If the earth be one half its present distance from the sun, how many days will present in one year on the surface of earth?
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Answer:
129 days
Explanation:
by Kepler's law
T2 α a3
T1²÷a1³3=T2²÷a2³
T2²=(a2³÷a1³3)×T1²
t2³=(1/2)³×T1²
t2²=1/8×1 year
t2=1/8×365
T22=(a2a1)3×T21
T22=(12)3×T21
T22=(18)×1year
T2=√1/8×365days
=129days
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