Answer :
The mass of the large star based on orbital period is 6912 x 10³⁰ kg.
We need to know about the orbital period to solve this problem. The orbital period depends on the semi-major axis, gravitational constant, and the mass of the massive object. It can be written as
T² = 4π².a³ / GM
where T is the orbital period, a is the semi-major axis, G is the gravitational constant and M is the mass.
From the question above, the given parameters are
a = 24 au = 24 . ³√(G.Msun yr²/ 4π²)
T = 2 years
Msun = 2 x 10³⁰ kg
By substituting the given parameters, we get
T² = 4π².a³ / GM
2² = 4π². (24 . ³√(G.Msun yr²/ 4π²))³ / GM
4 = 13824 Msun / M
M = 6912 x 10³⁰ kg
Find more on orbital period at: https://brainly.com/question/22247460
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