Solved Problem on One-dimensional Motion
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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:

Figure 1

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
\[ \begin{gather} C=\pi D \end{gather} \]
uaing, 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
\[ \begin{gather} \bbox[#99CCFF,10px] {v=\frac{\Delta S}{\Delta t}} \end{gather} \]
\[ \begin{gather} \Delta t=\frac{\Delta S}{v} \end{gather} \]
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
\[ \begin{gather} \Delta t=6264141\;\cancel{\mathrm{hours}}\times\frac{1\;\cancel{\mathrm{day}}}{24\;\cancel{\mathrm{hours}}}\times\frac{1\;\mathrm{year}}{365\;\cancel{\mathrm{days}}} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {\Delta t=715\;\mathrm{years}} \end{gather} \]

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.
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