The radius of the nearly circular orbit of mercury is 5.8×10^10 m and its orbital period 88 days if hypothetical planet has an orbital period of 55 days what is the radius of its circular orbit
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by kepler's law,
T²≡a³
Let x be the new distance
(88÷55)²=(5.8×10∧10÷x)³
⇒x=14.8×10∧10
T²≡a³
Let x be the new distance
(88÷55)²=(5.8×10∧10÷x)³
⇒x=14.8×10∧10
Answered by
2
Answer:
Explanation:
t1squre / t2 square = r1cube and r2 cube..... T1 = 88 ,T2 = 55, r1 = 5.8 *10 to the power ten and r2 =? Then, put the value in above formula and take tha answer...... 4.3 *10to the power 10m
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