2
$\begingroup$

In the Poynting Robertson effect, it says in general, the Poynting Robertson effect is caused by the fact that objects like dust re-emit the light they absorb. Is that correct? Is this effect weak? Where does it apply and where it does not?

$\endgroup$
5
  • 2
    $\begingroup$ Hello curious. Have you, for example, read the Wikipedia page en.wikipedia.org/wiki/Poynting%E2%80%93Robertson_effect It is good to have done so first, and then ask about what is unclear to you. This is called "doing prior research". $\endgroup$
    – James K
    Nov 4, 2023 at 7:08
  • $\begingroup$ I've already done so, I'm confused about where does it apply, how does it work, the formulas , and if it's a very weak or strong force . I did my prior research and read a LOT about it but still I'm confused , Like in the wiki page you cited. It says just the meaning in general that the cause of the Pointying Robertson effect is the fact that objects like dust reemit the light they absorb. Not due to absorption . Is that correct? $\endgroup$ Nov 4, 2023 at 7:51
  • $\begingroup$ Good, thanks for that. Now you'll need to edit your question to ask as specific question. "What is the P-R effect" is not specific (and the answer is given on the Wiki page) but in your comment there is a specific question, which is answerable. $\endgroup$
    – James K
    Nov 4, 2023 at 8:02
  • $\begingroup$ So I've replaced your question by the specific one. $\endgroup$
    – James K
    Nov 4, 2023 at 8:03
  • $\begingroup$ But it doesn't explain the strength of this force . Is it a very weak force/effect or what? $\endgroup$ Nov 4, 2023 at 8:35

1 Answer 1

2
$\begingroup$

The Poynting Robertson effect (or P-R drag), as explained in the wikipedia Essentially there is a small, Special Relativistic aberration effect whereby the light received from the Sun by an orbiting body appears to come at them from a slightly off-radial angle. This imparts a tangential force (in addition to the usual radial radiation pressure). The force acts to slow bodies down in their orbits, causing them to spiral inwards as they lose angular momentum.

The best way to compare this with gravitational force acting inwards and radiation pressure outwards is to write expressions for the magnitude of each of these forces in terms of the sizes and density of the orbiting body, the orbital distance of the body and the mass and luminosity of the star it orbits; and to express all the latter quantities with respect to some calibration object, which I will assume is a spherical, black object of density $\rho=5000$ kg m$^{-3}$ and radius $R=10^{-6}$m orbiting the Sun (mass $M=M_\odot$, luminosity $L=L_\odot$) at $r=1$ au.

Poynting Robertson drag (acting tangentially): $$F_{PR} = 1.4\times 10^{-21} \left(\frac{R}{10^{-6}\ {\rm m}}\right)^2 \left(\frac{L}{L_\odot}\right) \left(\frac{M}{M_\odot}\right)^{1/2} \left(\frac{r}{1\ {\rm au}}\right)^{-5/2}\ {\rm N}\ . $$ Radiation pressure (acting radially outwards): $$ F_R = 1.4\times10^{-17} \left(\frac{R}{10^{-6}\ {\rm m}}\right)^2\left(\frac{L}{L_\odot}\right) \left( \frac{r}{1\ {\rm au}}\right)^{-2}\ {\rm N}\ . $$ Graviational force (acting radially inwards): $$ F_G = 1.2\times 10^{-16} \left(\frac{\rho}{5000\ {\rm kg/m}^3}\right) \left(\frac{R}{10^{-6}\ {\rm m}}\right)^{3} \left(\frac{r}{1\ {\rm au}}\right)^{-2}\ {\rm N}\ . $$

From comparing these equations we can see that radiation pressure and gravity both depend on $r^{-2}$ so their ratio is independent of orbital distance. However, $F_R/F_G \propto R^{-1}$, so if our calibration object were as small as $R \sim 10^{-7}$ m, radiation pressure would exceed gravity and drive the body out of the Solar System. P-R drag would be completely irrelevant in this case.

For our calibration object, P-R drag is 4 orders of magnitude smaller than gravity so it has a weak influence. However the ratio of P-R drag to gravity for the Sun depends on $\rho^{-1} R^{-1} r^{-1/2}$. For P-R drag to become more important, the density of the body would have to be much (implausibly) lower; or the size of the object would have much smaller - but then radiation pressure would blow it out of the Solar System; or it would have to be much closer to the Sun.

The force would be much weaker/negligible if the size of the object were increased or it were further away from the Sun.

Note though, that even though it is a weak force in the calibration case considered, the force acts tangentially, so is in addition to the radial forces of gravitation and radiation pressure. Thus, if the object is bound to the Solar System then P-R drag will inevitably act on a timescale which becomes shorter as its size becomes larger with respect to $F_G$.

In the calibration case, a $F_{PR} = 1.4\times 10^{-21}$ N acting on the calibration body produces an acceleration of $7\times 10^{-8}$ m/s$^2$. With an orbital velocity of $\sim 30000$ m/s at 1 au, this suggests a timescale on which the P-R drag would act of $\sim 10,000$ years. As discussed above, this timescale would decrease if the body moved closer but increase linearly with the size of the body, and would exceed the age of the Solar System for $R> 0.3$ m.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .