# Effect of radiation pressure on Earths orbit in large timescales

I have seen some posts concerning the radiation pressure exerted by the Sun on Earths surface (Force on Earth due to Sun's radiation pressure). Though it is fairly small may it have a considerable effect on Earths orbit? Considering the time dependent kinetic energy $E_{\text{kin}}(t)$ of the photons over a large time scale $[0,T]$ (let's say billions of years) and the time dependent potential energy $E_{\text{pot}}(t)$ of the Earth w.r.t. the Sun my first thought was to equal their increments:

$$\dot{E}_{\text{kin}}(t)=\dot{E}_{\text{pot}}(t).$$ I consider the kinetic energy of Suns radiation pressure as time dependent because over the years the Sun became "brighter". If the rest mass intensity of the photons coming from the suns surface was $T$ years ago only a fraction (let's say $q$-times) of the intensity $\dot{m}_{\text{tot},0}$ now we would have

$$\dot{E}_{\text{pot}}(t)=\gamma m_{\text{E}}m_{\text{S}}\frac{\dot{d}_{\text{SE}}(t)}{d_{\text{SE}}(t)^2}$$ and $$\dot{E}_{\text{kin}}(t)=\frac{\dot{m}_{\text{tot}}(t)}{32}\left(\frac{d_{\text{E}}}{d_{\text{SE}}(t)}\right)^2c^2$$ with $\gamma$ the gravitational constant, $c$ the speed of light, $m_{\text{E}}, m_{\text{S}}$ the masses of Earth and Sun, $d_{\text{SE}}(t)$ the distance between Earth and Sun, $d_{\text{E}}$ the Earths diameter and $\dot{m}_{\text{tot}}(t)$ the total rest mass of photons hitting the Earth (seen as a disk). So equaling those two energy increments would lead to $$d_{\text{SE}}(t)=d_{\text{SE},0}+\frac{1}{2}R(1-q)\left(\frac{t^2}{T}-T\right)$$ with $$R=\frac{1}{32}\frac{d_{\text{E}}c^2}{\gamma_{\text{E}}m_{\text{S}}}\dot{m}_{\text{tot},0}$$ with $d_{\text{SE},0}$ is the distance of Sun and Earth now. All this would lead to a covered distance of the Earth within a time scale $T$ of 4.5 billion years of more than 40 million kilometers. This seems to be oddly too much. In another post (Radiation pressure question) I read the hint, that the acceleration of the Earth due to the radiation pressure has to be seen in proportion to the gravitational acceleration due to Suns gravitational field. Therefore the covered distance of the Earth due to solar pressure would be some meters at most. My actual question is where is the fault in my approach above and how can one calculate exactly the difference of the distance between Earth and Sun if the radiation pressure is "switched" on and off?

• Where does the factor 32 come from? What have you assumed for $q$? Dec 9, 2015 at 9:06
• The total rest mass of the photons that hits the Earth is $\dot{m}_{\text{tot}}$ times the ratio of the surface area of the disk representing the Earth with diameter and that of the ball with radius : $\dot{m}_{\text{tot}}\frac{1/4\pi d_{\text{E}}^2}{4\pi d_{\text{SE}}^2}$. Together with the factor $\frac12$ from the kinetic energy I get the 32. I have assumed $q$ so that the Sun would have an increase of 25% of radiation intensity over the last 4.5 Billion years. Dec 10, 2015 at 9:52
• The kinetic energy of photons is $pc$. Have you just used that $\dot{m_{tot}} = L_{\odot}/c^2$ ? The surface area of a disk is $\pi d_E^{2}/4$ Dec 10, 2015 at 12:15

One obvious problem with your approach is that if the Earth is struck by photons carrying energy, then the Earth also gains mass. Therefore your equation for the rate of change of potential energy should include the mass gained by the Earth and also the mass lost by the Sun and the change in the centre of mass position. A second problem is that the energy of the Earth's orbit has both potential and kinetic energy; both of which must change if the Earth's orbit changes. Finally, the kinetic energy of photons is $pc$, where $p$ is their momentum.

To first order you can tell that radiation pressure is negligible in the following way.

The force on the Earth due to the radiation from the Sun is roughly equal to the Poynting vector (the solar constant) multiplied by the disc area presented by the Earth, divided by the speed of light (neglecting albedo and such). $$F = \frac{1300}{c} \pi R_E^{2} \simeq \frac{L_{\odot}}{4\pi d_E^2\, c} \pi R_E^{2} \ \ \ N$$

This force is in the radial direction opposing the gravitational attraction between the two bodies and also depends on the inverse square of separation. There is thus a small(!) correction to the centripetal force, such that the Earth's orbit is larger than it would be if it we orbiting a dark object of the same mass.

This can be expressed by making an effective correction to the solar mass. i.e. We leave the expression for gravitational attraction the same, but replace the mass of the sun with $$M' = M_{\odot} - \frac{\kappa_0}{G},$$ where $$\kappa_0 = \frac{L_{\odot}}{4 \pi M_E c}\pi R_E^{2} = 2.17\times 10^{6} \ m^3 s^{-2}$$ and $M_E$ is the mass of the Earth. Notice that $\kappa$ is inversely proportional to the mass of the accelerated object, thus radiation pressure is more noticeable for less massive objects (of the same size).

To make progress we then note that the angular momentum of the Earth is conserved, so that $$J = M_E\, d_E\, v = M_E\, d_E\, \left( \frac{G M'}{d_E}\right)^{1/2}$$ is a constant. This means that $d_E\,M'$ is a constant.

Thus at time $t$ we can say that $$d(t) = d_E \left(\frac{M'_0}{M'(t)} \right)$$ So the orbital radius will change in response to the changing mass of the Sun and the changing radiation pressure.

If we just ignore the mass loss for the moment and say that in 5 billion years time the solar luminosity increases to $100 L_{\odot}$, then $$\frac{d(t)}{d_E} = \frac{M'_0}{M'(t)} \simeq 1 + \frac{(\kappa(t) -\kappa_0)}{GM_{\odot}} \simeq 1 + \frac{\kappa(t)}{GM_{\odot}}\ \tag{1}$$

So I get that the orbital radius increases by about 24 cm if the radiation pressure increased by a factor of 100.

Similarly, if you switched the radiation pressure off, then one would replace $M'(t)$ with $M_{\odot}$ and find that $$\frac{d(t)}{d_E} = \frac{M'_0}{M_{\odot}} = 1 - \frac{\kappa_0}{GM_{\odot}}$$ which tells us that the Earth's orbit would shrink by 0.24 cm.

These are negligible compared with the changes associated with mass loss from the Sun, both in terms of the solar wind and mass loss due to radiation. These amount to $\sim 1$ part in $10^{13}$ every year. So a fractional change in $M'$ of $\sim 5\times 10^{-4}$ over 5 billion years and hence an increase in orbital radius of $\sim 10^{8}$ m.

• Thank you very much for the explanation! There is just one unclarity on my side: Besides the neglection of the mass of the photons I have (falsely) assumed that the kinetic energy of the earth remains constant while the kinetic energy of the photons hitting the earth are equal to some potential energy-increase of the earth. Can I now likewise identify the "energy-increase" of the distance-difference between Earth and Sun (e.g. 24cm) with the kinetic energy of the photons that hit the earth in that period of time (e.g. $\frac{1}{2}mc^2$ with $m$ is the rest mass of all those photons)? Dec 10, 2015 at 9:40
• @SMatthes $0.5 m c^2$ is not the kinetic energy of a photon, and photons do not have a rest mass. I'll think about energy conservation. Dec 10, 2015 at 12:17