# While helping an astronomy professor, you discover a binary star system in which the two stars...

## Question:

While helping an astronomy professor, you discover a binary star system in which the two stars are in circular orbits about the system's center of mass. From their color and brightness, you determine that each star has the same mass as our Sun. The orbital period of the pair is {eq}21.3 {/eq} days, based on the oscillation of brightness observed as one star occludes (hides) the other. From this information you are able to ascertain the distance between the stars. Calculate the distance between the stars. Express your answer to two significant digits and include the appropriate units.

## Luminosity:

The luminosity of a star is defined as the quantity of energy emitted every second from the stars which tell about stars brightening power. It depends on the distance and size of a star. The color is a sign of surface temperature of any star. The hottest stars radiate blue or blue-white color.

Given data

• The orbital period is: {eq}T = 21.3\;{\rm{days}} {/eq}.

The expression for gravitational force balanced by linear speed is,

{eq}\dfrac{{G{M^2}}}{{{D^2}}} = \dfrac{{M{V^2}}}{r} {/eq}

Here, {eq}G = 6.67 \times {10^{ - 11}}\;{{{{\rm{m}}^{\rm{3}}}} {\left/ {\vphantom {{{{\rm{m}}^{\rm{3}}}} {{\rm{kg}} \cdot {{\rm{s}}^{\rm{2}}}}}} \right. } {{\rm{kg}} \cdot {{\rm{s}}^{\rm{2}}}}} {/eq} is gravitational constant, {eq}M = 1.99 \times {10^{30}}\;{\rm{kg}} {/eq} is mass of sun, {eq}r = \dfrac{D}{2} {/eq} is radius and D is diameter of star and {eq}V = \dfrac{{2\pi r}}{T} {/eq} is linear velocity of star.

Convert orbital period in seconds.

{eq}\begin{align*} T &= 21.3\,{\rm{days}} \times \dfrac{{1\;{\rm{day}}}}{{24\;{\rm{hr}}}} \times \dfrac{{1\,{\rm{hr}}}}{{3600\,{\rm{s}}}}\\ T& = 1840320\;{\rm{s}} \end{align*} {/eq}

Substitute all the values in above equation.

{eq}\begin{align*} \dfrac{{G{M^2}}}{{4{r^2}}} &= \dfrac{{m\dfrac{{4{\pi ^2}{r^2}}}{{{T^2}}}}}{r}\\ {r^2}& = \dfrac{{GM{T^2}}}{{16{\pi ^2}}}\\ r &= {\left( {\dfrac{{6.67 \times {{10}^{ - 11}} \times 1.99 \times {{10}^{30}} \times {{1840320}^2}}}{{16{\pi ^2}}}} \right)^{\dfrac{1}{3}}}\\ r &= 1.42 \times {10^{10}}\;{\rm{m}} \end{align*} {/eq}

The distance of star is,

{eq}\begin{align*} d &= 2r\\ d &= 2 \times 1.42 \times {10^{10}}\\ d &= 2.83 \times {10^{10}}\;{\rm{m}} \end{align*} {/eq}

Thus, the distance between the stars is {eq}2.83 \times {10^{10}}\;{\rm{m}} {/eq}. Kepler's Three Laws of Planetary Motion

from Basics of Astronomy

Chapter 22 / Lesson 12
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