Question

The asteroid belt circles the sun between the orbits of mars and jupiter. one asteroid has a period of 6.2 earth years. what are the asteroid's orbital radius and speed?

Answer 1

The orbital radius of the asteroid in the asteroid belt can be determined using Kepler's third law. The speed of the asteroid can be calculated using the formula v = 2 ×π × R / T.

Step-by-step explanation:

The orbital radius of an asteroid can be determined using Kepler's third law, which relates the square of the orbital period to the cube of the orbital radius: T^2 = k *× R^3, where T is the period in years, R is the orbital radius in astronomical units (AU), and k is a constant. Plugging in the values given in the question, we can solve for the orbital radius: 6.2^2 = k × R^3,

Since the asteroid is in the asteroid belt, we know that its orbital radius is between 2.2 and 3.3 AU, so we can choose a value within that range that satisfies the equation.

To calculate the speed of the asteroid, we can use the formula: v = 2 ×π × R / T., where v is the speed in km/s, R is the orbital radius in km, and T is the period in seconds. By plugging in the values we found for R and T, we can calculate the speed of the asteroid.

Answer 2

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