# The distance between the perihelion and aphelion of the sun is 10A.U. What is its orbital period in term of years?

Jan 31, 2016

The period will be 11.18 years.

#### Explanation:

The distance between perihelion and aphelion is twice the semi major axis distance. So the semi major axis $a = 5 A U$.

Kepler's third law relates the period $T$ to the semi-major axis by the formula ${T}^{2} = {a}^{3}$. So ${T}^{2} = {5}^{3} = 125$. Taking the square root gives $T = 11.18$ years.

Jan 31, 2016

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.

#### Explanation:

Approximately, the period is

(circumference) /( average orbital speed )

= (2 pi AU km)/(29.78 km/s}

=31563169 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.. .