diameter? (i.e. Gmm/D^2)?For the regular case, consider a satellite orbitting around earth, in this case, the centripetal force is equal to the force of gravity between the satellite and the earth, but the RADIUS is used in GMm/r^2 instead of the DIAMETERv = square root of (G m r / D^2)v=22371.299m/sv = 2 pi r / TT=2.8x10^7 s Is 7.9x10^7 s definitely the wrong answer? Is it also a conincidence that method 1's answer is exactly half the answer provided? Quote Link to comment Share on other sites More sharing options...

Jay-qu Posted June 7, 2006 Report Share Posted June 7, 2006 I think by looking into it to much you are confusing yourself. Dont bother with thinking about centripetal force, there is one force acting on each star and that is a gravitational force supplied by the other star, these are equal and opposite - in all cases. The earth applies the same force on you that you do on the earth. As for the D, it is just a matter of terminology. You would be used to problems where the mass of one object is negligible in comparrison to another (satellite around earth), in these cases the center of mass, which both objects orbit around, is the center of the larger object. So the distance between their centers is taken as radius of the circle of the orbit. In the case of 2 equal masses r is still the distance between the centers, but you may become confused and tempted to use r as radius of the mutual orbit, which it is not, it is actually the diameter. Quote Link to comment Share on other sites More sharing options...

kingwinner Posted June 7, 2006 Author Report Share Posted June 7, 2006 In this case, the centre has no object (no mass). Both stars are in the circular orbit. Both stars are moving in circular path, so centripetal force should be considered, right? Then, is it true that the force of gravity of star 1 on star 2 provides the centripetal force for star 2, and that the force of gravity of star 2 on star 1 provides the centripetal force for star 1? Quote Link to comment Share on other sites More sharing options...

Jay-qu Posted June 7, 2006 Report Share Posted June 7, 2006 yes that is correct - but you dont have to think of it as centripetal force, it is in fact just gravity - it just so happens that this force results in circular motion (which requires a force to keep it there..) Quote Link to comment Share on other sites More sharing options...

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