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The figure shows elliptical orbit of a planet 'M' about the Sun 'S', the shaded area SCD is twice the shaded area SAB. If t_(1) is the time for the planet to move from C and D and t_(2) is the time to move from A to B then. |
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Answer» Solution :ACCORDINGTO Kepler.slaw LAWOF areas `(DeltaA)/(Deltat) ` = constant `DeltaA prop Deltat` `(DeltaA_(SCD))/(DeltaA_(SAB)) = (t_(1))/(t_(2))` `(2A)/A = (t_(1))/(t_(2))` ` :. "" t_(1) = 2t_(2)`
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