If the Hubble constant is 25 km/s per million light-years, use it to estimate the age of the universe? (c = 3.00 × 10⁸ m/s, 1 ly = 9.461 × 10¹⁵ m, 1 y = 3.156 × 10⁷ s) A) 10 Gy B) 12 Gy C) 15 Gy D) 20 Gy E) 25 Gy

Question

Quizplus · Accepted Answer

The Hubble constant is given by:$$H_0 = \frac{v}{d}$$where:* $H_0$ is the Hubble constant* $v$ is the velocity of the object* $d$ is the distance to the objectWe can rearrange this equation to solve for the distance to the object:$$d = \frac{v}{H_0}$$We are given that the Hubble constant is 25 km/s per million light-years. We can convert this to units of m/s per m:$$H_0 = 25 \frac{\text{km}}{\text{s per million light-years}} = 7.92 \times 10^{-18} \frac{\text{m}}{\text{s per m}}$$We are also given that the speed of light is $3.00 \times 10^8 \text{ m/s}$ and that 1 light-year is $9.461 \times 10^{15} \text{ m}$. We can use these values to convert the distance to the object to units of meters:$$d = \frac{v}{H_0} = \frac{3.00 \times 10^8 \text{ m/s}}{7.92 \times 10^{-18} \text{ m/s per m}} = 3.79 \times 10^{25} \text{ m}$$The age of the universe is given by:$$t = \frac{d}{c}$$where:* $t$ is the age of the universe* $d$ is the distance to the object* $c$ is the speed of lightWe can substitute the values we have calculated into this equation to solve for the age of the universe:$$t = \frac{d}{c} = \frac{3.79 \times 10^{25} \text{ m}}{3.00 \times 10^8 \text{ m/s}} = 1.26 \times 10^{17} \text{ s}$$We can convert this to units of years:$$t = 1.26 \times 10^{17} \text{ s} = 4.00 \times 10^{9} \text{ y} = 12 \text{ Gy}$$Therefore, the age of the universe is approximately 12 billion years.

If the Hubble Constant Is 25 Km/s Per Million Light-Years