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![[Post New]](/templates/default/images/icon_minipost_new.gif) 23 May 2007 21:40:17 IST
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A double star is a system of two stars having masses m1 and m2 moving around the centre of inertia of the system due to gravitations. Then, ratio of sweeps of area of star of mass m1 to the star of mass m2 is : (A) m1/m2 (B)m2/m1 (C) m12/m2 (D)m22/m12 Please explain the answer. thanks.
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Guide to latex:
http://www.goiit.com/posts/list/community-shelf-a-guide-to-latex-48056.htm
JEE and OLYMPIA INFINATUM
http://iit-redefined.theforum.name/index.php
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23 May 2007 21:42:28 IST
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is it m1/m2...........
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23 May 2007 21:52:22 IST
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I tihnk it will b solved ny angular momentum conservation.
I m trtying the question on paper.
will tell u the answer in just a while fm now
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23 May 2007 21:54:06 IST
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The distance fm centre of inertia of each mass is not given
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Hi. See fig. Area swept =  /360*pir 2 so area swept by m1 = /360*pi(m 2l/m 1+m 2)^2 and area swept by m2 = /360*pi(m 1l/m 1+m 2)^2 a1/a2 = m22/m12
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23 May 2007 22:58:09 IST
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thanks a lot joy. that is the correct answer.. by the way, i got a doubt now.. u have implied in your solution that the angular velocities of m1 and m2 are equal right? does that mean that the earth and moon rotate about their common centre of mass with the same angular velocity?
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Guide to latex:
http://www.goiit.com/posts/list/community-shelf-a-guide-to-latex-48056.htm
JEE and OLYMPIA INFINATUM
http://iit-redefined.theforum.name/index.php
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10 Apr 2008 18:17:10 IST
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yes , i guess it is right to say so..
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