An asteroid orbits the Sun at an average distance of a= 4 AU. How long does it take to orbit the Sun?Use Kepler’s Third Law(p2= a3) to calculate the answer.

7982 seconds

Explanation:

Parameters given:

Distance of asteroid, a = 4 AU = 5.984 * 10^8 m

Mass of sun, M = 1.9891 * 10^30 kg

Orbital period, P, is given as:

P² = (4π²a³) /(GM)

Where G = 6.6774 * 10^(-11) m³/kgs²

P² = [4 * π² * (5. 984 * 10^8)³]/(6.674 * 10^(-11) * 1.9891 * 10^30)

P² = (8.459 * 10^27)/(1.328 * 10^20)

P² = 6.372 * 10^7

=> P = 7982 seconds

like poles repel so keep pushing north to north and repels funny to see this

probably not the "magnets"in the magnet, so to speak, are thought to be electrons orbiting in shells. vurrent loops two sides to the loop