Homework Help: physics

Posted by Sandhya on Tuesday, June 16, 2009 at 5:16am.

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.

physics - drwls, Tuesday, June 16, 2009 at 6:18am
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

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