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

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?

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Phúc Điền 1 year 2021-08-01T22:55:41+00:00 1 Answers 184 views 0

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    2021-08-01T22:57:02+00:00

    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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