Any units can be used because the 3rd law is couched in proportional terms. It's common to use astronomical units, with the Earth's distance equal to 1.
You can find the major axis, 0.5+31.5 or 32 AU. The semimajor axis is half that, 16 AU. Then you can use Keplers 3rd law to calculate the period, which is 161.5 or 64 years.
The major axis is the diameter across the widest part. The semimajor axis is half that, and for a planet it's the average of the maximum and minimum distances from the Sun .
That can be calculated from Kepler's 3rd law which says if the period is T years the semimajor axis must be T2/3 astronomical units. So for a period of 12 years the s/m axis is 5.421 AU or 784 million km.
jkjnl
The planet is at minimum and maximum distances when it is at either end of the major axis. In this case the major axis is 0.5 AU so the semimajor axis is 0.25 AU. The eccentricity is found by noting that the Sun is 0.15 AU displaced from the centre, so the eccentricity is 0.15/0.25 or 0.6. The semiminor axis is equal to the semimajor axis times sqrt(1 - e-squared) which in this case is equal to 0.2.
One of the parts of an ellipse is the length of its major axis. Half that is called the semimajor axis. Kepler's 3rd law says that the time to do one orbit is proportional to the 3/2 power of the semimajor axis. IF the semimajor axis is one astronomical unit the period is one year (the Earth). For a planet with a semimajor axis of 4 AUs the period would have to be 8 years, by Kepler-3.
You can find the major axis, 0.5+31.5 or 32 AU. The semimajor axis is half that, 16 AU. Then you can use Keplers 3rd law to calculate the period, which is 161.5 or 64 years.
The major axis is the diameter across the widest part. The semimajor axis is half that, and for a planet it's the average of the maximum and minimum distances from the Sun .
That can be calculated from Kepler's 3rd law which says if the period is T years the semimajor axis must be T2/3 astronomical units. So for a period of 12 years the s/m axis is 5.421 AU or 784 million km.
jkjnl
The major and minor axes of a circle are the same - either is any diameter. So a semimajor axis is half the diameter which is 12 cm.
One of the parts of an ellipse is the length of its major axis. Half that is called the semimajor axis. Kepler's 3rd law says that the time to do one orbit is proportional to the 3/2 power of the semimajor axis. IF the semimajor axis is one astronomical unit the period is one year (the Earth). For a planet with a semimajor axis of 4 AUs the period would have to be 8 years, by Kepler-3.
the period of revolution is related to the semimajor axis.... :)
(I'm going to assume that when you said "first" you meant "fastest," because otherwise the question is nonsense.) Because of Kepler's Third Law. The orbital period for a body is related to the semimajor axis of its orbit. Mercury's orbit has the shortest semimajor axis of all the Solar planets, and therefore it has the shortest orbital period.
Its average distance from the Sun (to be more precise, its semimajor axis) is about 46 AU.
The planet is at minimum and maximum distances when it is at either end of the major axis. In this case the major axis is 0.5 AU so the semimajor axis is 0.25 AU. The eccentricity is found by noting that the Sun is 0.15 AU displaced from the centre, so the eccentricity is 0.15/0.25 or 0.6. The semiminor axis is equal to the semimajor axis times sqrt(1 - e-squared) which in this case is equal to 0.2.
The coordinates of a point two units to the right of the y-axis and three units above the x-axis would be (2,3).