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A binary star has a time period `3` yeas (time period of earth is one year) while distance between the earth and the sun. Mass of one star is equal to mass of the sun and mass of other is `20n` times mass of the sun then calculate `n`.

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`M_(1)r_(1)omega^(2)=(GM_(1)M_(2))/((9r)^(2)),r_(1)+r_(2)=9r`
where `r` is dist between earth and sun
`impliesM_(1).(M_(2)9r)/(M_(1)+M_(2))((2pi)/3)^(2)=(GM_(1)M_(2))/((9r)^(2))`………(1)
for the earth
`M_(epsilon)r((2pi)/1)^(2)=(GM_(epsilon)M_(S))/(r^(2))`.........(2)
from (1) and (2)
`implies (M_(1)+M_(2))=81Ms`
If `M_(1)=M_s`
`M_(2)=80M_(s)=20mM_s`
`impliesn=4`

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