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german
3 years ago
15

A main-sequence star at a distance of 20 pc is barely visible through a certain telescope. The star subsequently ascends the gia

nt 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?
Physics
1 answer:
Anika [276]3 years ago
6 0

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

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