Question

Suppose a supernova explosion emits 10^(51) ergs spread over a year. How far away from Earth would a typical supernova have to be to be as bright as our Sun if the Sun emits 10^(33) ergs/sec? Chinese observers record seeing a supernova 1000 years ago that appeared 10 times less bright than the Sun. How far away was this supernova?

Answer 1

In order to answer this question, you need to know that the surface area of a spherical volume of radius r is , and that the cross-sectional area of a sphere is just a circle, and thus has an area of just .

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Part a)

Step 1) In order to compare the intensities of the Sun versus supernova, they should both be expressed in the same units. The power for the supernova is given as ergs over a whole year, whereas that for the Sun is given as ergs per second. So convert the power of the supernova to ergs/s by dividing by all the seconds in a year:

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Step 2) The power emitted by the Sun is spread out over a spherical area, with the Earth only capturing a small portion of that. So the total power received P_TS is just equal to the power emitted by the Sun P_S, times the ratio of the cross-sectional area of the Earth CA_E, and the surface area of the spherical volume defined by the distance between the Earth and the Sun SA_ES,

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Step 3) Fill in that the surface area covered by the Sun is that of the surface area of a sphere of radius R_ES (Earth to Sun distance), and that the cross-sectional area of the Earth is just a circle of radius R_E (radius of the Earth):

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Step 4) Produce a similar equation for the total power received from the supernova P_Ts, just changing the power from the Sun P_S to that of the power from the supernova P_s, and the distance from the Earth to the Sun R_ES to that of the distance between the the Earth and the supernova R_Es:

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Step 5) Now just equate the two total powers, using the expressions from the previous two steps:

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Step 6) Cancel the common terms:

So equating them produces an expression that only depends on the relative powers and distances involved, with the area of the Earth dropping out.

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Step 7) Rearrange the equation to solve for the ratio of the distance between the Earth and supernova R_Es to that of the Earth and Sun R_ES:

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Step 8) Evaluate the expression by filling in the 3.17x10⁴³ erg/s found earrlier for the power emitted by the supernova P_s, and the 10³³ erg/s given for the power emitted by the Sun P_S:

So the distance from the Earth to the supernova would have to be about 180,000 times the distance between the Earth and the Sun (for a total of 180,000 AU) for the brightness of the two of them to be equal.

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Part b)

Step 1) Since the supernova witnessed by the Chinese astronomers was only 10% as bright as the Sun, that means that the total power received was only 10% as that from the Sun:

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Step 2) Repeat the procedure in the previous part to get an expression for the ratio of the distance between the supernova and Earth, and that of between the Sun and Earth:

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Step 3) Fill in the same values as before for the power emitted by the supernova and for the Sun:

So the distance between the the supernova witnessed by the Chinese astronomers and Earth is about 560,000 times the distance between the Earth and the Sun (for a total of about 560,000 AU).