I have a planet, let's call it Davy-Tim. Davy-Tim has 2 times the mass of the Earth and has a magnetic field. Davy-Tim has a negligible atmosphere and orbits an average millisecond pulsar named Dad, at a distance of 0.8 AU. Dad has a certain tilt and poles so that its radiation beams are aligned with Davy-Tim's orbit. This means that the radiation beams always hit within Davy-Tim's orbit and hit Davy-Tim every time the beams come around.

My question is, what types of/how much radiation will Davy-Tim be receiving?

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    $\begingroup$ At 0.8 AU a planet cannot complete an orbit in milliseconds, so the pulsar's beam cannot be always hitting the planet. $\endgroup$
    – L.Dutch
    Oct 11, 2022 at 12:17
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    $\begingroup$ I meant that the beams hit the planet every rotation(every time they come around they hit the planet). $\endgroup$
    – KaffeeByte
    Oct 11, 2022 at 12:18
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    $\begingroup$ This is gonna be one of those research-intensive questions, I can feel it. $\endgroup$
    – Tmartin
    Oct 11, 2022 at 13:13
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    $\begingroup$ This question is better suited for physics/astronomy stackexchange since it is asking a genuinely scientific question (names of the pulsar and planet are irrelevant). $\endgroup$
    – XYZT
    Oct 11, 2022 at 13:29
  • $\begingroup$ @L.Dutch I think it would be possible if the magnetic pole was on the equator of the neutron star. Whether the magnetic pole can be there or not I have no idea. $\endgroup$ Oct 12, 2022 at 1:58

2 Answers 2


Note: The 10 Sunsworth figure was wrong for this particular pulsar but is realistic in general. It is not the luminosity of the pulsar. it is the "thermal luminosity" of the pulsar's orbiting star.

The luminosity for pulsars in general ranges from the order of $10^{29}$ ergs/second to the Crab pulsar which is $10^{38}$ ergs/second. The Sun is about $10^{33}$ ergs/second. So we see pulsars range from $0.0001$ to $100000$ Sunsworth. 10 Sunsworth falls in that range. But the range is so big you can really make the pulsar as powerful as you like.

The figures on the table are not the exact measured luminosity. They are the so-called "spin-down luminosity". This is a theoretical upper bound based on the period and mass of the pulsar, and how quickly it is losing kinetic energy.

520 Sunsworth

If you stand on the planet then 520 times as much energy hits you from the pulsar jet, compared to how much would hit you from the Sun on Earth.

HOWEVER: The beam will hit the planet for a very brief amount of time before it rotates out of focus again. So you will not have time to be cooked alive. It is an exercise for the reader to compute if you will be cooked alive over several rotations.

The pulsar PSR J0952–0607 weighs about 3 Suns. The luminosity is about 10 Suns' worth. The following method works the same if you substitute figures from your favorite pulsar.

Luminosity is the amount of energy released by the star. For a normal star the energy is radiated symmetrically. For a pulsar the energy goes out the two jets. So each jet shoots out 5 Suns worth of energy.

For simplicity let's say the emission shoots out in a cone of 10 degrees. That's consistent with this answer on Physics Exchange

For simplicity we'll suppose your planet is exactly 1 AU from the pulsar. Consider the sphere of all points 1 AU from the pulsar. The jet sprays into "some spherical cap" (the blue part of the sphere surface)

enter image description here

If that cap is big then the energy is more spread out and less hits the planet. If the cap is small then the beam is focused and the planet gets more energy.

Fortunately the area is easy to compute. The formula is $A = S (1-\cos \theta)$ for $S$ the area of the whole sphere. For $\theta = 10^\circ$ we have $\cos \theta =0.985$ and $A = 0.015 S$.

That means we have 5 Sunsworth of energy compressed into an area $0.015$ the size of the sphere. So the density of the energy passing through the cap is $5$ times $1/0.015 \simeq 66.666\ldots $ times the energy density of the Sun on the Earth. The total is $5/0.015 = 333.333$ Sunsworth.

Move the planet to 0.8 AU and the area of the cap shrinks by $0.8^2 = 0.64 $. So multiply the density by $1/0.64 = 1.5625$ to get about $520$ Sunsworth.

If you shrink the angle to 1 degree as mentioned in the same Physics answer, then we have $\cos \theta = 0.9998$ and the final answer goes up by a factor of 100. So it would be

52,000 Sunsworth.

  • $\begingroup$ You make an error in your hypothesis when you mention "The [pulsar] luminosity is about 10 Suns' worth." 10 suns is the luminosity of the whole system: pulsar and companion. Most of the luminosity comes from the companion of the pulsar. The companion's luminosity is mostly coming from the tremendous level of radiation it receives from the pulsar itself. So 10 sun is only a very small amount of energy the pulsar emits. $\endgroup$
    – Uriel
    Oct 23, 2022 at 20:06
  • $\begingroup$ @Uriel I don't follow. Is the luminosity coming mostly from the pulsar or mostly from the companion? $\endgroup$
    – Daron
    Oct 23, 2022 at 20:37
  • $\begingroup$ See wikipedia: 10 suns is the luminosity of "PSR J0952–0607 B", the companion. And this luminosity comes from energy received from PSR J0952–0607 A (the pulsar), and re-emited as light. The pulsar itself emits in more energetic wavelengths (gamma rays), not in visible light. $\endgroup$
    – Uriel
    Oct 23, 2022 at 20:47
  • $\begingroup$ @Uriel Luminosity is a measure of the total amount of energy radiated by a star or other celestial object per second. Not just the visible light. $\endgroup$
    – Daron
    Oct 23, 2022 at 21:39
  • $\begingroup$ My point was about your assumption of '10 suns luminosity' that is incorrect. $\endgroup$
    – Uriel
    Oct 24, 2022 at 6:37

Davy-Tim receives Nova-level radiation

Neutron stars may be small, but they are really hot. As in millions of kelvins hot. This means that most of the radiation emitted by a neutron star will be in HARD X-RAYS, and Gamma Rays.

The fact that Dad has a radiation beam proves that it too has a accretion disk. Neutron stars are very similar to black holes in accretion characteristics i.e. only a tiny fraction of the accreted matter actually reaches the star, and the rest is blasted away. Neutron stars are really sloppy eaters.

Most pulsars rotate hundreds of time a second. And since Dad's radiation beam are in perfect alignment with Davy-Tim, you can expect to be hit hundreds of time a second. We don't know how much radiation you would get though, as there are really vague sources of information which are not clear, but from what I have read, Davy-Tim would receive supernova-level radiation every single second. And that too in hard x-rays and gamma rays, with traces of ultraviolet.

Dad's magnetic field would make the scenario even worse, as neutron stars tend to rotate so fast that their magnetic fields tangle, which releases massive amounts of energy.

Supernova with ultra-solar-flare combo.

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    $\begingroup$ Thank you for using the names :) $\endgroup$
    – KaffeeByte
    Oct 19, 2022 at 14:12
  • $\begingroup$ I find supernova-level radiation hard to believe. That would lead to the star supernova-ing itself out in a matter of seconds. $\endgroup$
    – Daron
    Oct 21, 2022 at 20:55
  • $\begingroup$ @Daron well, not exactly a supernova, but something really close. Don't forget that neutron stars are simply super-dense solid lumps of neutrons and heavy-nuclei, and that even the radiation would be hardly sufficient to break the star apart, as the atoms/neutrons are packed too tightly too budge even a bit. $\endgroup$ Oct 22, 2022 at 5:03
  • $\begingroup$ After doing the calculations, it looks like if the beam has an angle of 1 degree, then it sprays over about 1/10,000 of the surface. That is a ratio of one second to three hours. So if a supernova takes 100 seconds to happen then a pulsar spraying a supernova beam will burn itself out over about two weeks. $\endgroup$
    – Daron
    Oct 23, 2022 at 14:25

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