The distance between the perihelion and aphelion of the sun is 10A.U. What is its orbital period in term of years?
The period will be 11.18 years.
The distance between perihelion and aphelion is twice the semi major axis distance. So the semi major axis
Kepler's third law relates the period
As of now, AU = 149597871 km, nearly, and the distance referred is 2 AU. If AU becomes 5 AU, for the same orbital speed, the period might become 5 years.
Approximately, the period is
(circumference) /( average orbital speed )
= (2 pi AU km)/(29.78 km/s}
= 365.3 d, nearly
= 1 year...
When AU becomes 5 AU, for the same orbital speed, period
becomes 5 years. This phenomenon is virtual.,
Theoretically, we can have infinite number of orbits for a = 5 AU.
The parameter eccentricity e of the orbit is arbitrary. Also, Jupiter
has a = 5.20 AU. Its period is 11.87 years. When the distance is in
AU units and time in Earth-year units,
the dimensional constant of proportionality
in Kepler's third law ( in the normalized form),
(Period in year unit )^2 = (Semi-major axis 1 AU )^3
= 1 (year.year)/(AU.AU.AU).
Indeed, this ( Orbit Period ) - (Mean Sun-Earth distance) near-
exactitude relation is quite handy for use.
By dimensional analysis, the period, of dimension T (time), can be
related to any distance a of dimension L, by including a constant of
proportionality. This is necessary, for dimensional homogeneity.. .