Respuesta :
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 orbital period of the given asteroid is 7982 s. The square of the orbital period is directly proportional to the major axis radius of the planet.
What does Kepler’s Third Law state?
The square of the orbital period of the planet is directly proportional to the major axis radius of the planet.
[tex]P^2 = \dfrac {4\pi ^2a^3}{GM}[/tex]
Where,
[tex]P[/tex] - period
[tex]a[/tex] - distance = 4 AU =[tex]\bold{ 5.984 \times 10^8\ m}[/tex]
[tex]G[/tex] - gravitational constant = [tex]\bold{6.6774 \times 10^{-11} m^3/kgs^2}[/tex]
[tex]M[/tex] - mass of the sun = [tex]\bold { 1.9891 \times 10^30 \ kg}[/tex]
Now put the values in the formula,
[tex]P^2 = \dfrac {4 \pi^2 \times (5. 984 \times 10^8)^3}{(6.674 \times 10^{-11} \times 1.9891 \times 10^{30})}\\\\P = 7982 \rm \ s[/tex]
Therefore, the orbital period of the given asteroid is 7982 s.
Learn more about Kepler’s Third Law state:
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