Question

Calculate the orbital period of an asteroid with a semimajor axis of a=5.0E11 meters in seconds and years.

Answer 1

Solution :

The semi-major axis of any terrestrial body may be defined as equal to mean distance of the body, so that the cube of its mean distance of the planet is directly proportional to the square of the sidereal period.

It is given by Kepler's third law,

$p^2 = a^3$

where p is the orbital period

           a is the semi major axis

Given :

a = $5 \times 10^{11}$ m

Therefore, a = $5 \times 10^{11}$ m

                     = 3.3 AU

So, $p^2=3.3^{3}$

     $p^2=35.93$

     p = 6 years (approximately)

or   p = $1.8 \times 10^8$ seconds    ( since, 1 year = $3.1 \times 10^7$ seconds )