A planet of mass m = 2.55 × 10^24 kg orbits a star of mass M = 5.95 × 10^29 kg in a circular path. The radius of the orbit is R = 1.65 × 10^7 km. What is the orbital period Tplanet of the planet in Earth days?
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So, r = 1.65 × 10^7 km = 1.65 × 10^7 × 10^3 m = 1.65 × 10^10 m. The gravitational force between the star and the planet provides the centripetal force that keeps the planet in its circular orbit. This force is given by F = GmM/r^2, where G is the gravitational Show more…
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