The asteroid belt circles the sun between the orbits of mars and jupiter. one asteroid has a period of 5.0 earth years. assume a 365.25-days year and msun = 1.99 * 1030 kg. what is the asteroid's orbital radius?

Using Kepler's 3rd law which is: T² = 4π²r³ / GM

Solved for r:

r = [GMT² / 4π²]⅓

Where G is the universal gravitational constant, M is the mass of the sun, T is the asteroid's period in seconds, andr is the radius of the orbit.

Change 5.00 years to seconds:

5.00years = 5.00years (365days/year) (24.0hours/day) (6 ... = 1.58 x 10^8s

The radius of the orbit then is computed:

r = [ (6.67 x 10^-11N∙m²/kg²) (1.99 x 10^30kg) (1.58 x 10^8s) ² / 4π²]⅓ = 4.38 x 10^11m