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For the solution to question $a$ of this problem, I'm confused with what that $3.0\times 10^8 m/s$ conversion factor is. I thought that light year is a measurement of distance, so why are we multiplying it by m/s where $s$ is a unit of time?

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There is a subtle unit conversion going on.
Recall that $(1 {\rm\ ly})=(3\times 10^8 {\rm\ m/s})(1 {\rm\ y})$, the distance traveled by light in one year.

By substitution, which may be what they meant but didn't do correctly [so your complaint and confusion is warranted] \begin{eqnarray*} (2.3 \times 10^4 {\rm\ ly}) &=& 2.3 \times 10^4\ (1 {\rm\ ly}) \\ &=& 2.3 \times 10^4\ (3\times 10^8 {\rm\ m/s})(1 {\rm\ y})\\ &=& 6.9 \times 10^{12} {\rm\ (m/s)y} \\ &=& 6.9 \times 10^{12} {\rm\ m\ (y/s)} \end{eqnarray*} Note that this is a distance [in some number of meters] because $(y/s)$ is a dimensionless number.

Essentially, you are trading
the "l" for "light-" (in light-year)
for "$(3\times 10^8{\rm\ m/s})$" in "$(3\times 10^8{\rm\ m/s})\cdot y$".


Another way to do this is by "fancy-multiplication by 1".
Since $(1 {\rm\ ly})=(3\times 10^8 {\rm\ m/s})(1 {\rm\ y})$, then $$1 = \frac{ (3\times 10^8 {\rm\ m/s})(1 {\rm\ y})}{ \rm\ ly}.$$

So, by "fancy-multiplication by 1"

\begin{eqnarray*} (2.3 \times 10^4 {\rm\ ly}) &=& (2.3 \times 10^4 {\rm\ ly})(1) \\ &=& (2.3 \times 10^4 {\rm\ ly}) \frac{ (3\times 10^8 {\rm\ m/s})(1 {\rm\ y})}{ \rm\ ly} \\ &=& (2.3 \times 10^4 ) (3\times 10^8 {\rm\ m/s})(1 {\rm\ y}) \\ \end{eqnarray*}

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  • $\begingroup$ The "fancy multiplication by 1" trick can at times be an extremely handy trick. $\endgroup$ – David Hammen May 6 at 5:35
  • $\begingroup$ As can be the "fancy adding 0" trick. $\endgroup$ – David Hammen May 6 at 5:43

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