Force from the sun (MN)

[Leader Bee]

http://en.wikipedia.org/wiki/Orders_of_magnitude_(force)

 

The wiki article says that the force of light on earth from the sun is 570 MegaNewtons but I was wondering what this actually means and if there is any practical applications or useful information we can gather from this?

 

The same article says the force from the space shuttle equals 1.8 MN at lift off and 570 seems a huge number yet I dont feel myself being crushed or incinerated when I go out in the daytime. What's the actual meaning of this expression.

 

I assume it has some application in solar energy but i really dont understand what the expression means? could someone explain?

 

Cheers,

[timo]

A ray of light has a certain momentum. When the ray of light is absorbed by earth, it must take up this momentum so that the total momentum is conserved. A transfer of momentum can be considered the effect of a force working: the sunlight pushes earth outward on its orbit around the sun. The reason the number is so large compared to the shuttle is because the shuttle force is concentrated on a relatively tiny area (the size of the shuttle shape). The force caused by sunlight given is the sum over half of earth's surface (the part being lit) - a tiny force per area sums up to a reasonable value that way.

 

For the actual meaning or significance of the sunlight force: I doubt there is one. The force is 10^14 times smaller than the gravitational attraction (*), so it's simply irrelevant for earth's orbit around the sun. For solar energy, the relevant parameter would be the amount of power, not the force Power and exerted force of a light ray are closely related, of course. But force in the sense of pushing things seems pretty irrelevant for solar energy applications to me. Perhaps you should ask on the discussion page of the article what the idea behind adding this seemingly-pointless value was (my guess is "we had no better example for this slot").

 

(*): I have not bothered checking the numbers from the Wikipedia article, but simply assumed they are correct.

Edited September 22, 2010 by timo

[swansont]

When you divide by the area the earth presents to the sun, you get a pressure about 5 microNewtons per square meter.

[Leader Bee]

Do solar sails then work on the same principal being discussed? That photons emit a % of pressure, or is this more related to solar wind, which i'm not sure works in the same way?

[Klaynos]

Yes, solar sails use the same physics that photons have momentum and momentum is conserved.

[swansont]

Solar sails also try and leverage the fact that reflection gets you about twice the momentum kick as absorption.

[between3and26characterslon]

Solar sails also try and leverage the fact that reflection gets you about twice the momentum kick as absorption.

 

Can you give some explanation here because that makes no sense to me (not saying it's wrong, just makes no sense to me). Wouldn't that mean its wavelength is changed by being reflected?

 

As I understand it when a photon is absorbed its energy converts into kinetic energy but when it's reflected there is no transfer of energy. Like in those little glass vacuum jars with the black and white sails in them.

[swansont]

Can you give some explanation here because that makes no sense to me (not saying it's wrong, just makes no sense to me). Wouldn't that mean its wavelength is changed by being reflected?

 

As I understand it when a photon is absorbed its energy converts into kinetic energy but when it's reflected there is no transfer of energy. Like in those little glass vacuum jars with the black and white sails in them.

 

The Crookes radiometers you can buy use heating, not radiation pressure — they rotate the wrong way for radiation pressure to be the cause.

 

Absorption gives a momentum kick of p = E/c to whatever absorbed the photon. But if it's reflected, it gets an additional kick, since the photon goes back in the direction from which it came. This second kick will be slightly smaller since a tiny amount of momentum and energy are transferred to the target. But as an approximation, it will be 2*p

[lemur]

For the actual meaning or significance of the sunlight force: I doubt there is one. The force is 10^14 times smaller than the gravitational attraction (*), so it's simply irrelevant for earth's orbit around the sun.

 

But how much of this energy is translated into kinetic energy adding to the revolution speed of the planet? In the short terms this would be negligible, but what about over billions of years?

 

 

 

[Mr Skeptic]

Force is not energy. I'd imagine the force from sunlight and solar wind would cancel a tiny fraction of the gravitational force. In any case, over a billion years the above quoted 570 MegaNewtons force would accelerate earth by 3 meters per second.

http://www.google.com/search?sclient=psy&hl=en&q=570+MN%2Fmass+of+the+earth&aq=f&aqi=&aql=&oq=&gs_rfai=&pbx=1&cad=cbv#sclient=psy&hl=en&q=570+MN%2Fmass+of+the+earth+*+1+billion+years&aq=f&aqi=&aql=&oq=&gs_rfai=&pbx=1&psj=1&fp=5a997d1e90bc09a

[swansont]

However, the effect is noticeable on smaller objects — the forces and torques on satellites and asteroids has a measurable effect.

 

http://en.wikipedia.org/wiki/Yarkovsky–O'Keefe–Radzievskii–Paddack_effect

http://www.sciencemag.org/cgi/content/abstract/131/3404/920

http://repositories.lib.utexas.edu/handle/2152/6623

[Leader Bee]

How would the effect of this change at the earths apehelion & perihelion, or is the distance minute enough to be negligible?

[D H]

http://en.wikipedia....agnitude_(force)

That link doesn't work. And oh fudge, it can't be corrected. The new underlying software is "too smart" (read: Stupid).

 

The same article says the force from the space shuttle equals 1.8 MN at lift off and 570 seems a huge number yet I dont feel myself being crushed or incinerated when I go out in the daytime. What's the actual meaning of this expression.

That simplistic radiation pressure calculation (which is low, BTW) represents for force spread over the entire Earth. The Earth is huge in comparison to the Shuttle. There is a clue in the article as to why you don't feel anything. The cited number is referenced in a footnote, which says that that force is "1.63 x 10[sup−14[/sup] x gravitational attraction between Earth and Sun, assuming total absorption of sunlight." The force exerted by solar radiation pressure on the Earth is many, many orders of magnitude smaller than the gravitational force that the Sun exerts on the Earth.

 

That gravitational is an easy number to calculate: [math]F_{\text{grav}} = \mu_{\odot}M_{\oplus}/R_{\odot,\oplus}^{\,\,2}[/math] where

• μ_⊙ is the standard gravitational parameter for the Sun (132,712,440,018 km³/s),
• M_⊕ is the mass of the Earth (5.9736×10²⁴ kg), and
• R_⊙,⊕ is the distance between the Earth and the Sun (1.00000261 AU = 149,598,261 km).

Grinding out the numbers, the gravitational force exerted by the Sun on the Earth is 3.5 yottanewtons.

 

I assume it has some application in solar energy but i really dont understand what the expression means? could someone explain?

Yes and no. Yes because the calculated result (the one in the wiki page) is based on the solar constant, and the amount of power available from the Sun is also based on the solar constant. No because force and energy are different things.

 

 

Solar sails also try and leverage the fact that reflection gets you about twice the momentum kick as absorption.

Can you give some explanation here because that makes no sense to me (not saying it's wrong, just makes no sense to me). Wouldn't that mean its wavelength is changed by being reflected?

Conservation of momentum. Suppose the solar sail is oriented normal to the Sun (which typically is not a good idea). When the solar sail absorbs a photon, the immediate change in the solar sail's momentum is exactly equal to the momentum of the photon. Absorbing a photon transfers the energy carried by the photon to the solar sail in the form of heat. The solar sail will re-radiate that absorbed energy into space in the form of thermal radiation. If the sail was thick, the sunlit side would be a lot warmer than the dark side, so the thermal radiation would generate a force away from the Sun. However, solar sails have to be very thin, so there isn't going to be that big a temperature difference between the sunlit and dark sides of the sail. The thermal radiation away from the Sun will be more or less equal to the thermal radiation toward the Sun. The net force from that thermal radiation will be very small. Thus the total force generated by absorbing photons is the product of the photon flux (the number of photons hitting the solar sail per second) and the average momentum of a single photon.

 

Now for reflected photons. A reflected photon will go straight back to the Sun. To conserve momentum, the change in the momentum of the solar sail has to be twice that of momentum of the incoming photon. There is no thermal energy change in the solar sail from this reflection, so the total force generated by reflecting photons is twice that of absorbing photons.

Edited September 23, 2010 by D H

[between3and26characterslon]

 

 

Now for reflected photons. A reflected photon will go straight back to the Sun. To conserve momentum, the change in the momentum of the solar sail has to be twice that of momentum of the incoming photon. There is no thermal energy change in the solar sail from this reflection, so the total force generated by reflecting photons is twice that of absorbing photons.

 

 

What effect, if any, does this have on the photon?

[swansont]

What effect, if any, does this have on the photon?

 

It will have a slightly smaller energy, because of the energy it has transferred to the target, which is p^2/2m, and since the photon's momentum is E/c, the energy reduction is E^2/2mc^2, so the fractional reduction in the photon's energy is its energy divided by twice the mass energy of the object it hits.

[between3and26characterslon]

It will have a slightly smaller energy, because of the energy it has transferred to the target, which is p^2/2m, and since the photon's momentum is E/c, the energy reduction is E^2/2mc^2, so the fractional reduction in the photon's energy is its energy divided by twice the mass energy of the object it hits.

 

Will this change its wavelength?

[D H]

What effect, if any, does this have on the photon?

The incoming photon and the outgoing reflected photon are not the same photon. That said, the frequency of the outgoing reflected photon is slightly red shifted. Conservation of energy comes into play as well as conservation of momentum. For a normal reflection (the solar sail is perpendicular to the Sun), the change in wavelength in the solar sail's rest frame is approximately [math]\Delta \lambda \approx \frac{2 h}{m c}[/math] where m is the mass of the solar sail.

Edited September 23, 2010 by D H

[timo]

But how much of this energy is translated into kinetic energy adding to the revolution speed of the planet? In the short terms this would be negligible, but what about over billions of years?

I find the reverse direction much more interesting: If a non-measurable effect could in principle have an impact over some large time span that directly means that you can't even know if earth will still be orbiting sun the way it does now - irrespective of this additional force.

[lemur]

I find the reverse direction much more interesting: If a non-measurable effect could in principle have an impact over some large time span that directly means that you can't even know if earth will still be orbiting sun the way it does now - irrespective of this additional force.

 

How is this the opposite of what I'm saying? It sounds like you're saying the same thing.

[timo]

It's not the opposite, it's the reverse (direction of thought). You're asking about the effect. I'm saying that if it has an effect then the effect is "visible" at a point for which no prediction exists in the first place (and what do you want to compare the effect with, then?).

Edited September 23, 2010 by timo

[between3and26characterslon]

The incoming photon and the outgoing reflected photon are not the same photon. That said, the frequency of the outgoing reflected photon is slightly red shifted. Conservation of energy comes into play as well as conservation of momentum. For a normal reflection (the solar sail is perpendicular to the Sun), the change in wavelength in the solar sail's rest frame is approximately [math]\Delta \lambda \approx \frac{2 h}{m c}[/math] where m is the mass of the solar sail.

 

 

Does this apply to only photons which are reflected directly back on themselves or does it apply to all reflected photons. I'm wondering what happens inside a fibre optic cable for instance. There are transatlantic fibre optic cables, is there measurable red shift over this distance because that's a lot of reflections?

[D H]

The effect of solar radiation pressure on the Earth's orbit is very, very small. Assuming perfect absorption and uniform emission (the simplistic calculation in the source for the number in the wiki table), the effect is 1.6×10^-14 smaller than the gravitational force exerted by the Sun on the Earth. That is three and a half orders of magnitude smaller than the uncertainty in the acceleration of the Earth due to the Sun's gravitational field. In any physical modeling endeavor, one generally ignores effects that are two or more orders of magnitude smaller than uncertainties for the simple reason that the uncertainties will overwhelm those tiny effects.

 

The net result of this effect (the simplistic one, anyhow) is to very slightly reduce the effective gravitational force of the Sun on the Earth. Let's see what the impact is. Imagine some alternate universe where the Earth (same mass as our Earth) is still orbiting the Sun (same mass as our Sun) but that Sun is not shining. We'll assume that that other Earth has the same orbital velocity as does ours. That means that that other Earth will be orbiting at a slightly different distance from its dead Sun than is our Earth from our shining Sun.

 

How much of a different distance? About 1.6×10^-14 AU, or about 2.4 millimeters.

[Mr Skeptic]

What might be a more interesting effect is the fact that the earth is spinning and stays warm for a while, so that it re-radiates in a slightly different direction. But this effect would still be minuscule.

[D H]

That's the Yarkovsky effect (Wiki).

 

Since the high temperature typically occurs around 3 or 4 PM local time and the low around 6 AM, the Earth is going to be subject to this effect. This will result in a small secular effect (while specular and diffuse reflection will just have the effect of altering the effective gravitational force from the Sun).