An object is $\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^4\mathbf{ly}$fromus and does not have any motion relative to us except for the motion due to the expansion of the universe. If the space between us and it expands according to Hubble’s law, with $\mathbf{H}\mathbf{=}\mathbf{21}\mathbf{.}\mathbf{8}\frac{\mathrm{mm}}{s}\cdot \mathbf{ly}$,

(a) how much extra distance (meters) will be between us and the object by this time next year and

(b) what is the speed of the object away from us?

Current distance of object from us

$r_i=1.5\times {10}^4\text{\hspace{0.17em}ly}$

According to Hubble's law, galaxies at a distance$r$ from Earth is moving away from Earth at a velocity

$\mathbf{v}\mathbf{=}\mathbf{Hr}$ ..... (I)

Here,$H$is the Hubble's constant with value

role="math" localid="1663087577539" $\mathbf{H}\mathbf{=}\mathbf{21}\mathbf{.}\mathbf{8}\mathbf{mm}\mathbf{/}\mathbf{s}\cdot \mathbf{ly}$

Assume the velocity to be constant in one year. From equation (I)

$\begin{array}{c}v=21.8\frac{\text{mm}}{\text{s}}\cdot \text{ly}\times 1.5\times {10}^4\text{\hspace{0.17em}ly}\\ =32.7\times {10}^4\cdot (1\text{\hspace{0.17em}mm}\times \frac{1\text{\hspace{0.17em}m}}{1000\text{\hspace{0.17em}mm}})\cdot \left(\frac{1}{1\text{\hspace{0.17em}s}}\right)\\ =327\frac{\text{m}}{\text{s}}\end{array}$

Distance receded by the object in one year.

$d=vt$

Substitute the values and solve as:

$\begin{array}{c}d=327\frac{\text{m}}{\text{s}}\times 1\text{\hspace{0.17em}y}\\ =327\cdot \left(1\frac{\text{m}}{\text{s}}\right)\cdot (1\text{\hspace{0.17em}y}\times \frac{3.154\times {10}^7\text{\hspace{0.17em}s}}{1\text{\hspace{0.17em}y}})\\ =1.03\times {10}^{10}\text{\hspace{0.17em}m}\end{array}$

Thus, the required distance is .$1.03\times {10}^{10}\text{\hspace{0.17em}m}$

(b)

The velocity has been considered constant in this small time interval. So the velocity will be$327\frac{\text{m}}{\text{s}}$ as obtained in the previous step.