Question

If the sun expanded to a radius 100 times its present radius, what would its density be? (Hint: The volume of a sphere is 43πR3.)\frac{4}{3} \pi R^3.)

Solution

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The density formula is \text{\textcolor{#4257b2}{The density formula is }}

Density(ρ)=mass(m)/volume(V)\boxed{Density (\rho) = mass(m) / volume(V)}

The mass of the Sun is \text{\textcolor{#c34632}{The mass of the Sun is }} M=1.99×1030M_{\odot}=1.99 \times 10^{30} kg =1.99×1033=1.99 \times 10^{33} grams

The radius of the Sun is\text{\textcolor{#c34632}{The radius of the Sun is}} R=6.96×105R_{\odot}=6.96 \times 10^5 km =6.96×1010=6.96 \times 10^{10} cm,

V=43πR3=(43)(π)(6.96X1010cm)3V =\frac{4}{3} \pi R_{\odot}^3 = (\frac{4}{3})(\pi)(6.96 X 10^{10}\hspace{2mm}cm)^3

.\text{\color{white}.}      =1.41×1033cm3= 1.41 \times 10^{33}\hspace{2mm} cm^3

ρ=M/V=(1.99×1033gm)/(1.41×1033cm3\rho = M / V = (1.99 \times 10^{33} \hspace{2mm} gm) / (1.41 \times 10^{33} \hspace{2mm} cm^3

=1.41gm/cm3= 1.41\hspace{2mm} gm/cm^3

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