Answer: Jupiter's mass
Explanation:
From Kepler's third law:
where T is the orbital period of a satellite, a is the average distance of the satellite from the Planet, M is the mass of the planet, G is the gravitational constant.
If the average distance of one of Jupiter's moons to Jupiter and its orbital period around Jupiter is given then mass of the Jupiter can be found:
A type O star is likely to appear blue.
Given: Mass of earth Me = 5.98 x 10²⁴ Kg
Radius of earth r = 6.37 x 10⁶ m
G = 6.67 x 10⁻¹¹ N.m²/Kg²
Required: Smallest possible period T = ?
Formula: F = ma; F = GMeMsat/r² Centripetal acceleration ac = V²/r
but V = 2πr/T
equate T from all equation.
F = ma
GMeMsat/r² = Msat4π²/rT²
GMe = 4π²r³/T²
T² = 4π²r³/GMe
T² = 39.48(6.37 x 10⁶ m)³/6.67 x 10⁻¹¹ N.m²/Kg²)(5.98 x 10²⁴ Kg)
T² = 1.02 x 10²² m³/3.99 x 10¹⁴ m³/s²
T² = 25,563,909.77 s²
T = 5,056.08 seconds or around 1.4 Hour
