T² = 4π²rÂł / GM Solved for r : r = [GMT² / 4π²]â…“ Where G is the universal gravitational constant, M is the mass of the sun, T is the asteroid's period in seconds, and r is the radius of the orbit. Covert 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 : r = [(6.67 x 10^-11Nâ™m²/kg²)(1.99 x 10^30kg)(1.58 x 10^8s)² / 4π²]â…“ = 4.38 x 10^11m The orbital speed can be found from the circular velocity formula : v = âš[GM / r] = âš[(6.67 x 10^-11Nâ™m²/kg²)(1.99 x 10^30kg) / 4.38 x 10^11m = 1.74 x 10^4m/s
Answer: 3 to 6 years Explaination: Most of them occupy a great ring, known as the asteroid belt, at mean distances of 2.2 to 3.3 AU from the Sun and with orbital periods of 3 to 6 years.
