Answer to Question #276628 in Physics for Hutchii

Question #276628

Solve the problem involving the third law of Kepler's planetary motion with complete solutions. The mean distance of an object A from the Sun is 19.6 x 10^9 km and the mean distance of object B from the Sun is 57.9 x 10^8 km. The period of Earth's revolution is 1 year, what is the period of objects B's revolution?

1
Expert's answer
2021-12-21T12:17:18-0500

Let's apply the Third Kepler's Law of planetary motion:


PB2aB3=PE2aE3,\dfrac{P_B^2}{a_B^3}=\dfrac{P_E^2}{a_E^3},PB=PE2aB3aE3,P_B=\sqrt{\dfrac{P_E^2a_B^3}{a_E^3}},PB=(1 yr)2×(57.9×108 km)3(1.5×108 km)3=240 yr.P_B=\sqrt{\dfrac{(1\ yr)^2\times(57.9\times10^8\ km)^3}{(1.5\times10^8\ km)^3}}=240\ yr.

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