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?

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