Finding the temperature change of Earth's surface over 12 hours

Click For Summary

Homework Help Overview

The discussion revolves around calculating the temperature change of Earth's surface over a 12-hour period, specifically focusing on the energy received from the sun at different times of the day. The original poster attempts to determine the velocity of air moving from the equator to the poles due to temperature-induced pressure changes, utilizing solar flux values and albedo effects.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of an integral to calculate total energy received over time based on solar flux, questioning the function of the angle θ in relation to time and the assumptions made regarding the equator's position relative to the sun.

Discussion Status

Some participants provide guidance on integrating the solar flux equation and suggest considering the angle as a function of time. There is a divergence in results regarding the calculated energy, prompting further examination of the angle's measurement units in calculations.

Contextual Notes

Participants are operating under the assumption of the longest day at the equator and are discussing the implications of using degrees versus radians in their calculations. There is a focus on ensuring the correct application of mathematical principles in the context of the problem.

leonmate
Messages
81
Reaction score
1
Ok, this question goes on a little further. I want to find the velocity of air moving from the equator to the poles due to the change in pressure caused by differing temperatures.

I've found the flux density from the sun, so I have a value of flux (hope I am using the right term here - whatever W/m^2 is) for the equator and the N pole and I factored in different albedos too :approve:

But now I'm stuck.

I have a flux on the equator, so energy/s for 1m^2 essentially. I want to find how much the surface heats up and use that to find how much that heats the column of air above it.

If we consider at sunrise the angle between the surface and the sunlight is 0°, at midday 90°, and sunset 180°. So we are going from minimum>maximum>minimum. I've drawn myself a graph, the maximum flux is 1270 W/m^2

As the sun moves across the sky, the 1m^2 of ground we are considering has a differing amount of flux on it as F = (Fmax) sinθ so we wind up with a graph that looks like the sin curve. What I want to know is how can I work out the total energy on this 1m^2 of ground over 12 hours (43200 seconds if that's easier)? I guess it's an integral but so far nothing has worked for me :cry:
 
Physics news on Phys.org
well, you've got the equation F = (Fmax) sinθ which is the power on this 1m^2, right. So yes, to get the total incident energy, it is going to be an integral over time. So now, you need to find out what is θ as a function of time? hint: it is fairly simple, think about the rotation of the earth.

edit: also, I'm guessing you are making the assumption that it is the longest day on the equator, so that the equator is the point closest to the sun. (or equivalently, just ignoring the effect of the seasons).
 
Check out Introduction to Dynamic Meteorology by Holton
 
Ok, so I subbed 180*t/43200 as a value for θ into the integral:

E = ∫ Fmax*sin(t/240) dt and used the limits 0 - 43200

I found an answer to be 609600J.. seems like a good answer! Thanks :biggrin:

Just trying to figure out the change in air temperature because of that... need some help with this bit!

I jumped a step and rearranged bernoulli's theorem to find the change in velocity due to the change in pressure
 
hmm. your method looks good. But I get a different answer than you did. You are using Fmax=1270, right? Maybe you accidentally used pi=180 instead of the radians answer?
 
yeah, i used 1270.

I worked the whole thing through with degrees, shouldn't make a difference?

E = ∫ Fmax*sin(t/240) dt
E = 240*Fmax*(-cos(t/240)

once you stick in limits of 0 and 43200 i get:

E = [240*Fmax*(-cos(180))] - [240*Fmax*(-cos(0))]
E = [240*Fmax*1] - [240*Fmax*-1] = 480*Fmax = 609600J

How did yours differ?
 
It is not OK to use degrees when the angle comes out of the sin function (when you did the integration). If the angle was only inside the sin function, then you can use degrees or radians, it doesn't matter. But when you're using an angle outside of a sinusoid function, you have to use radians.

For example, with radians the arc length = radius * angle But this equation only works when the angle is in degrees. And other similar formulas require radians, not degrees.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
Replies
14
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
Replies
2
Views
2K
Replies
4
Views
2K
Replies
12
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 13 ·
Replies
13
Views
2K
Replies
7
Views
2K
Replies
9
Views
2K