# The diameter of the solar system is approximately: 7,500,000,000 miles. How long would it take to drive this distance if traveling 60 mph?

Nov 3, 2017

14.26 millenia, or 125,000,000 hours.

#### Explanation:

When we're dealing with numbers this large, it can help to convert them to scientific notation before performing calculations with them. $7 , 500 , 000 , 000$ is $7.5 \times {10}^{9}$ in scientific notation, and $60$ is simply $6 \times 10$. To find the time it would take to travel $7.5 \times {10}^{9}$ miles, we divide it by the rate of $6 \times 10$ mph, obtaining:

(7.5times10^9 \ "mi")/(6times10\ "mi/hr")=7.5/6times10^8\ "hr"

We find that $\frac{7.5}{6}$ gives us $1.25$, leaving us with $1.25 \times {10}^{8}$ or $125 , 000 , 000$ hours. We could stop there, but to get a feel for just how long this is, it'd help to convert it to a more sensible timescale.

Let's first convert those hours to years. To make the conversion, we'll use the unit rates of $\left(1 \setminus \text{day")/(24\ "hr}\right)$ and $\left(1 \setminus \text{yr")/(365\ "days}\right)$:

Hours to days:

1.25times10^8\ "hr"*(1\ "day")/(24\ "hr")=
=(1.25times10^8\ cancel("hr"))/(24\ cancel("hr"))*1\ "day"
$= \frac{1.25}{24} \times {10}^{8} \setminus \text{days}$

Days to years:

1.25/24times10^8\ "days"*(1\ "yr")/(365\ "days")=
=1.25/24times10^8\ cancel("days")*1/(365\ cancel("days"))*1\ "yr"
$= \frac{1.25}{24 \cdot 365} \times {10}^{8} \setminus \text{yr}$

We can rewrite $24$ and $365$ in scientific notation as $2.4 \times 10$ and $3.65 \times {10}^{2}$, obtaining

$\frac{1.25}{2.4 \times 10 \cdot 3.65 \times {10}^{2}} \times {10}^{8} \setminus \text{yr} =$
$= \frac{1.25}{2.4 \cdot 3.65} \times {10}^{8} / {10}^{3} \setminus \text{yr}$
$= \frac{1.25}{8.76} \times {10}^{5} \setminus \text{yr}$

Finally, we can use the unit rate of $\left(1 \setminus \text{millennium")/(1000\ "yr}\right)$ (or equivalently, $\left(1 \setminus \text{millennium")/(10^3\ "yr}\right)$) to get our answer in millennia:

1.25/8.76times10^5\ "yr"*(1\ "millennium")/(10^3\ "yr")=
=1.25/8.76times10^5\ cancel("yr")*1/(10^3\ cancel("yr"))*1\ "millennium"
$= \frac{1.25}{8.76} \times {10}^{5} / {10}^{3} \setminus \text{millennia}$
$= \frac{1.25}{8.76} \times {10}^{2} \setminus \text{millennia}$

$\frac{1.25}{8.76} \approx 0.1426$, so our answer in millennia is approximately $0.1426 \times {10}^{2} = 14.26 \setminus \text{millennia}$.