alleryStart learning →Mercury orbits the sun with a period of 88.0 days at an average radius of 5.791 x 10^10 m. Using r^3/T^2 = GM/(4 pi^2), what is the mass of the sun?
A3.43 x 10^19 kg
B1.99 x 10^30 kg
C2.56 x 10^29 kg
D1.48 x 10^40 kg
Answer & Solution
Correct answer: B. 1.99 x 10^30 kg
1. Convert the period: T = 88.0 d x 86,400 s/d = 7.603 x 10^6 s, so T^2 = 5.781 x 10^13 s^2.
2. Cube the radius: r^3 = (5.791 x 10^10)^3 = 1.942 x 10^32 m^3.
3. Rearranged, M = 4 pi^2 r^3 / (G T^2) = (39.48)(1.942 x 10^32) / [(6.674 x 10^-11)(5.781 x 10^13)].
4. M = 7.667 x 10^33 / 3.859 x 10^3 = 1.99 x 10^30 kg.
_Source: OpenStax Physics (CC BY 4.0), Ch 7 "Newton's Law of Gravitation", section 7.1 Kepler's Laws of Planetary Motion_
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