Answer:
The diameter is [tex]\approx[/tex][tex]10^{11}m[/tex] and [tex]r_{sun} = 10^{9}[/tex]m
Explanation:
Given:
[tex]m_{sun} = 10^{30} kg[/tex]
[tex]\\p_{sun}= 10^{3}\frac{kg}{m^{3}}[/tex]
Required:
[tex]r_{sun}[/tex]=?
d= ?
We know,
[tex]{v_{sun} = \frac{m_{sun}}{p_{sun}}[/tex]
= [tex]\frac{10^{30}}{10^{3}}[/tex]
= [tex]10^{27}m^{3}[/tex]
[tex]v_{sun} = \frac{4}{3}\times\pi r^{3_{sun}}[/tex]
[tex]\approx[/tex][tex]r^{3}_{sun}[/tex]
[tex]r_{sun} = 10^{9}[/tex]m
From the figure attached,
[tex]sin\theta=\frac{r_{sun}}{d}}[/tex]
[tex]d=\frac{r_{sun}}{sin\theta}}[/tex]
=[tex]\frac{10^{9}}{sin5}}[/tex]
[tex]\approx[/tex][tex]10^{11}m[/tex]
The diameter is [tex]\approx[/tex][tex]10^{11}m[/tex] and [tex]r_{sun} = 10^{9}[/tex]m