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.
On Earth, the power per square meter of sunlight, known as the solar constant, is approximately 1361 watts per square meter (W/m²) when measured at the top of the atmosphere. At the surface, this value is reduced due to atmospheric absorption and scattering, averaging around 1000 W/m² on a clear day at solar noon.
The power per square meter of sunlight decreases with the square of the distance from the star. This means that if a planet is twice as far from the star as another planet, it will receive only one-fourth the power per square meter. This relationship is described by the inverse square law.
Several factors influence the power per square meter of sunlight received on a planet's surface, including the planet's distance from the star, the star's luminosity, the planet's atmospheric composition and thickness, the angle of sunlight incidence, and local weather conditions such as cloud cover.
The power per square meter of sunlight is typically measured using instruments called pyranometers or radiometers. These devices measure the solar irradiance, which is the power of solar radiation received per unit area. Measurements can be taken at various altitudes and locations to account for atmospheric effects.
The power per square meter of sunlight varies significantly with latitude. Near the equator, sunlight strikes the surface more directly, resulting in higher power per square meter. As you move towards the poles, sunlight arrives at a more oblique angle, spreading the energy over a larger area and reducing the power per square meter.