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

the simple machine has three main features. they are:

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever