The Reason Why

There are odd things about the orbits of trans-Neptunian objects that suggest ( to some) that there might be an undiscovered super-Earth-sized planet  a few hundred AU from the Sun..

We haven’t seen it, but then it would be very hard to see. The underlying reason is simple enough, but I have never seen anyone mention it: the signal from such objects drops as the fourth power of distance from the Sun.   Not the second power, as is the case with luminous objects like stars, or faraway objects that are close to a star.  We can image close-in planets of other stars that are light-years distant, but it’s very difficult to see a fair-sized planet a few hundred AU out.

 

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26 Responses to The Reason Why

  1. Anonymous says:

    fourth power
    Because the light emitted from the sun (second power) has to reflect off this hypothetical planet to arrive back to us on Earth?

  2. engleberg says:

    If we find it we have to call it Pluto.

  3. Shouldn’t astronomers be able to determine the approximate position of this hypothetical planet by carefully measuring the orbital perturbations of the trans-Neptunian objects?

    I’m guessing that if astronomers can’t do this it’s because even measuring the orbits of those trans-Neptunian objects is hard to do with precision.

  4. dearieme says:

    Why not just consult the wee chaps in Area 51?

  5. Анисимов Дмитрий says:

    If it’s Jupiter-sized planet, then it could emit much more heat than it receives from Sun, so fourth power doesn’t apply…

    • protokol2020 says:

      This is very true. Every Jupiter size (even less) object can’t be very cold. At least not yet. So many radioactive atoms in such a big mass and so relatively small surface for that energy produced inside to escape through it — makes this surface relatively hot.

      There are other contributors, too. Like compressing under gravity. Every gold atom falling slowly toward the center makes heat. Every hydrogen molecule, falling into a small hole beneath its previous position, makes heat. Every place swap between a lighter and heavier isotope also makes heat.

      For big objects, 1/r^2 still applies.

      • gcochran9 says:

        Generally, people haven’t been talking about a Planet X as big as Jupiter.

        • just a lurker says:

          According to Michael Brown (the infamous Pluto killer) this is the best case for hypothetical Planet Nine.

          https://physicstoday.scitation.org/doi/10.1063/PT.3.4172

          In a sophisticated comparison of solar-system observations to numerical simulations, we find a best match to be a putative Planet Nine that is approximately six times the mass of Earth, inclined with respect to the ecliptic by a little less than 20 degrees, and in a moderately eccentric orbit about 400 times as distant from the Sun as Earth.

          • zimriel says:

            Much of that mass will be ice, lower density than Earth’s iron and silicon. So it might be six times the mass but, oh, eight times the radius. Sixteen times the surface-area as viewed head-on – which it will be, since we’re only going to see the sunny side.

  6. ckp says:

    Let’s see if I’m getting this right:

    A perfectly flat, perfectly reflective Sun-facing mirror orbiting at a distance x would produce a coherent image of the Sun at an apparent distance of 2x, so its signal strength would go as 1/(2x)^2 = 1/4x^2, i.e. still an inverse-square.
    But a planet(oid) isn’t a flat mirror, it gives diffuse reflections. Incident radiation is reflected back in all directions in an expanding (hemi)sphere, so we get another inverse-square factor for the journey back which stacks on top of the original inverse-square, yielding an overall inverse-fourth relationship.

    • protokol2020 says:

      You are getting it mostly right, I guess, but in fact, it’s much more complicated. Assume, that we have quite a perfect sphere in the great distance from the Sun. Assume also, that the Sun is above this sphere’s north pole. At the north pole, the incident light would be reflected back to the Sun (a 180 degrees turn), but at a small distance from the north pole, the incident light would reflect slightly away, let say 179 degrees turn). At every sphere’s point, the angle of incidence rule applies. So at 45 degrees north parallel on that sphere, every ray (photon) from the Sun would be reflected 90 degrees from its previous course. Near the sphere’s equator, those rays would be almost unscattered.

      Fortunately, we are very close to the Sun, so only the light reflected from a thin ring around the sphere’s north pole would ever reach us. It’s a bit worse than 1/r^4.

      This is purely geometrical consideration. But there are also small albedos for distant objects. Not only small reflections of very diluted sunlight but also less and less reflective stuff when you are going away from the Sun. As some deep space black “mold” sets in when there is not enough light, but a lot of cosmic rays instead.

  7. deliciousprions says:

    For analogous reasons, the equation for detection range of radar has an inverse fourth power term in it as well. Ever wonder why the F-22 and F-35 are advertised as having such teeny weeny radar cross sections in the approximately .001M^2 region, or whatever it is? Because it has to be that small to get tactically useful reductions in detection range.

    You wanna halve the range at which a given radar system picks up an airplane (or missile or whatever)? RCS has to go down by a factor of 16. You want to cut the range to a third? RCS has to go down by a factor of 81. And so on. This type of engineering is possible, but spendy.

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