Blog Post 4, Worksheet 2.1, Problem 3

3. You observe a star and you measure its flux to be \( F_*\). If the luminosity of the star is \(L_*\), 

(a) Give an expression for how far away the star is.

The flux of an object is the total energy that crosses a unit area per unit time, and has a unit of energy per area per time. The luminosity of an object is a measure of the object's intrinsic brightness, and has units of energy per time. The relationship between flux and luminosity is expressed in the following equation: \[ F = \frac{L}{4πd^2},\] where F is the flux, L is the luminosity, and d is the distance between the object and the location at which the flux is being observed. From this equation, we can solve for d, how far away the star is.
\[ F_* = \frac{L_*}{4πd^2}\] \[  4πd^2 = \frac{L_*}{F_*}\] \[  d^2 = \frac{L_*}{4πF_*}\] \[  d = (\frac{L_*}{4πF_*})^{1/2}\]
(b) What is its parallax?

The parallax of an object is the angle the object appears to move against more distant, "fixed" stars as the Earth moves a distance of 1 AU in its orbit.

The relationship between the parallax, given in arcseconds, and the distance, given in parsecs, is shown by the following equation: \[ \text{parallax [arcsecond]} = (\frac{1}{\text{distance [pc]}})\]
Using this equation, the parallax of this star is given by: \[p = (\frac{L_*}{4πF_*})^{-1/2},\] where p is equal to the parallax, \( F_*\) is the observed flux of the star, and \( L_*\) is the star's luminosity.

(c) If the peak wavelength of its emission is at \( \lambda_0\), what is the star's temperature?

In question 1c on Worksheet 2.1, we derived an equation for the wavelength \( \lambda_{\text{max}}\) corresponding to the peak of the intensity distribution of a blackbody at a given temperature T: \[ \lambda_{\text{max}} = \frac{hc}{5kT},\] where h is Planck's constant, 6.6 * \(10^-27\) ergs*s, c is the constant for the speed of light, k is the Boltzmann's constant 1.4 * \(10^{-16} \text{ ergs K}^{-1}\), and T is the temperature of the blackbody in Kelvin.

Using this relationship, if the peak wavelength of the star's emission is at \( \lambda_0\), the relationship between the this wavelength and the star's temperature can be solved for temperature:
\[ \lambda_{\text{max}} = \frac{hc}{5kT}\] \[ 5kT = \frac{hc}{\lambda_0}\] \[ T = \frac{hc}{5k\lambda_0}\]
(d) What is the star's radius, \( R_*\)?

To solve this problem, we look first to the information that we already know.

- The flux emitted from a black body as a function of temperature is given by the equation: \[ F(T) = \sigma T^4,\] with \( \sigma = \frac{2πk^4}{c^2h^3}\int_{0}^{\infty}\frac{u^3}{e^u-1}du\), \(u = \frac{h\nu}{kT}\), and \( \nu\) as the frequency of emission. The units of flux are energy per time per area of the object, or \( \frac{\text{energy}}{\text{time * area}}\).

- The luminosity of the star is given as \(L_*\), and has units of \(\frac{\text{energy}}{\text{time}}\).

- The temperature of a black body as a function of wavelength, as shown in part (c), is given by: \[ T = \frac{hc}{5k\lambda_0}\]
By looking at the units for flux and luminosity, we can see that the relationship between flux and luminosity is given by: \[ F = \frac{L}{A},\] with F = flux, L = luminosity, and A = area. By assuming that the star is spherical, we can plug in the formula for the surface area of a sphere, SA = \(4πr^2\), for the area of the object: \[ A = \frac{L}{F} = 4πr^2,\] with r as the radius of the sphere. From this, we can solve for the radius of the object: \[r^2 = \frac{L}{4πF}\] \[r = (\frac{L}{4πF})^{1/2}\] Now, we can plug in the values known for this star to get the radius of the star, \(R_*\):
\[R_* = (\frac{L_*}{4πF_*})^{1/2}\] \[R_* = (\frac{L_*}{4π\sigma T^4})^{1/2}\] \[R_* = (\frac{L_*}{4π\sigma (\frac{hc}{5k\lambda_0})^4})^{1/2}\]