Question

E asteroid belt circles the sun between the orbits of mars and jupiter. one asteroid has a period of 6.0 earth years. part a what is the asteroid's orbital radius?

Answer 1

To solve the problem, use Kepler's 3rd law :

T² = 4π²r³ / GM

Solved for r :

r = [GMT² / 4π²]⅓

but first covert 6.00 years to seconds :

6.00years = 6.00years(365days/year)(24.0hours/day)(6...
= 1.89 x 10^8s

The radius of the orbit then is :

r = [(6.67 x 10^-11N∙m²/kg²)(1.99 x 10^30kg)(1.89 x 10^8s)² / 4π²]⅓
= 6.23 x 10^11m