# Does the Earth experience air resistance?

Consider the earth body excluding the atmosphere, undergoing circular motion around the sun. Does it experience air resistance due to the atmosphere?

• First you say to exclude the atmosphere, then you ask about resistance due to atmosphere? The bottom line is no, the space its orbiting through is a vacuum. Commented Jun 5, 2023 at 0:33
• @SeñorO: It seems perfectly reasonable to ask "Does the air that forms the atmosphere exert air resistance on the parts of Earth that are not the atmosphere?" Commented Jun 5, 2023 at 9:25
• The next logical question has already been answered: en.wikipedia.org/wiki/Luminiferous_aether Commented Jun 5, 2023 at 11:04
• Commented Jul 22, 2023 at 7:51
• @SeñorO: Yes 2 hypothesis are fundamentally contradictory. There is a missing piece in this question. Perhaps bring a still atmosphere instantaneously around the Earth? Commented Jul 22, 2023 at 8:11

You are free to consider the Earth without its atmosphere as a system and ask what are the forces the wind exerts on the Earth. But it isn't simple.

The atmosphere orbits around the Sun with about the same velocity as the Earth. Orbital velocity is $$30$$ km/sec. So it isn't a headwind like on a car's windshield. The atmosphere doesn't get left behind in the orbit. Earth's gravity keeps it held down against the surface.

Of course the wind blows in various directions a few km/hour. And this exerts a drag force on the Earth in various directions. The atmosphere doesn't blow toward or away from any spot on the Earth on the average. We know this because air doesn't pile up leave at any spot.

You might think if the wind blows predominantly around the Earth in one direction, it might slow or speed the rotation of the Earth. Such drag would slow the wind to a stand still unless there was another force speeding it back up. That comes from solar heating that drives the weather. On the average, there is as much slowing as speeding. So the average effect on Earth comes out $$0$$.

• It's worth also mentioning the reason why the slowing and speeding average each other out: they have to, because the total angular momentum of the Earth-atmosphere system is conserved. Commented Jun 5, 2023 at 9:38
• Does the Coriolis Force sap the Earth's rotational momentum? Over millions of years? Commented Jun 5, 2023 at 13:38
• @MindwinRememberMonica The Coriolis force is fictitious and cannot "cause" anything. It is a bookkeeping device needed because the surface of the earth is not an inertial reference frame. In a non-rotating inertial frame, the Coriolis force doesn't even exist. Commented Jun 5, 2023 at 13:54
• @N.Virgo - Conservation of angular momentum means that at every instant to total angular momentun of earth + atmosphere adds to the same value. What I mean here is a little different. The earth and atmosphere may trade momentum or angular momentum back and forth, but the average value in each over time does not change. The atmosphere is about as windy last year as this year. Commented Jun 5, 2023 at 15:34
• @MindwinRememberMonica - If it did, the angular momentum the Earth lost would have to increase the angular momentum of the atmosphere. Winds would have to steadily increase around the Earth. So no. Keep in mind as the wind blows south from the north pole, it is deflected west. The air at the pole must be replaced by wind blowing from the south, or there would soon be none left at the pole. Air blowing from south to north is deflected east. Commented Jun 5, 2023 at 15:45

Not at the moment (at least not in any significant way), but it will as soon as the sun enters its red giant phase roughly six billion years from now. Whether the sun will grow enough in diameter to actually "swallow" Earth is currently poorly understood. We know that it will engulf both Mercury and Venus. It does not seem to be able to get to Mars and the outer planets. Earth is just about the only planet that has a somewhat uncertain fate. Should Earth be engulfed by the sun, though, then it's a fairly quick end game. While red giant atmospheres have a very low density (on the order of 1e-8 to 1e-6g/cm^3 I believe), moving through even such a thin gas at an orbital velocity of almost 30km/s will start stripping off material from the outer layers of the planet. Think about a plasma etching process that will lead to a slow but steady loss of mass. I believe even at the lower end of that pressure scale the steady ablation alone would dissolve earth's entire mass in less than a million years and that's without any loss of orbital velocity. The latter will accelerate that process exponentially, of course, because the planet will sink into ever denser layers. (Disclaimer: that's a back of the envelope that I did in my head just now, I might be completely wrong about it.). Since a solar mass red giant lasts for approx. one billion years, there is plenty of time for this process to play itself out. Should Earth survive, however, then it will find itself in an orbit that is much further away from the center of mass of the solar system (all else being equal, of course) because the sun will lose roughly half of its mass during the red giant phase, which will allow the planet to migrate out.

But even today the only thing that prevents the solar wind from stripping Earth's atmosphere is the magnetic field of the planet, so we can say that while "friction" is not a major problem for the planet as a whole, it is a major concern for the biosphere.

• What a great post ! Commented Jun 5, 2023 at 12:38
• So what you are saying is the outer gas layers will not be in rotation, or at least slower? Because would there still be resistance if the gas moves at the same speed as the earth? Commented Jun 5, 2023 at 13:06
• @kutschkem There will be a rotation, but it will be very slow. Angular momentum conservation increases the rotation period as the star expands. The relative surface velocity might be similar to what it is today (which is around 1.3km/s). That is small compared to the orbital velocity of the planet (and has to be, otherwise the gas of the sun would accumulate in a bulge/disk rather than be confined to an almost spherical shape). Commented Jun 5, 2023 at 18:14
• @FlatterMann Ok, so it has velocity lower than the orbital velocity, so it is kept from falling in by the radiation pressure? Commented Jun 6, 2023 at 6:04
• Radiation pressure? No. That's way too small. The atmosphere of a star is just hot gas (roughly ten times the temperature of the gases of Earth's atmosphere). The pressure is caused by atomic collisions (and in deeper, hotter layers it's made out of plasma). Commented Jun 6, 2023 at 9:22

For drag to happen, there has to be relative motion between the solid body and the fluid. The atmosphere and the Earth all move together.

For momentum to be exchanged between two bodies, they have to possess a net difference in velocity to begin with. The Earth and its atmosphere have the same velocity on the whole.

Now there is low density plasma in the solar system primarily generated by the Sun. So yes that creates drag on the Earth, but it is exceedingly small.

No. The definition of air resistance is:

The resistance to motion experienced by an object moving through the air caused by the flow of air over the surface of the object.

The Earth is not moving 'through' it's atmosphere causing a flow of air over the surface. For that to be the case the air would all have to be flowing over the surface of the Earth in the direction opposite to the direction of Earth's travel through the solar system, and that is not the case. The solid Earth and the atmosphere can be treated as a single system for this purpose.

The interplanetary medium is too tenuous for its interaction with the Earth to have an appreciable effect on Earth's orbit.

As a rule of thumb, air resistance can be ignored while the mass of displaced air is much less than the mass of the moving object. This is why air resistance can be ignored when you are juggling golf balls, but air resistance is important when you are playing golf.

The density of Earth is larger than the density of the interplanetary medium by a factor of roughly $$10^{23}$$. Earth's orbit is about $$10^{4}$$ Earth diameters in circumference, so it would take something like $$10^{19}$$ years, optimistically, for Earth's magnetosphere to displace one Earth mass of interplanetary plasma. That's a billion times the lifetime of the Sun. No big deal.

You might possibly get the density ratio down to $$10^{20}$$-ish if you include the volume of Earth's magnetosphere instead, which interacts with the interplanetary plasma more directly. I still don't think interplanetary drag matters on the timescale if the Sun's lifetime, but I haven't got a second envelope to do the calculation.

Earth's rotation about its axis is actually subject to "water drag," thanks to the tidal action of the Moon. The Moon's gravity attracts the water in the oceans, which then "wants" to rotate about Earth's axis once per month rather than once per day. Friction between the oceans and the solid Earth steals rotational energy from the Earth-Moon system, making both the day and the month longer. It's not only the ocean that experiences this effect (there are "rock tides" as well) but the ocean tides are the most familiar.