The discussion revolves around calculating the power per square meter received from a star by a planet located at a distance of 1.5 x 10^11 m. The subject area includes concepts from astrophysics and geometry, particularly relating to the distribution of solar power over a surface area.
Participants are actively engaging with the problem, questioning the assumptions and interpretations of the original question. Some have provided guidance on how to approach the calculation, while others express uncertainty about the clarity of the problem statement and the assumptions that should be made.
There is ambiguity in the problem statement regarding whether to consider the total power or the power per square meter, leading to varied interpretations among participants. The discussion also touches on the effects of the planet's rotation and the averaging of power across its surface.
What should I do now?GlubbyJug said:Well the area of a sphere is 4pir^2 so inputing the numbers I got around 2.83*10^23
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?GlubbyJug said:What should I do now?
This is a meaningless number. If you have dimensionful quantities you always need to specify the units. No exceptions.GlubbyJug said:Well the area of a sphere is 4pir^2 so inputing the numbers I got around 2.83*10^23
No. As others have said, to find the energy per area you simply should divide the energy by the area it is spread over.GlubbyJug said:Does this effect what I should do?
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.DaveC426913 said: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.
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.Chestermiller said:What is the angle subtended by the planet?
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 said: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
Good point. I'm not clear that that is what is being asked for, but it's still a good point.Chestermiller said: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##.
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.phinds said: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.
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.Chestermiller said: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##.
Yes, that was why I interpreted it that way instead of thinking about it in the more sophisticated way that others did.Orodruin said: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.Orodruin said: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.
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 said:What the planet does with it (could be tidally locked) is another matter.
GlubbyJug said: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.
How did you get 2.83*10^23 (W/m2) from 1.5*10^11 m and 4.01*10^27 WGlubbyJug said:the area of a sphere is 4pir^2 so inputing the numbers I got around 2.83*10^23
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.haruspex said: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.
Which is why those reading it like that started talking about averages over the planet I guess.Ibix said: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.
Ok, yeah. I thought it was asking for total power. Per square metre is much simpler.phinds said: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.