Question

A main-sequence star at a distance of 20 pc is barely visible through a certain telescope. The star subsequently ascends the giant branch, during which time its temperature drops by a factor of three and its radius increases a hundredfold. What is the new maximum distance at which the star would still be visible in the same telescope?

Answer 1

Answer:

Explanation:

The surface area of a star estimated by the energy emitted per sq meter yields the overall luminosity, which can be represented mathematically as:

$L= 4 \pi R^2 \sigma T^4 --- (1)$

where;

L ∝ R²T⁴

and;

R = radius of the sphere

σ = Stefans constant

T = temperature

Also; The following showcase the relationship between flux density as well as illuminated surface area as:

$F = \dfrac{L}{A}$

where

A = 4πd² and L ∝ R²T⁴

$F = \dfrac{R^2T^4}{4 \pi d^2} \\ \\ F \alpha \dfrac{R^2T^4}{ d^2} --- (2)$

Given that:

distance d₁ = 20 pc

Then, using equation (2)

$F_1 \ \alpha \ \dfrac{R^2_1T^4_1}{ d^2_1}$

However, we are also being told that there is a temp. drop by a factor of 3;

So, the final temp. $T_2 = \dfrac{T_1}{3}$; and the final radius is $R_2 = 100R_1$ since there is increment by 100 folds.

Now;

$F_2 \ \alpha \ \dfrac{R^2_2T^4_2}{ d^2_2}$

SInce;

$F_1 = F_2$

It implies that:

$\dfrac{R^2_1T^4_1}{ d^2_1 } = \dfrac{R^2_2T^4_2}{ d^2_2} \\ \\ d_2 = \sqrt{\dfrac{R_2^2T_2^4}{R_1^2T_1^4}}(d_1)$

Replacing all our values, we have:

$d_2 = \sqrt{\dfrac{(100R_1)^2 \times (\dfrac{T_1}{3})^4}{R_1^2T_1^4}}(20 ) \\ \\ d_2 = \sqrt{\dfrac{(100)^2 }{3^4}}(20 ) \\ \\ d_2 = \sqrt{\dfrac{(100)^2 }{3^4}}(20 ) \\ \\ d_2 =222 \ pc$