InterviewSolution
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InterviewSolution
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One of the satellite of jupiter, has an orbital period of `1.769` days and the radius of the orbit is `4.22 xx 10^(8) m`. Show that mass of jupiter is about one thousandth times that of the radius of the sun. (Take `1` year `= 365.15` mean solar day). |
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Answer» For a satellite of jupiter period, `T_(1) = 1.769 days = 1.769 xx 24 xx 60 xx 60 s` Radius of the orbit of satellite, `r_(1) = 4.22 xx 10^(8) m` Mass of the jupiter, `M_(1)` is given by `M_(1) = (4 pi^(2) r_(1)^(3))/(GT_(1)^(2)) = (4pi^(2) xx (4.22 xx 10^(8))^(3))/(G xx (1.769 xx 24 xx 60 xx 60)^(2))` ..(i) We know that the orbital period of Earth around the sun, `T = 1 year = 365.25 xx 24 xx 60 xx 60 s`, Orbitial radius, `r = 1 A.U. = 1.496 xx 10^(11) m` Mass of the sun is given by `M = ( 4 pi^(2) r^(3))/(GT^(2)) = (4pi^(2) xx (1.496 xx 10^(11))^(3))/(G xx (365.25 xx 24 xx 60 xx 60)^(2))` ..(ii) Dividing (2) by (1), we get `(M)/(M_(1)) = (4pi^(2) xx (1.496 xx 10^(11))^(3))/(G xx (365.25 xx 24 xx 60 xx 60)^(2))xx (G xx (1.769 xx 24 xx 60 xx 60)^(2))/(4pi^(2) xx (4.22 xx 10^(8))^(3)) = 1046` or `(M_(1))/(M) = (1)/(6400) ~~(1)/(1000)` or `M_(1) ~~ (1)/(1000) M`, which was to be proved. |
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