What is the power per square meter of a sun on a planet?

[GlubbyJug]

Homework Statement

Using your answer from the previous question, find how much power per square meter is arriving at a planet 1.5*10^11 m away

Relevant Equations

No clue what equation to use

From the previous question I got the star is giving off around 4.01*10^27 W. I am unsure how to find the answer it asks for. Please help.

 

[phinds]

how many square meters ARE there on the surface of a sphere 1.5*10^11 m in radius?

EDIT: OOPS --- see post #12

 

[GlubbyJug]

Well the area of a sphere is 4pir^2 so inputing the numbers I got around 2.83*10^23

 

[GlubbyJug]

Well the area of a sphere is 4pir^2 so inputing the numbers I got around 2.83*10^23

What should I do now?

 

[GlubbyJug]

Update: I checked my calculations for the first part and got the star is really giving off around 3.19*10^26 Watts. Does this effect what I should do?

 

[DaveC426913]

Step 1:
At Earth orbit, the sun's rays are effectively parallel. The spherical shape of the Earth is not the correct area to compute. What you really want to compute is the surface area of the disc of the Earth, which is simply a disc whose diameter is that of the Earth.

Step 2:
If you have a power output, and you have an area, and you are asked "how much power per square meter is arriving at the planet", do you think you can divine what the appropriate calculation might be?

 

[Ibix]

What should I do now?

You're being asked to compute the power per unit area passing outward through a sphere of radius $1.5\times 10^{11}\mathrm{m}$. You know the total power and you know the total area. How do you go about computing power per unit area?

 

[Orodruin]

Well the area of a sphere is 4pir^2 so inputing the numbers I got around 2.83*10^23

This is a meaningless number. If you have dimensionful quantities you always need to specify the units. No exceptions.

Does this effect what I should do?

No. As others have said, to find the energy per area you simply should divide the energy by the area it is spread over.

 

[phinds]

At Earth orbit, the sun's rays are effectively parallel. The spherical shape of the Earth is not the correct area to compute. What you really want to compute is the surface area of the disc of the Earth, which is simply a disc whose diameter is that of the Earth.

Dave, since is asking for the power per square meter, not the total power, the area of the Earth is irrelevant. One has to assume that the answer is to be based on a fairly central square meter, else there is no way to determine the answer.

EDIT: OOPS --- see post #12

 

[Chestermiller]

What is the angle subtended by the planet?

 

[phinds]

What is the angle subtended by the planet?

Irrelevant since the question doesn't ask for the TOTAL power but the power per square meter. As I pointed out directly above to Dave, you HAVE to assume that the question is based on a square meter that is facing the sun square-on, else you can't find an answer without knowing the angle at which the square meter faces the sun and then the math gets nasty.

EDIT: OOPS --- see post #12

 

[phinds]

OOPS ! I now see why everyone but me IS considering the total area of the Earth's disc. One has to compute the power to the disc and then average it out.

 

[Chestermiller]

Irrelevant since the question doesn't ask for the TOTAL power but the power per square meter. As I pointed out directly above to Dave, you HAVE to assume that the question is based on a square meter that is facing the sun square-on, else you can't find an answer without knowing the angle at which the square meter faces the sun and then the math gets nasty.

EDIT: OOPS --- see post #12

If the planet is rotating, the time- and spatial average power per square area is equal to the total power divided by the total area $4\pi R^2$.

 

[phinds]

If the planet is rotating, the time- and spatial average power per square area is equal to the total power divided by the total area $4\pi R^2$.

Good point. I'm not clear that that is what is being asked for, but it's still a good point.

 

[Orodruin]

OOPS ! I now see why everyone but me IS considering the total area of the Earth's disc. One has to compute the power to the disc and then average it out.

This is unclear from the problem statement. I’d tend to favor your original interpretation simply because it seems more appropriate for OP’s level.

 

[Orodruin]

If the planet is rotating, the time- and spatial average power per square area is equal to the total power divided by the total area $4\pi R^2$.

As said above, the formulation by the OP is ambiguous as it does not specify which area. It certainly does not stipulate a time average or area average and the typical thing for irradience would be to specify power per area orthogonal to the radiation direction.

 

[phinds]

This is unclear from the problem statement. I’d tend to favor your original interpretation simply because it seems more appropriate for OP’s level.

Yes, that was why I interpreted it that way instead of thinking about it in the more sophisticated way that others did.

 

[haruspex]

This is unclear from the problem statement. I’d tend to favor your original interpretation simply because it seems more appropriate for OP’s level.

If the quoted wording can be trusted, I see no ambiguity. It says "power arriving at", I.e. reaching the specified distance. What the planet does with it (could be tidally locked) is another matter.

 

[Ibix]

What the planet does with it (could be tidally locked) is another matter.

And there isn't a unique answer if we start talking about power per square meter of planet surface, because it varies across the planet.

 

[haruspex]

how much power per square meter is arriving at a planet 1.5*10^11 m away
the star is giving off around 4.01*10^27 W.

the area of a sphere is 4pir^2 so inputing the numbers I got around 2.83*10^23

How did you get 2.83*10^23 (W/m²) from 1.5*10^11 m and 4.01*10^27 W

 

[Orodruin]

If the quoted wording can be trusted, I see no ambiguity. It says "power arriving at", I.e. reaching the specified distance. What the planet does with it (could be tidally locked) is another matter.

Well ... some people that replied read it differently so there is certainly ambiguity even if one particular person cannot read it another way. (For the record, I would also read it as asking for the irradiance.) If it was not ambiguous then there would not have been a reason for people to start talking about planetary geometry and time averages.

 

[Orodruin]

And there isn't a unique answer if we start talking about power per square meter of planet surface, because it varies across the planet.

Which is why those reading it like that started talking about averages over the planet I guess.

 

[DaveC426913]

Dave, since is asking for the power per square meter, not the total power, the area of the Earth is irrelevant. One has to assume that the answer is to be based on a fairly central square meter, else there is no way to determine the answer.

Ok, yeah. I thought it was asking for total power. Per square metre is much simpler.

Apologies @GlubbyJug , we've had you on more than one wild goose chase.