Based on an XKCD comic, XKCD 1626 which has the image:

enter image description here

Do any of the active duty ICBM's have the ability to actually target the Sun?

Considering that the R-7 booster that is the basis used by Soyuz/Progress was an ICBM, as were the original Atlas and Titan programs boosters, perhaps orbital. But Earth escape, and then towards the Sun?

  • 4
    $\begingroup$ Note that the extra text that pops up on hover says 'It took a lot of booster rockets, but luckily Amazon had recently built thousands of them to bring Amazon Prime same-day delivery to the Moon colony'. So Randall covered himself, possibly to avoid persistent comments from his NASA buddies. $\endgroup$
    – kim holder
    Jan 6 '16 at 14:56
  • $\begingroup$ Obviously the AI figured out how to greatly improve the performance of the rockets. $\endgroup$ Jan 8 '16 at 3:40

The delta-v needed from low Earth orbit to a Hohmann transfer orbit with a periapse inside the Sun is actually "just" 21,300 m/s. But there is a better option. A bi-elliptic transfer to just hit a central body is better when the ratio between the orbital radius and the radius of the central body is larger than 4.82. The orbital radius of the Earth divided by the radius of the Sun is larger than 200. Therefore, accelerating up to solar system escape velocity, and then do a minuscule manoeuvre at a great distance is only going to cost 8750 m/s of delta-v from LEO. Even that is impossible for an ICBM. A fly by of Jupiter is even better, and combined with a multi fly hit-the-sun trajectory, it is only going to cost around 6300 m/s.

The theoretically lowest is however not going to cost more than a transfer orbit towards the Moon (3120 m/s). With three Moon flybys, you can escape the Earth-Moon system with a little $v_\text{inf}$. After a long time waiting for the then long synodic period to sync up, you can do a flyby of the Earth-Moon system to increase the $v_\text{inf}$. (You can not increase the $v_\text{inf}$ just with a flyby of a planet, however, you can if it is a two body system. That is the same effect as a binary star flyby, only much weaker.) Repeat this over again several times for the next century, and the warhead would hit the Sun.

3120 m/s is maybe achievable if you replace parts of the warhead with an upper stage, as a regular ICBM has trouble even going into orbit. Even then, major upgrades are required. But at least, it is within the same magnitude.

Of course, why are you going to nuke the Sun? I assume you have a good reason for that. (Kind of not assuming that actually)

  • $\begingroup$ Yes, the 21300 m/s number is correct, even if the Earth’s orbital velocity is 29800 m/s. $\endgroup$ Jan 6 '16 at 20:08
  • $\begingroup$ Superman 4: the Quest for Peace en.wikipedia.org/wiki/Superman_IV:_The_Quest_for_Peace $\endgroup$
    – geoffc
    Jan 6 '16 at 20:19
  • 24
    $\begingroup$ We're not nuking the sun, we're sunning the nukes. $\endgroup$ Jan 6 '16 at 22:03
  • 2
    $\begingroup$ It only takes a little more $\Delta V$ to get to Venus as compared to the Moon, and then it will take less than ten years of Venus flybys to hit the Sun. Solar Probe Plus is doing seven Venus flybys to get down to 8.5 solar radii. $\endgroup$
    – Mark Adler
    Jan 8 '16 at 2:32

Borrowing from Wikipedia's article on modern ICBMs:

One particular weapon developed by the Soviet Union (Fractional Orbital Bombardment System) had a partial orbital trajectory, and unlike most ICBMs its target could not be deduced from its orbital flight path. It was decommissioned in compliance with arms control agreements, which address the maximum range of ICBMs and prohibit orbital or fractional-orbital weapons.

Orbital and fractional-orbital weapons were prohibited by the SALT II agreement in 1979, and the FOBS missile was phased out in early 1983.

Remember that the "B" in ICBM stands for ballistic; other than the boost phase, ICBM flight is largely unpowered and suborbital.

Given that breaking free of Earth's gravity, which would be necessary in order to reach the Sun, requires a higher velocity than attaining Earth orbit, and by implication of the SALT II agreement no current ICBMs are able to attain even Earth orbit, we can conclude that no matter the transfer orbit trickery involved, no current ICBMs have the capability to target the Sun.


No, nowhere near. Hohmannfan's answer describes various time-consuming and complex routes to the sun which take over 3200m/s ∆v beyond that needed to get to LEO, and ICBMs can't generally get to LEO. They are built to go from point A to point B on Earth's in about half an hour, and don't have substantial performance margin beyond that.

This Q&A touches on the topic a little bit.

  • $\begingroup$ I was pretty sure. Stupid XKCD/Superman4. $\endgroup$
    – geoffc
    Jan 6 '16 at 14:47
  • 2
    $\begingroup$ Well, "all" you need to do is get to Jupiter, then do a gravity assist retroburn to remove as much velocity and increase eccentricity to later fall by Mars, Earth, Venus and Mercury... Dave Lister could do it :) $\endgroup$
    – TildalWave
    Jan 6 '16 at 15:30
  • 1
    $\begingroup$ @gerrit We've already consolidated our Jupiter energy debt into one-off crash-dive of the Galileo spacecraft into it, repaying our mortgage at Pu-238 decay rate. $\endgroup$
    – TildalWave
    Jan 6 '16 at 17:38
  • 2
    $\begingroup$ I would expect that solar heat would cause the warhead to simply fall apart long before any significant nuclear reaction occurred. Sun's hot, and nuclear warheads don't detonate without very precise things happening. $\endgroup$ Jan 7 '16 at 2:21
  • 3
    $\begingroup$ @rolfedh: it wouldn't, at least not into nuclear explosions. About all contemporary nukes are implosion-type, meaning a central rather large subcritical plutonium sphere, surrounded by conventional charges. It requires extremely precisely timed ignition mechanism to create a spherical blast wave that compresses the plutonium to supercritical levels. If the timing of ignition is off (e.g. the charge explodes starting from the most heated side) you're getting a big dirty bomb, but no nuclear explosion - plutonium won't be uniformly compressed. $\endgroup$
    – SF.
    Jan 7 '16 at 2:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.