explain question no 5
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By Kepler's law
T2αa3T2αa3
T21a31=T22a32T12a13=T22a23
T22=(a2a1)3T22=(a2a1)3×T21×T12
T22=(12)3T22=(12)3×T21×T12
T22=(18)T22=(18)×1year×1year
T2=18–√T2=18×365days×365days
=129days
T2αa3T2αa3
T21a31=T22a32T12a13=T22a23
T22=(a2a1)3T22=(a2a1)3×T21×T12
T22=(12)3T22=(12)3×T21×T12
T22=(18)T22=(18)×1year×1year
T2=18–√T2=18×365days×365days
=129days
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