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An asteroid has a mean radius equal to 17 times that of Earth. Using Kepler's 3rd law find its period. T^2=a^3

Sagot :

Answer:

If T^2 = a^3

T1^2 / T2^2 = a1^3 / a2^3

Or T2^2 = T1^2 * ( a2^3 / a1^3)

Given T1 = 1 yr and a2 / a1 = 17

Then T2^2 = 1 * 17^3

or T2 = 70 yrs

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