Mercury orbits the Sun (mass = 1.99 x 1030 kg) at a distance = of 5.79 x 1010 m. What is its orbital period (in days)?

Answer 1

GMm/r² = mv²/r, where G is gravitational constant, M mass of Sun, m is mass of Mercury, r is distance from

Mercury to Sun, v is velocity of Mercury. From here v = (GM/r)^1/2. = (6.67408·10^-11·1.99·10³⁰/(5.79·10^10))^1/2 =

4.79·10⁴m/s

Then orbital period of Mercury T = 2πr/v = 2π·5.79·10¹⁰m/(4.79·10⁴m/s) = 7.595·10^{6 s} = 88 days

Answer 2

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