As a planet moves through the solar system, a bow shock is formed as the solar wind is decelerated by the magnetic field of the planet. Presumably the creation of this shock wave would cause drag on the planet, certainly in the direction of orbit but possibly rotation as well.

Is there an estimate for the amount of drag on the Earth as it orbits the Sun? Based on the drag, how long would it take before the orbital velocity slows to the point that we spiral slowly into the Sun? Would any planets fall into the Sun prior to the Sun expanding into a Red Giant, gobbling them up?

  • $\begingroup$ Assuming the sun is far enough away that we can basically treat it as a point source of solar wind, then doesn't the solar wind slow the Earth down whenever it is moving towards the sun and speed us up as we travel away? From symmetry can we argue that the net effect is basically zero? $\endgroup$ Commented Aug 19, 2013 at 4:47
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    $\begingroup$ I think the bow shock is in the direction of the orbit, so it is perpendicular to the sun. My understanding is the solar wind in the radial direction from the sun is relatively slow moving, particularly in relation to the orbital velocity of the planets. So the bow shock should be perpendicular, or slightly towards the Sun away from perpendicular. Which means there could be a radial force outward which might just offset the slowing factor. But that's all speculation, I don't know enough about astrophysics which is why I asked! $\endgroup$
    – tpg2114
    Commented Aug 19, 2013 at 4:52
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    $\begingroup$ Worth mentioning that this is basically a not-too-relativistic version of Poynting-Robertson drag. $\endgroup$
    – user10851
    Commented Aug 19, 2013 at 17:45
  • $\begingroup$ @ChrisWhite Good thought. I forgot to include radiation pressure in my calculation... wonder how much that affects it? $\endgroup$
    – Michael
    Commented Aug 19, 2013 at 23:11
  • $\begingroup$ Hmm...how does solar mass loss through out-gassing factor in here? $\endgroup$ Commented Sep 28, 2013 at 1:58

2 Answers 2


This is a really rough calculation that doesn't take into account the realistic direction of the bow shock, or calculation of the drag force. I just take the net momentum flow in the solar wind and direct it so as to produce the maximum decceleration and see what happens.

Apparently the solar wind pressure is of the order of a nanoPascal. As I write this it's about $0.5\ \mathrm{nPa}$. You can get real time data from NASA's ACE satellite or spaceweather.com (click through "More data" under "Solar wind"). During periods of intense solar activity it can get up to an order or magnitude or so more than this. Let's take this worst case and assume, unrealistically, that all of the pressure is directed retrograde along the Earth's orbit. This will give the maximum deccelerating effect. I get a net force of $\sim 10^6\ \mathrm{N}$. Dividing by the Earth's mass gives a net acceleration $2\times 10^{-19}\ \mathrm{m/s^2}$. Let's fudge up again and call it $10^{-18}\ \mathrm{m/s^2}$. The time it would take for this to make a significant dint the the Earth's orbital velocity ($30\ \mathrm{km/s}$) is of the order of $10^{15}\ \mathrm{yr}$. I think we're safe.

For the other planets there is a $1/r^2$ scaling of the solar wind with the distance from the sun (assuming the solar wind is uniformly distributed) and an $R^2$ scaling with the size of the planet. So for Mercury the former effect gives an order of magnitude increase in drag and the latter effect takes most of that increase away again. There is an additional $R^{-3}$ increase in effect due to the decreased mass of a smaller body (assuming density is similar to the Earth). Then there is the $r^{-1/2}$ increase in orbit velocity due to being closer to the sun. So the total scaling factor for the time is $ R r^{3/2} $, which for Mercury is about 0.1. So the end result is not much different for Mercury.

This site always causes me to learn new Mathematica features. It made really quick work of this since it has all sorts of astronomical data built in:

enter image description here

Note that the number of digits displayed in the final column is ludicrous. :)

  • $\begingroup$ +1 Was just about to try this kind of order of magnitude estimate myself. $\endgroup$ Commented Aug 19, 2013 at 5:02
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    $\begingroup$ I think the highlight of the answer is the Wolfram Alpha link because now I know the Sun exerts the same force on the Earth as 140 American Alligators biting down. And that the space shuttle launching directly in the orbital direction actually applies 10 times more force. $\endgroup$
    – tpg2114
    Commented Aug 19, 2013 at 5:09
  • $\begingroup$ What about tides of the earth on the sun? shouldn't that work in the opposite direction, increasing the distance as the moon is receding from earth due to the tides? $\endgroup$
    – anna v
    Commented Aug 19, 2013 at 5:35
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    $\begingroup$ @JackAidley The sun will turn into a red giant in ~5bn years. The most detailed model of what will happen then created so far has the Earth being tidally dragged into the sun just before the end of the red giant phase. However since the Earth is right on the cusp of being destroyed/surviving the result shouldn't be considered definitive. $\endgroup$ Commented Aug 19, 2013 at 13:10
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    $\begingroup$ Using Poynting-Robertson force of $F=vL_\odot R^2/4r^2c^2$ and a timescale of $\tau=v/a=vM_\oplus/F$, I get a timescale for this alternative effect of $10^{17}\ \mathrm{yr}$, scaling as $r^2R\rho$. Assuming my arithmetic is right, this means we can neglect PR for Earth. Note that PR is not the same as raw outward-directed radiation pressure, the latter being an even smaller effect in most cases. $\endgroup$
    – user10851
    Commented Aug 19, 2013 at 23:58

This question is different from, but related to another question How is it that the Earth's atmosphere is not “blown away”?.

In answering that question with respect to solar wind, I remarked that the orbital speed of Earth is 30 km/s while the speed of the solar wind varies between 300 km/s and 800 km/s in a nearly orthogonal directions (fully orthogonal if the orbit is considered circular). Hence the apparent wind is mostly a side wind, slightly in front (a slightly close reach in sailing terms). As a first small angle approximation, the dragging effect of the solar wind on the planet orbital speed does not come from the solar wind speed, but only from the planet own speed, which is at best a tenth of the solar wind speed.

Hence the actual effect of the solar wind on braking down Earth orbital speed is at best one tenth of the effect computed by Michael Brown, which makes it even less significant.

Another point is that the pressure due to the speed of the solar wind itself is pushing the planet outwards, away from the Sun. I am not sure how this should be analyzed, I mean to give the best insight. One way to do it is to consider that it reduces the centripetal force towards the Sun due to gravity. Furthermore, its effect must also decrase like the solar wind density in proportion to the square of the distance from the Sun, as does gravity. However the effect is proportional to the cross-section surface of the Earth, rather than its mass.

The energy output of the Sun is thought to have increased by about 30% since it formation (some 4.6 billion years ago). So the pressure from the solar wind should have increased in proportion, being equivalent to a minute reduction of the centripetal force that keeps the Earth in orbit. But it also increases the orbital drag in the same proportion. This energy output should continue to increase slowly.

Note that I assumed in these last remarks that the increased output is due to a greater amount of particles being output at the same speed. Some of the energy could be due to a greater speed of the solar wind which would increase the outward push, but not the orbital drag. I do not know which actually occurs.

More detailed calculations, which I have not done, should tell which of the two effects dominates, though they are probably both very negligible.


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