Given,
The orbital diameter is
[tex]d=1.34\times10^9m[/tex]
The radius is
[tex]r=\frac{d}{2}=\frac{1.34\times10^9}{2}=0.67\times10^9m[/tex]
The time period is T=3.55 days=3.55x86400 sec.
We know,
[tex]\begin{gathered} T=\frac{2\pi r^{\frac{3}{2}}}{\sqrt[\square]{GM}}^{}^{} \\ \Rightarrow3.55\times86400=\frac{2\pi\text{ (0.67}\times10^9)^{\frac{3}{2}}}{\sqrt[]{6.67\times10^{-11}\times M}} \\ \Rightarrow9.4\times10^{10}=\frac{1.18\times10^{28}}{6.67\times10^{-11}M} \\ \Rightarrow M=1.89\times10^{27}kg \end{gathered}[/tex]
The answer is
[tex]1.89\times10^{27}kg[/tex]
