Whizzing stars

Formulas:

$v_{average} = \frac{displacement}{time}$

$C = 2 \pi r$

Information bank

$t_{period} = 15.5y$

$y = 3.15 \times 10^7 s$

$r = 1.84 \times 10^{13} m$

$v_{light} = 3 \times 10^8 m/s$

Substitution

$C = 2 \pi r = 2\pi (1.84 \times 10^{13} m) = 1.15 \times 10^{14} m$

$t = 15.5y = (15.5)(3.15 \times 10^7) = 4.88 \times 10^8s$

$v_{S2} = \frac{1.15 \times 10^{14}}{4.88 \times 10^8} = 2.36 \times 10^5 m/s$

$\frac {v_{S2}}{v_{light}} = \frac {2.36 \times 10^5}{3 \times 10^8} = 7.85E-4$

total distance travelled in one revolution= 2.pi.R = D (where R=1.84x10^13 meters) total time taken=15.5 years= 15.5 x 3.15x10^7 seconds=T therefore; v = D/T and hence answer= v/speed of light = v / 3x10^8 = 0.00079

The ratio $\dfrac{v_\text{star}}{v_\text{light}}$ can be written as $\dfrac{2\pi R}{T v_\text{light}}$ , where $R$ is the star's orbit radius and $T$ is its period. Since $T$ is being expressed as $17 \times 3600 \: \text{seconds} \times v_\text{light}$ , we can simply the expression to $\dfrac{2\pi R}{T}$ . By placing the adequate and correct numeric values, we find that our desired ratio is: $\dfrac{2\pi \times 17 \times 3600}{15.5 \times 3.15 \times 10^7} = 0.00079$

Orbital Radius of S2(R) = 17 light-hours = 1.84*(10**13) m

Therefore, Circumference of the orbit = 2*pi*R

Orbital period(t) = 15.5 years = 15.5*(3.15*(10**7)) s

c = 3*(10**8) m

Therefore, required ratio = ((2*pi*R)/t)/c = ((3.141592*2*(1.84*(10**13)))/(15.5*(3.15*(10**7))))/(3*(10**8)) = 0.00078928544529783238