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If you know any two of the following you can calculate the third: -Period of planet's revolution -Orbital Speed -Distance from sun
According to Kepler's law, the expression P2/a3 is approx equal to 4pi2/GM where P = period,a = average orbital distance = 0.39 AU = 58,343,169,871 metresG = universal gravitational constant = 6.67408*10^-11 m3 kg-1 s-2M = mass of the Sun = 1.989*1030 kilograms.Substituting these values into the expression and solving givesP2 = 59061382597774 seconds2 which implies that P = 7685140 seconds. This equals 88.9 Earth days.
Quaoar's orbital period is 287.97 years or 105,101
Yes, the equation p2 = a3, where p is a planet's orbital period in years and a is the planet's average distance from the Sun in AU. This equation allows us to calculate the mass of a distance object if we can observe another object orbiting it and measure the orbiting object's orbital period and distance.
Uranusus orbital period is 32.234 Earth days
Here is the calculation:P squared = A cubedP squared = 3.36 cubedP squared = 37.9330P = 6. 1589
If you know any two of the following you can calculate the third: -Period of planet's revolution -Orbital Speed -Distance from sun
Orbital information. You need to know the size of the "semi-major axis". Then you can calculate the orbital period, using Kepler's Third Law.
According to Kepler's law, the expression P2/a3 is approx equal to 4pi2/GM where P = period,a = average orbital distance = 0.39 AU = 58,343,169,871 metresG = universal gravitational constant = 6.67408*10^-11 m3 kg-1 s-2M = mass of the Sun = 1.989*1030 kilograms.Substituting these values into the expression and solving givesP2 = 59061382597774 seconds2 which implies that P = 7685140 seconds. This equals 88.9 Earth days.
Its orbital period is 10,832.327 days long.
According to Kepler's law, the expression P2/a3 is approx equal to 4pi2/GM where P = period,a = average orbital distance = 0.39 AU = 58,343,169,871 metresG = universal gravitational constant = 6.67408*10^-11 m3 kg-1 s-2M = mass of the Sun = 1.989*1030 kilograms.Substituting these values into the expression and solving givesP2 = 59061382597774 seconds2 which implies that P = 7685140 seconds. This equals 88.9 Earth days.
The orbital period of Jupiter is 4332.71 days.
Venus' orbital period is 224.7 earth days.
2007or10's orbital period is 552.52 years
Haumea's orbital period is 283 or 103,468 days
Quaoar's orbital period is 287.97 years or 105,101
Yes, the equation p2 = a3, where p is a planet's orbital period in years and a is the planet's average distance from the Sun in AU. This equation allows us to calculate the mass of a distance object if we can observe another object orbiting it and measure the orbiting object's orbital period and distance.