You observe two gravitationally bound stars (a binary pair). One is blue and one is yellow. The yellow star is six times brighter than the blue star. Qualitatively compare their temperature and radii, i.e. which is hotter, which is smaller? Next, quantitatively compare their radii (to 1 significant figure).
Let's think about this system quantitatively first. We know that hotter objects tend to be bluer in color, since they emit lower frequency of light, like a blue hot coal verses a red hot coal. So, this means that the blue star is hotter.
Next let's consider size. We know that the yellow star is 6 times brighter than the blue star. Intuitively, this means that it should be larger, since it emits light that is less energetic than the blue star. It needs to emit more of the less energetic light, so it must be larger to do so.
So we have this rough picture:
In order to compare their radii, we first need to quantitatively compare their temperatures.
In a previous blog post we derived the formula for the Wein's Displacement Law which expresses a star's peak wavelength of emitted light as a function of its temperature and fundamental constants.
Let's do the blue star first, knowing that a typical blue wavelength of light is about 475nm.
\[\lambda =2.898\times 10^{-3}m \times T\]
\[T_b=\frac{2.898\times 10^{-3}m}{\lambda}=\frac{2.898\times 10^{-3}m}{475nm}=\boxed{6.1\times 10^3 K}\]
This process can be repeated for the yellow star of peak wavelength 570nm.
\[\lambda =2.898\times 10^{-3}m \times T\]
\[T_y=\frac{2.898\times 10^{-3}m}{\lambda}=\frac{2.898\times 10^{-3}m}{570nm}=\boxed{5.1\times 10^3 K}\]
Next we can apply the Luminosity formula to each star to calculate their relative radii.
Since the Luminosities are relative, let's let \(L_y=6L_b\)
For the blue star:
\[L_b=4\pi R^2 \sigma T_b^4\]
\[R=\left(\frac{L_b}{4\pi \sigma T_b^4}\right) ^{\frac{1}{2}}=\left(\frac{L_b}{4\pi (5.67\times 10^{-8}Wm^{-2}K^{-4}) (6.1\times 10^3K)^4}\right) ^{\frac{1}{2}}=\boxed{\sqrt{L_b}\times 3\times10^{-5}m/W}\]
This may looks somewhat confusing, but because we were never given a numerical value for \(L_b\) we must leave the radius in terms of \(L_b\).
Let's repeat this for the yellow star.
\[L_y=4\pi (R)^2 \sigma T_y^4\]
\[R=\left(\frac{L_b}{4\pi \sigma T_b^4}\right) ^{\frac{1}{2}}=\left(\frac{6L_b}{4\pi (5.67\times 10^{-8}Wm^{-2}K^{-4}) (5.1\times 10^3K)^4}\right) ^{\frac{1}{2}}=\boxed{\sqrt{L_b}\times 1\times10^{-4}m/W}\]
When we calculate the ratio of these values we find that the radius of the yellow star is 3.3 times the radius of the blue star, so we can confidently declare that while our qualitative reasoning is correct, our picture is not to scale.
Let's think about this system quantitatively first. We know that hotter objects tend to be bluer in color, since they emit lower frequency of light, like a blue hot coal verses a red hot coal. So, this means that the blue star is hotter.
Next let's consider size. We know that the yellow star is 6 times brighter than the blue star. Intuitively, this means that it should be larger, since it emits light that is less energetic than the blue star. It needs to emit more of the less energetic light, so it must be larger to do so.
So we have this rough picture:
In a previous blog post we derived the formula for the Wein's Displacement Law which expresses a star's peak wavelength of emitted light as a function of its temperature and fundamental constants.
Let's do the blue star first, knowing that a typical blue wavelength of light is about 475nm.
\[\lambda =2.898\times 10^{-3}m \times T\]
\[T_b=\frac{2.898\times 10^{-3}m}{\lambda}=\frac{2.898\times 10^{-3}m}{475nm}=\boxed{6.1\times 10^3 K}\]
This process can be repeated for the yellow star of peak wavelength 570nm.
\[\lambda =2.898\times 10^{-3}m \times T\]
\[T_y=\frac{2.898\times 10^{-3}m}{\lambda}=\frac{2.898\times 10^{-3}m}{570nm}=\boxed{5.1\times 10^3 K}\]
Next we can apply the Luminosity formula to each star to calculate their relative radii.
Since the Luminosities are relative, let's let \(L_y=6L_b\)
For the blue star:
\[L_b=4\pi R^2 \sigma T_b^4\]
\[R=\left(\frac{L_b}{4\pi \sigma T_b^4}\right) ^{\frac{1}{2}}=\left(\frac{L_b}{4\pi (5.67\times 10^{-8}Wm^{-2}K^{-4}) (6.1\times 10^3K)^4}\right) ^{\frac{1}{2}}=\boxed{\sqrt{L_b}\times 3\times10^{-5}m/W}\]
This may looks somewhat confusing, but because we were never given a numerical value for \(L_b\) we must leave the radius in terms of \(L_b\).
Let's repeat this for the yellow star.
\[L_y=4\pi (R)^2 \sigma T_y^4\]
\[R=\left(\frac{L_b}{4\pi \sigma T_b^4}\right) ^{\frac{1}{2}}=\left(\frac{6L_b}{4\pi (5.67\times 10^{-8}Wm^{-2}K^{-4}) (5.1\times 10^3K)^4}\right) ^{\frac{1}{2}}=\boxed{\sqrt{L_b}\times 1\times10^{-4}m/W}\]
When we calculate the ratio of these values we find that the radius of the yellow star is 3.3 times the radius of the blue star, so we can confidently declare that while our qualitative reasoning is correct, our picture is not to scale.
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