The biggest known star (until June 2019) is VY Canis Majoris in the constellation of Canis Major, with an
estimated diameter of 1,975,000,000 km. Making the (absurd) assumption that a commercial jet could fly close
to the surface of the star at a constant speed of 990 km / h. How long would the jet take for a spin on the
star. Answer in years.
Problem data:
- star Diameter: D = 1,975,000,000 km;
- jet Speed: v = 990 km/h.
Problem diagram:
Solution:
The distance that the airplane should travel is the circumference of the star. The length of a circle is given by
\[
\begin{gather}
C=2\pi r
\end{gather}
\]
where, r is the radius of the circumference, the diameter D of a circle being equal to
2r, the length of a circle can also be calculated by
\[
\begin{gather}
C=\pi D
\end{gather}
\]
using, π = 3.14 the circumference of the star will be
\[
\begin{gather}
C=3.14 \times 1975000000\;\mathrm{km} \\[5pt]
C=6201500000\;\mathrm{km}
\end{gather}
\]
As the plane speed is constant, we have that travel time is given by
\[
\begin{gather}
\bbox[background:#99CCFF; padding:10px; border-radius:10px;]
{v=\frac{\Delta S}{\Delta t}}
\end{gather}
\]
\[
\begin{gather}
\Delta t=\frac{\Delta S}{v}
\end{gather}
\]
using the speed value given in the problem and the distance being the length of the circumference calculated above
\[
\begin{gather}
\Delta t=\frac{6201500000\;\mathrm{\cancel{km}}}{990\;\mathrm{\frac{\cancel{km}}{h}}} \\[5pt]
\Delta t=6264141\;\mathrm{h}
\end{gather}
\]
Converting this value to years as it asks for the problem
\[
\begin{gather}
\Delta t=6264141\;\cancel{\mathrm{hours}}\times\frac{1\;\cancel{\mathrm{day}}}{24\;\cancel{\mathrm{hours}}}\times\frac{1\;\mathrm{year}}{365\;\cancel{\mathrm{days}}}
\end{gather}
\]
\[
\begin{gather}
\bbox[background:#FFCCCC; padding:10px; border-radius:10px;]
{\Delta t=715\;\mathrm{years}}
\end{gather}
\]
Note: for comparison, the Sun is 1,391,000 km in diameter, under the same conditions, the
airplane would take "only" 184 days to go around the Sun.