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Binary stars of comparable masses rotate under influence of each others gravity about their common centre of mass,P If the product of gravitational field at P due to smaller star and the inverse of centripetal acceleration of smaller star is 16/27 then find the ratio of the mass of smaller star to bigger star...

Asked by sandhyamsani 24th November 2017, 11:35 AM
Answered by Expert
Answer:
 
Gravitatational force balance the centripetal force for both the stars so we have

begin mathsize 12px style fraction numerator G m subscript 1 m subscript 2 over denominator d squared end fraction space equals space m subscript 1 omega squared r subscript 1

fraction numerator begin display style G m subscript 1 m subscript 2 end style over denominator begin display style d squared end style end fraction space equals space m 2 omega squared r subscript 2



end style
where d = r1+r2

from above begin mathsize 12px style m subscript 1 over m subscript 2 equals space fraction numerator r 2 over denominator r 1 end fraction end style
 
begin mathsize 12px style fraction numerator G r a v i t a t a t i o n a l space f i e l d space d u e space t o space m 1 space a t space P over denominator C e n t r i p e t a l space f o r c e space o f space m 1 end fraction space equals space fraction numerator fraction numerator G m subscript 1 over denominator r subscript 1 superscript 2 end fraction over denominator m subscript 1 omega squared r subscript 1 end fraction space equals space fraction numerator G m subscript 1 over denominator r subscript 1 superscript 2 end fraction space cross times fraction numerator d squared over denominator G m subscript 1 m subscript 2 end fraction space equals space fraction numerator d squared over denominator r subscript 1 superscript 2 m subscript 2 end fraction end style
begin mathsize 12px style equals space fraction numerator r subscript 1 superscript 2 open parentheses 1 plus begin display style fraction numerator m 1 over denominator m 2 end fraction end style close parentheses squared over denominator r subscript 1 superscript 2 m 2 end fraction space equals fraction numerator space open parentheses 1 plus fraction numerator m 1 over denominator m 2 end fraction close parentheses squared over denominator m 2 end fraction space equals space 16 over 27 end style
from mthe above equation we try to get m1/m2 ratio. we try to get better solution if possible
Answered by Expert 27th November 2017, 5:05 PM
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