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Help with Luminosity question please.

A sensor of cross sectional area 4.0*10^-4 m^2 mounted on a satellite orbiting the Earth is used to gather the electromagnetic radiation from the star Antares.
Antares is 550 light years from the Earth. The radiant power entering the sensor from Antares is 2.6*10^-11 W.

Calculate the luminosity L of Antares.

The only equation for luminosity I need to know for my spec is L=4pi*r^2*sigma*T^4...

However I don't understand how to attempt this question - could someone explain it in simple terms? So far I've just done 550*9.5*10^15 to get 5.225*10^18 as the distance but I don't know what to do with it.
Original post by MrToodles4
A sensor of cross sectional area 4.0*10^-4 m^2 mounted on a satellite orbiting the Earth is used to gather the electromagnetic radiation from the star Antares.
Antares is 550 light years from the Earth. The radiant power entering the sensor from Antares is 2.6*10^-11 W.

Calculate the luminosity L of Antares.

The only equation for luminosity I need to know for my spec is L=4pi*r^2*sigma*T^4...

However I don't understand how to attempt this question - could someone explain it in simple terms? So far I've just done 550*9.5*10^15 to get 5.225*10^18 as the distance but I don't know what to do with it.


Stars are isotropic radiators. That is to say, they are spherical objects and radiate energy equally in all directions.

The surface area of a sphere is:

A=4πr2A = 4\pi r^2

The energy reaching the detector will be in the same ratio as the surface area of the detector vs the surface area of a sphere of radius equal to the distance from the origin star.

i.e. you need to convert the distance from the star to metres and use that as the radius for calculating the surface area of the sphere that the output energy from the star has uniformly expanded to encompass at that distance. Your distance looks OK. :smile:



Once you have that ratio, it's simply a case of multiplying the incident energy at the detector to get the energy radiated by Antares.

The answer will be circa L = 2.2x1031 W

I've included a diagram so you can hopefully understand the concept more easily. (NB ignore the other equations and stuff in the diagram, it does not concern us and is not part of the A-level syllabus) Concentrate on the surface area of the sphere and the surface area of the detector only. The unit area in the diagram represents the surface area of the detector

(edited 5 years ago)
Reply 2
Original post by uberteknik
Stars are isotropic radiators. That is to say, they are spherical objects and radiate energy equally in all directions.

The surface area of a sphere is:

A=4πr2A = 4\pi r^2

The energy reaching the detector will be in the same ratio as the surface area of the detector vs the surface area of a sphere of radius equal to the distance from the origin star.

i.e. you need to convert the distance from the star to metres and use that as the radius for calculating the surface area of the sphere that the output energy from the star has uniformly expanded to encompass at that distance. Your distance looks OK. :smile:



Once you have that ratio, it's simply a case of multiplying the incident energy at the detector to get the energy radiated by Antares.

The answer will be circa L = 2.2x1031 W

I've included a diagram so you can hopefully understand the concept more easily. (NB ignore the other equations and stuff in the diagram, it does not concern us and is not part of the A-level syllabus) Concentrate on the surface area of the sphere and the surface area of the detector only. The unit area in the diagram represents the surface area of the detector



Thank you for your reply,

So I've worked out the surface area using the distance to be 3.43*10^38.

This means the ratio is (4.0*10^-4) vs (3.43*10^38). I don't fully understand why you multiply by the incident radiation now?

Is it that I do (3.43*10^38)/(4.0*10^-4) * (2.6*10^-11) and I get 2.2*10^31?

I can't see the diagram you posted.
Original post by MrToodles4
Thank you for your reply,

So I've worked out the surface area using the distance to be 3.43*10^38.

This means the ratio is (4.0*10^-4) vs (3.43*10^38). I don't fully understand why you multiply by the incident radiation now?

Is it that I do (3.43*10^38)/(4.0*10^-4) * (2.6*10^-11) and I get 2.2*10^31?

I can't see the diagram you posted.


Here's the link which you should be able to see:

http://www.cdt21.com/parts/guide_image/guide_g308e.gif

The detector receives a small proportion of the total energy radiated by Antares.

At 550 ly distance, that received power is given as 2.6x10-11 Watts.

We are also given the detector surface area as 4.0x10-4 m2

i.e. the power arriving at the detector is PA=2.6x10114.0x104=6.5x108 Wm2\frac{P}{A} = \frac{2.6 \text{x}10^{-11}}{4.0 \text{x}10^{-4}} = 6.5\text{x}10^{-8} \text{ Wm}^{-2}

But the surface area of the sphere at 550 ly is 3.39x1038 m2 (remember Antares radiates power in all directions) means that if we multiply the Watts per square metre calculated at the detector by the total surface area of the sphere at 550 ly, we end up with the luminosity (total power) radiated by Antares:

L = 3.39x1038 x 6.5x10-8

L = 2.2x1031 Watts
(edited 5 years ago)

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