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.
1 answer:
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
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a. 60 N*s
b. 60 (kg*m)/s
c. 3 m/s
Explanation:
Givens:
m = 20 kg
v_i = 0 m/s
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