Calculating Period of Mars Using Kepler's Law: Appendix C

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To calculate the orbital period of Mars using Kepler's Law, the mean distance of Mars from the Sun, which is 1.52 times that of Earth's distance, is essential. By applying the simplified version of Kepler's Law, T^2 = a^3, where 'a' is the semi-major axis in astronomical units, the calculation can be performed. This results in a period of approximately 1.88 years for Mars, which aligns with the value provided in Appendix C. Understanding the reduction of the original equation is crucial for solving the problem effectively. This approach allows for a clear comparison of calculated and referenced values.
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The mean distance of Mars from the Sun is 1.52 time that of Earth from the Sun. From Kepler's law of periods, calculate the number of years required for Mars to make one revolution around the Sun; compare your answer with the value given in Appendix C.


Kepler's Law of periods is: T^2=[4 x (pi)^2 x r^3]/[GM]

Appendix C says the Period=1.88 y

I have no idea how to begin this problem. Please help set me in the right direction.
 
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