This one's been bugging me for years.

I want to know the total mass of photons that hit the surface of the Earth in a single moment in time. Or for the duration of one second, whatever is convenient. As far as I can tell photons don't actually have mass, but momentum, but an equivalent measure of weight will do. They must be pushing on the Earth.
I'm assuming the photons from stars is negligible in comparison to the Sun?

I've tried finding the answer on the internet, but I get anything from 8 pounds to 'the mass of an aircraft carrier'. I realize there is such a thing as reflection and diffussion in the atmosphere, but if I search for solar energy on wikipedia I can see how many Watts per square meter that the Earth gets on average, so it must be possible to use that figure and the energy of a photon to figure out how many there are and how much they're pushing on us.

Can someone help?

Search for 'Solar Wind' does that help? They can calculate how much force will be imparted on a "solar sail", you should be able to scale that and get a very rough estimate. As to the number of photons to hit the earth? unless i'm wrong: Lets just say my vocabulary isn't good enough to describe that number.

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textureglitch;958834 try watts per square meter, times the surface area of a hemisphere....to get total watts. A watt is a joule/second.
To convert energy to mass use E=mc²....but photon momentum is E/c = hv/c, and you can pretend the sun is monochromatic to get a ballpark figure....you need enough of those photons to match the joules/second

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A second rate theory explains after the fact.
A first rate theory predicts.
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Here's my guess.

The Sun's luminosity is 3.827e26W. Therefore, its flux at Earth's distance is:
3.827e26/(4*pi*149597870691^2)=1360 Watts/square meter

Earth presents a crossection of pi*6378000^2=127796483130631 square meters.

Therefore, it intercepts 1360 * 127796483130631 = 1.73906845174188E+17 watts.

A watt is a joule per second, and you want to know how much sunlight in 1 second.

1.73906845174188E+17 joules/s x 1 s = 1.73906845174188E+17 joules.

E=mc^2, therefore, m=E/c^2.

1.73906845174188E+17 joules / (299792000 m/s)^2 = 1.93498052253782 joule / m^2 / s^2 =

1.93498 kg

So the Earth gains 1.98498 kilograms every second from solar photons.

Of course, the Earth will re-radiate much of this.

Mass of photons hitting the Earth ..O. Photons have no mass. We can determine the out pouring of energy from the solar mass. A large impressive number. We know the size of the target. Earth. We can calculate the energy that must be impacting Earth. Photons are not mass. They seem to be the visible part of the spectrum and then theres the scientific explanation of what a photon actually is. Photon radiation is only a fraction of the suns out pouring of energies. Thats as close as I am going to get to your answer, good luck.

I was going to go with the "photons have no mass" explanation as well. But they do have energy, and E=mc^2, which implies if they have energy, that they have mass. What would be the correct way to phrase this? The Sun loses mass when it emits a photon. Even if the photon is massless, does the Earth gain mass by absorbing the photon?

I think you didn´t mean that. Photons are mediators for electromagnetic interactions. All forms of light [EM radiation] are comprised of photons.

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I think the real info the OP is going for is how much force is imparted onto the earth from photons. Probably thinking along the same lines as the "Gravity Pushes" theories. At least, that's the angle I was coming from when I asked a similar question as a new member. *shrugs*

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I'm like one of those idiot savants...well, except for the savant part.
"In order to increase awareness of the homeless, security have been given binoculars."

For force and acceleration:


The force would be 3.827e26*6378000^2/(4*149597870691^2*299792000) = 580091680.812658 Newtons.

The acceleration would be 3.827e26*6378000^2/(4*149597870691^2*299792000*5.97e24) = 9.71677857307635E-17 meters per second.

But keep in mind that this would not cause the Earth to spiral out in its orbit. 6 months from now, it will experience an equal acceleration magnitude in the opposite direction.

But the Earth doesn't completely absorb all that energy. For example, some of that energy eventually get re-radiated as IR back into space. So, I would think, whatever mass gain you got would have to be adjusted for this mass loss.

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The force will always be away from the Sun.

Yep. I'm not really sure where he was going with that either.

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I'm like one of those idiot savants...well, except for the savant part.
"In order to increase awareness of the homeless, security have been given binoculars."

If you want to spiral out from the Sun, you must increase your semi-major axis. Applying thrust away from the Sun will not do this. (At least not on small scales). It will only alter the shape of your orbit by making it more elliptical. But any change in eccentricity done today will be cancelled in 6 months by an equal but opposite change in the opposite direction. To spiral away from the sun, you must have a prograde component to your acceleration vector.

I made that point in post #4. Earth will also directly reflect a lot of the energy as well.

Ah, okay, I see what you're saying. I thought you meant any force acting upon the earth from the sun (photon/emission not gravity) would act on the opposide side of the earth in 6 months. Or, in other words, that the sun would push us away from it for 6 months then toward it for 6 months.

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I'm like one of those idiot savants...well, except for the savant part.
"In order to increase awareness of the homeless, security have been given binoculars."

I would assume planets release more energy than they gain from solar energy, otherwise they would keep getting hotter instead of cooler. I am ignoring GW, assuming it is still legal to do so.

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Lighten up! This is a stellar board!

The topic of this thread brings to mind a real effect due to the emission of infrared photons from warm bodies. The "Yarkovsky effect" can actually change the orbits of small asteroids by amounts we can detect within a single human lifetime.

It's been in the news lately, so there are plenty of articles describing it stashed in Google's cache. I suggest people go look it up.

Not important for the Earth, of course.

George. I agree. The total bolometric magnitude of the Earth is such that it is a net radiator of heat from it's interior. More is radiated away from the dark side than is absorbed from the sunside daily. Global cooling. Part of that darkside radiation is from neutrinos, (not photons), landing on the Earth, or interacting in Terra Firma's interior. pete.

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A third rate theory forbids.
A second rate theory explains after the fact.
A first rate theory predicts.
A. Lomonosov

Is the effect this large? I thought it was only used to explain how certain asteroids could migrate into different orbits over very long timescales.

I guess it is. Wikipedia has this:

Best info I have indicates 9.4 micronewtons per square meter for perfect reflectivity at Earth orbit. For less than perfect reflectivity, the average reemitted photon (heat) has about half the energy so the average force on a square meter is %reflectivity x 4.7 + 4.7 micronewtons
per square metre.

The albedo of the Earth is about 30% so we have 1,561,911,432 newtons so far.

That is at a 90 degrees angle of incidence. The incident angle of all 255,631,985,598,995 square meters must be counted or averaged for a bit over a hemisphere. This will downgrade our total force by around 30% so my guess is the Sun imparts 1,093,338,000 newtons on the Earth.

This is 245,792,160 pounds of force.

The space shuttle uses 32,500,000 newtons to lift off, so the Earth gets 33 2/3 times that at all times just from the sun (except for eclipses).

Now that you mention it, this must mess with the Earth/Moon dance just a little bit.

Best I can do at this hour.

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Neutrino pressure? I thought these little things mostly blow right through us, and if they don't [or still apply some measurable pressure] then wouldn't the sunny side receive more due to the Sun's output?

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Actually, as was pointed out to me as well, the radiation field will indeed make the orbit more elliptical, but the aberration of the light caused by the orbit itself will also create a retrograde component of acceleration that acts on the body and will actually cause the orbit to slowly decay, contrary to intuition. A large enough outward force would cause it to spiral outward (light pressure from the sun is nowhere near large enough for a large body, though, as you said) and at some point in between, it will be stable. Here is a link Nereid provided me for this.

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I think some of us are getting mixed up between rectangular coordinates and polar coordinates.

In our idealized model, the radiation pressure is always away from the sun, and the resulting force vector will always have the same bearing relative to the object's orbital motion. There will be no reversal of the effect as it comes around to the other side of the orbit. That would require that the light be going toward the sun on that side, which most definitely is not the case.

If the resultant pressure vector is pointing precisely away from the sun, my exercise in geometry tells me the result is slightly prograde. There would be no interference with the unperturbed forward orbital velocity, so the result of the upward push should be an increase in the object's kinetic energy at the moment. Thus it should spiral out gradually as a result of the steady state push.

If the dynamics are such that the pressure vector is pointing further to the rear, the actual result could be retrograde, and the object would spiral in instead of out. This happens in the counterintuitive Poynting-Robertson effect on tiny dust particles. I found numerous detailed explanations of this one via a Google search.

With larger objects the effect can be prograde or retrograde, depending on the object's thermal characteristics and the direction it is spinning, if any. In any case, I am 100% confident that the result is a smooth spiral. If I am missing something, please show me a rigorous mathematical calculation of your argument so I can see the error, if any, in my line of thought.

Thanks for the link, Grav. That's very interesting. And it comes with an easy-to-use formula.

The light is not going towards the sun on the opposite side of the orbit. It is also travelling away from the sun, as it must. The reversal happens because of the effect that an outward push has on an orbit that was originally perturbed by an outward push 180 degrees away.

Consider the following. Let's say we have 2 objects in orbit around the sun, a green test particle and a purple test particle. They are 180 degrees apart. They are both in round orbits with a semi-major axis of 0.1 AU. They both have a velocity of 94.19 kilometers per second perpendicular to their sun line, and 0.0 kilometers per second of radial velocity.


When the green object gets to point A, I give it an instantanous boost of 10 kilometers per second directly away from the sun. It's velocity vectors are now 94.19 kilometers per second perpendicular to its sun line, and 10 kilometers per second of radial velocity away from the sun. Its semi-major axis is boosted to 0.101 AU, and its eccentricity is boosted from 0 to 0.106.


Look a point A. At this point, what is preventing the green object, currently 0.1 AU from the sun, from having a circular orbit? It's the extra 10 kilometers per second of radial velocity directed away from the Sun. What would you do to circularize your orbit at this point? You would apply an instantaneous boost of 10 kilometers per second directly towards the sun.

The green object then spends 1/4 of its orbit climbing to aphelion. It spends the next 1/4 of its orbit dropping back down to 0.1 AU, which occurs at point B. At this point, its velocity vectors are now 94.19 kilometers per second tangent to its sun line, and 10 kilometers per second of radial velocity towards the sun. At point B, what is keeping the green object, currently 0.1 AU from the sun, from having a circular orbit? It's the extra 10 kilometers per second of radial velocity directed towards from the Sun. What would you do to circularize your orbit at this point? You would apply an instantaneous boost of 10 kilometers per second directly away from the sun.

This is why sunlight can have a reversing effect on the opposite side of the orbit, despite always pushing away from the sun.

So, let's give it this 10 km/s boost away from the Sun:

The green object's orbit is re-circularized. Despite being pushed away from the sun, its semi-major axis decreases. Its eccentricity drops to 0, and its velocity vectors are restored to 94.19 km/s perpendicular to its sun line, and 0.0 km/s radial.

Here's a screenshot from another simulation. There are 4 objects in orbit around the sun. Purple, green, blue, and red, at 0.1, 0.11, 0.12, and 0.13 AU from the sun. All 4 objects are instructed to accelerate at 1 millimeter per second². The directions of their accelerations are: purple=retrograde, green=towards the sun, blue=away from the sun, red=prograde. Their orbits are shown after 4 months of accelerating. The retrograde object smoothly spirals in. The prograde object smoothly spirals out. The towards and away objects are virtually unaltered. Even allowing the simulation to continue for several centuries at the heavy acceleration (compared to acceleration from light) of 1 mm/s², there is no significant change in their orbits.


If you want to do the "rigorous mathematical calculation", you can compute the energy of the orbit prior to the original 10 km/s boost away from the sun, and after the 2nd 10 km/s boost away from the sun, half an orbit later, and compare that they are the same. The position and velocity vectors below (14959787069 meters = 0.1 AU) are for the orbit after the 2nd boost. To get the original values, simply change the signs of the position and velocity vectors.

But without even plugging in any numbers, you can see that they all get squared. And since a negative number squared gives the same answer as its positive counterpart squared, they will be equal.

Using the formula from the Wikipedia article you link to, I've added a Poynting-Robertson force calculator to my collection of astronomy-related calculators available here: www.orbitsimulator.com/formulas

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This thread is a wake-up call about how rusty my orbital dynamics math is after nearly 40 years of neglect. I definitely will back off from my previous remark about 100% confidence.

This illustration appears to show the effects of two momentary outward pushes on opposite sides of the orbit, and the combination of changing eccentricity and unchanged semimajor axis makes perfect sense as such. However, this is not the hypothetical case of the same total impulse applied continuously around the orbit.

Is this the program grav used in the atmospheric drag on satellites thread a couple of weeks ago, or something similar to it? If so, it is plenty of mathematical rigor for me at the present time. That one did a great job on real-world satellites as well as some hypothetical cases with point masses surrounded by thin, extended atmospheres.

Are the blue and green orbits actually remaining constant, or is there some fluctuating eccentricity that is too small to show at this image scale? The red and purple spirals are just what I would expect for purely prograde and retrograde thrusts. I can see now how radial pushes would be very different.

Now we appear to be back to momentary shoves rather than a continuous gentle push. A thorough mathematical treatment of a different model. Can this be broken down into some sort of rough-and-dirty calculus, with progressively shorter step sizes? What would happen if we give half as much push at the start, then another one after 1/4 of a revolution, and so on? What happens at 1/8 revolution intervals, and so on?

This has been an interesting discussion, and I sincerely welcome everyone's contributions. I certainly do not claim to get it right on the first attempt every time.

Both eccentricity and semi-major do change after the initial push. But they are restored to their original values after the second push.

No. This is Gravity Simulator, which is a program I wrote. This is the same program that produced the rotating moon plane orbit I posted in "18 year cycle of the Moon" thread. I just put the link to the website is in my signature line. If you want to do "continuous orientation" you'll need the latest beta version which can be downloaded on top of a 2.0 install. The link for the beta version is on the Gsim forum board under the "Systemic" thread. There's no atmospheric drag in this program. Including "continuous orientation" was my first step in implementing atmospheric drag, since direction of acceleration is important here. Future versions may have an eval box so the user can give a formula describing acceleration as a function of some of the program's variables, as opposed to a constant acceleration. But at the moment, the real fun with "continuous orientation" combined with "thrust" is watching migrating planets capture test particles into resonances.

My intuition told me the same thing yours did, that the continuous outward push should cause an outward spiral, just like continuously thrusting prograde. When it did not, I started debugging my program. It took me an hour to realize that there was nothing wrong with my code, only my intuition.

The latter. The ecc and sma are fluctuating and it is barely large enough to see. Notice how the blue orbit is a little thicker, peaking around the 5 o'clock position? That's because its new orbit under the accelerating conditions is overplotted onto its original circular orbit. So this object was at around the 11 o'clock position when it began thrusting. For the green object, its around the 3 o'clock and 9 o'clock positions. That's why I used the words "virtually unaltered." I often use words like "virtually" and "insignificantly" to keep me out of trouble. See the graphs below for an illustration of these fluctuating values. My guess would be that the amplitude of the fluctuations would decrease as the amount of acceleration decreases. 1 millimeter per second squared is a pretty strong acceleration for anything other than a rocket, and the amplitudes are still not very large.

Yes, this is back to 2 momentary shoves 180 degrees apart. Doing it as you describe with progressivelty shorter intervals never really solves the integral, it just sets up a Riemann sum with a small delta x. Which is exactly what the Gravity Simulator simulation is doing using Newton's laws. It does something very similar to the math in my last post if you have the "Orbital Elements" interface visible. The orbits in these illustrations were generaged using either 16 or 32 second time steps over the course of an orbit whose period is about 12 days. The graphs below were generated with 1 second time steps, as I wanted a high-resolution graph.

Here's the 2 graphs. The data were also produced with the beta version of Gravity Simulator, although the non-beta can make them as well, and the graphs produced by importing this data into Excel. This is only a single particle, starting out in a circular orbit at 0.1 AU (14959787.34 km). It is instructed to accelerate away from the sun at 1 mm/s². This simulation is 1 year, which is about 32 orbits.


I would also guess that if the acceleration were gently applied (I believe acceleration of acceleration is called jerk), say starting out at 0.001 mm/s² and gradually increased to 1 mm/s², that the amplitudes would be much smaller. As it is now, there is one instant where an acceleration of 0 is immediately followed by an acceleration of 1, which creates a blip in its orbit, showing up as these flucuations, that it will never recover from.

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Thanks, Tony, for the prompt response. It is interesting and enlightening.

I thought of an easier way to explain this. An outward push by light is basically negating gravity. The light's intensity even even falls off as in inverse square, just like gravity. So for the same reason that the inward pull of gravity does not cause an inward spiral, the outward push of light does not cause an outward spiral.

One subtracts from the other, and basically leaves you with what acts like a less massive star.




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In my examples from the previous posts, with an object orbiting at 0.1 AU, and accelerating away from the sun at 1 mm/s, it is no different than simply subtracting M=ar²/G = 0.001 * 14959787340²/6.672e-11= 3.354e+27 kg from the sun's mass of 1.989e30 kg, giving the sun a new mass of 1.986e30kg. Instantly subtracting this from the mass of the sun should produce a blip in the orbits of everything circling the sun. But... once subtracted, everything should immediately make a one-time jump to their new orbits, and there should be no additional fluctuations in sma and ecc.

So why does my graph show fluctuations? 2 reasons. One is that any numerical simulation is only an approximation, and with Excel's auto ranging graphs, it can zoom in to show the numerical artifacts. See the new graph for an illustration. But the biggest reason is that in my simulations, the acceleration is constant. It does not fall off as an inverse square.

The first graph shows the error produced by the numerical simulation. Throughout the course of this graph, the time step is increased from 1 to 2 to 4 seconds, The test particle was originally orbiting a full-mass sun, and then mass was subtracted off the sun to lessen the acceleration produced by the sun's gravity by 1 mm/s². This transition is not shown on this graph. So sma and ecc should not fluctuate at all. The fact that they do is from numerical simulation error. But as you can see, it is very small, well less than 1 km of sma at time step =1.


The next graph shows the transition jump at about 50 days, when the sun magically loses 3.354E+27 kilograms. The eccentricity flucuations are barely noticable at a time step of 1 second, and increase as the time step is raised to 2 and then to 4 seconds. The sma does as well, but it is not visible at the resolution of this graph.

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