1.125 x 10⁶ g/cm³.
Given:
Question:
What will represent the density of our sun at the end of its lifetime? (in g/cm³)
The Process:
In the beginning, we calculate the volume of our sun which will end up like a white dwarf.
Let's assume the sun as a perfect sphere.
Prepare the radius, i.e., [tex]\boxed{ \ R = \frac{1}{2} \times diameter \ }[/tex]
[tex]\boxed{ \ R = \frac{1}{2} \times 15,000 \ km \ } \rightarrow \boxed{ \ 7,500 \ km \ }[/tex]
Volume of sphere [tex]\boxed{ \ V = \frac{4}{3} \pi R^3 \ }[/tex]
[tex]\boxed{ \ V = \frac{4}{3} \pi (7,500)^3 \ }[/tex]
We deliver the volume of the sun at the stage, i.e., [tex]\boxed{ \ V = 1.767 \times 10^{12} \ km^3 \ }[/tex]
Let us convert km³ to cm³ by multiplying [tex]\boxed{ \ (10^3)^5 \rightarrow 10^{15} \ }[/tex]
[tex]\boxed{ \ V = 1.767 \times 10^{12} \times 10^{15} \ cm^3 \ } \rightarrow \boxed{ \ V = 1.767 \times 10^{27} \ cm^3} \ }[/tex] \ }[/tex]
After preparing the volume, then we proceed with calculating its density. The formula of density is provided by [tex]\boxed{ \ Density = \frac{mass}{volume} \ }[/tex]
[tex]\boxed{ \ Density = \frac{1.989 x 10^{30} \ kg}{1.767 \times 10^{27} \ cm^3} \ }[/tex]
Let us convert kg to gram by multiplying 10³.
[tex]\boxed{ \ Density = \frac{1.989 x 10^{33} \ g}{1.767 \times 10^{27} \ cm^3} \ }[/tex]
Thus, the density of our sun at the end of its lifetime approximately will be [tex]\boxed{\boxed{ \ 1.125 \times 10^6 \ g/cm^3 \ }}[/tex]
Keywords: density, our sun will end up as a white dwarf, reduced to about 15,000 km in diameter, mass, volume of the sphere, in about 5 billion years, at the end of its lifetime
