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Suppose that the Milky Way galaxy is a typical size, containing say 10¹¹ stars, and that galaxies are typically separated by a distance of one megaparsec. Estimate the density of the Universe in SI units. How does this compare with the density of the Earth? 1M. – 2 x 10³⁰ kg, 1 parsec – 3 x 10¹⁶, m.

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Final answer:

The estimated density of the Universe is approximately 7.4 x 10^-26 kg/m^3, much lower than the density of the Earth 5,513 kg/m^3\, indicating the vast emptiness of intergalactic space.

Step-by-step explanation:

To estimate the density of the Universe, we consider a galaxy with 10ⁱ¹ stars and assume each star has the mass of our Sun, which is approximately 2 x 10³⁰ kg. Next, we calculate the volume that each galaxy would occupy if galaxies are spaced one megaparsec apart, which is about 10⁶ light-years. Since 1 parsec is 3 x 10¹⁶ m, a megaparsec is 3 x 10¹⁶ meters multipled by 10⁶, equaling 3 x 10²² m.

The volume of a sphere with a radius of one megaparsec is (4/3)π×(3 x 10²² m)³. We then divide the total mass of galaxies (10ⁱ¹ stars × 2 x 10³⁰ kg/star) by this volume to find the average density of matter in the Universe.

Comparing this with the density of the Earth (which is approximately 5.51 x 10³ kg/m³), we find that the cosmic density is much less than the Earth's density. This implies that, on a cosmic scale, matter is incredibly spread out, making the Universe predominantly empty space.

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