Answer:
0.03906 Au
Explanation:
T = Time period
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
M = Mass of planet
From Kepler's third law
[tex]T^2=\frac{4\pi^2 R^3}{GM}\\\Rightarrow R^3=\frac{GMT^2}{4\pi^2}\\\Rightarrow R=\left(\frac{GMT^2}{4\pi^2}\right)^{1/3}\\\Rightarrow R=\left(\frac{6.67\times 10^{-11}1.1\times 1.99\times 10^{30}(2.7\times 24\times 60\times 60)^2}{4\pi^2}\right)^{1/3}\\\Rightarrow R=5860332948.35716\ m[/tex]
[tex]1\ Au=1.5\times 10^{11}\ m[/tex]
[tex]1\ m=\frac{1}{1.5\times 10^{11}}[/tex]
[tex]\\\Rightarrow 5860332948.35716\ m=\frac{5860332948.35716}{1.5\times 10^{11}}=0.03906\ Au[/tex]
The distance between the star and the planet is 0.03906 Au.
