Answer:
21.54 × 10³⁵ kg
Explanation:
Given:
Period = 5 years
semi-major axis = 4.5 × 10⁸ km
Now,
According to the Kepler's third law
we have,
[tex]T^2=\frac{4\pi^2}{GM}r^3[/tex]
Here,
T is the period
G is the gravitational force constant = 6.67 × 10⁻¹¹ m³/kg⋅s²
M is the mass of the star
on substituting the respective values, we get
[tex]5^2=\frac{4\pi^2}{6.67\times10^{-11}M}(4.5\times10^8)^3[/tex]
or
M = 21.54 × 10³⁵ kg
