Energy received by a planet from the Sun

[Jahnavi]

Homework Statement

Homework Equations

The Attempt at a Solution

If P is the power radiated by sun and R is the radius of the planet then power per unit area at a distance d will be P/4πd² .

I am not sure about how to calculate the power received by the planet .Do I need to multiply the above term with the surface area of the hemisphere 2πR² because that would be the area of the planet exposed to the Sun ?

 

[Orodruin]

You should not multiply by half ghe surface area. Some parts of the planet are not perpendicular to the line to the Sun and so do not receive as much radiation. (Consider the limiting case where the sunrays are parallel to the area ...)

You can compute the exposure area (i.e., the area projected on a plane perpendicular to the sunrays) or just assume it is some number A. It should not affect the argument for what you want to show.

 

[Jahnavi]

Are we to assume that Sun's radiation are reaching almost parallel on the planet ? Or does it not matter ?

I am asking this because I have considered the radiations from Sun to be radially outwards .

 

[Orodruin]

What do you think? Would it be a good approximation?

 

[Jahnavi]

Yes , it would be a good approximation . But what confuses me is the radially outward radiations from the Sun .

Can both these things i.e radially emanating radiations from Sun , and , parallel rays reaching the planet , be considered simultaneously in this problem ?

 

[Jahnavi]

@Orodruin , please reply .

 

[haruspex]

Yes , it would be a good approximation . But what confuses me is the radially outward radiations from the Sun .

Can both these things i.e radially emanating radiations from Sun , and , parallel rays reaching the planet , be considered simultaneously in this problem ?

At typical distances of planet from star, you can treat the rays as parallel here.

 

[Orodruin]

It is a question of scales. At the planetary scale level, it is indeed a good approximation that the rays are parallel. At the solar system scale, i.e., when computing the 1/d^2 falloff, it is not.

@Orodruin , please reply .

I strongly suggest against making this kind of messages, in particular so soon after your previous one. Helpers here are here on a pro bono basis and are providing help for free. We have a life outside of PF and you do not know what is going on in them that could affect our ability to give a prompt reply.

 

[Jahnavi]

At typical distances of planet from star, you can treat the rays as parallel here.

Thank you .

 

[haruspex]

At the solar system scale, i.e., when computing the 1/d^2 falloff, it is not.

I took the question as being in respect of the solid angle subtended at the star by the planet. I.e. it can be approximated as πr²/(D²), where r is the radius of the planet and D is distance of the planet's centre from the star.

 

[Orodruin]

I took the question as being in respect of the solid angle subtended at the star by the planet. Strictly speaking it is a little less than r²/(4D²), where r is the radius of the planet and D is its distance from the star.

Yes, what I was trying to say was that the 1/d^2 falloff is due to the rays not being parallel on the solar system scale. If they were parallel on the solar system scale there would be no decrease in intensity.

Strictly speaking it is a little less than r2/(4D2)

The subtended solid angle is actually a little bigger than $\pi r^2/d^2$. The exact result for the solid angle is
$$
\Omega = \int_0^{\theta_0} d\Omega = 2\pi \int_{\cos(\theta_0)}^1 d\cos(\theta) =2\pi[1 - \cos(\theta_0)] = 2\pi \left[1 - \sqrt{1 - \frac{r^2}{d^2}}\right].
$$
In the limit of $r \ll d$, the solid angle becomes
$$
\Omega = \pi \frac{r^2}{d^2}\left( 1 + \frac{r^2}{4d^2}\right) + \mathcal O(r^6/d^6).
$$
The ratio to the full sphere solid angle $4\pi$ is therefore
$$
\frac{\Omega}{4\pi} = \frac{r^2}{4d^2} \left( 1 + \frac{r^2}{4d^2}\right) + \mathcal O(r^6/d^6)
$$

 

[haruspex]

little bigger than

Yes, I already edited my post.

 

[jbriggs444]

It seems to me that all of this discussion is irrelevant to the problem as posed. It is an extremely simple problem. One need not worry over details about whether the area of the planet goes as pi r² or 4 pi r² or some other figure due to incidence angle. One need not worry about the radius of the planet or its albedo.

One does not need any numbers at all.

What one needs are two proportionalities. One for how the heat transfer rate from the sun scales according to planetary distance. Another for how heat leaving the planet scales with temperature.

 

[Orodruin]

Clearly this is the case. This should not stop us from answering the OP’s questions or other statements made in the thread.

 

[Jahnavi]

One does not need any numbers at all.

What one needs are two proportionalities. One for how the heat transfer rate from the sun scales according to planetary distance. Another for how heat leaving the planet scales with temperature.

You are right . Nice suggestion

Thanks !

 

[OmCheeto]

You are right . Nice suggestion

Thanks !

I spent several hours yesterday solving this problem. Ultimately, I had to know that power/area = some constant * temperature to the 4th power.
From there I used the fact that Earth is 1 astronomical unit(au) from the sun, plugged in a temperature of 44°F that I found somewhere on the internet, and receives around 1000 watts per square meter. I then created a hypothetical planet at 4 au from the sun and solved for how much power that planet should be getting per square meter. From there I was able to come up with both "n" and the constant of proportionality "k":
Temp = k * d^-n

I double checked my answer by plugging in the temperatures and distances of all the real planets. Except for Venus, everything came out to the right temperature to within 5%. I thought that was pretty interesting. Venus has a thick atmosphere, which boogers things quite a bit. There are also LOTS of lists of planet, which all list different temperatures, so I'm sure if someone picks a list that I didn't use, the 5% error range will no longer be true.

In any event, please let us know what answer you get, as I'm dreadful at maths. Thanks!

 

[Jahnavi]

In any event, please let us know what answer you get

n = 1/2

 

[OmCheeto]

n = 1/2

Yay!