posted by Sandhya on .
Find the period of revolution of an artificial planet if the semi-major axis of the planet’s elliptic orbit is greater than that of the Earth’s orbit by 24 x 10^6 km. It is given that the semi-major axis of earth orbit round the sum is 1.5 x 10^8 km.
The ratio of orbit semimajor axis lengths is
a2/a1 = 1.524*10^8/1.50*10^8 = 1.016
For the earth, the period is P1 = 365.25 days.
Use Kepler's third law to get the period for the artificial planet.
a^3/P^2 = constant
P2/P1 = (a2/a1)^3/2 = 1.0241
P2 = 374 days