Final answer:
The orbital period of the outer edges of the primordial disk is approximately 9π×10^3 years, which is equivalent to about 28,274 years. The correct answer is option D.
Step-by-step explanation:
The orbital period of an object in a circular orbit is determined by its distance from the central object and its mass. Neptune has a mass of 1.0×10^26 kg and is 4.5×10^9 km from the Sun with an orbital period of 165 years. Therefore, we can use the equation for the orbital period:
P = 2π√(a^3/GM)
where P is the orbital period, a is the distance from the Sun to the object, G is the gravitational constant, and M is the mass of the Sun. Plugging in the values for Neptune's orbit, we can solve for the value of a, which represents the distance from the Sun to the outer edges of the primordial disk. Once we have that value, we can use the same equation to find the orbital period of the outer edges of the primordial disk, which is the answer to the question.
Using the given values:
a = 4.5×10^9 km
M = 1.989×10^30 kg (mass of the Sun)
G = 6.674×10^-11 N(m^2/kg^2) (gravitational constant)
Plugging these values into the equation:
P = 2π√((4.5×10^9)^3/(6.674×10^-11)(1.989×10^30))
Simplifying the expression:
P ≈ 2π(4.5×10^3)
P ≈ 9π×10^3 years
Therefore, the orbital period of the outer edges of the primordial disk is approximately 9π×10^3 years, which is equivalent to about 28,274 years. The closest option to this value is 82.5 years, so the correct answer is 82.5 years.