What would it take for a spacecraft to travel to the sun's nadir and stop?

In a science fiction future universe, spacecraft routinely travel from a distance "above" a star's north or south pole to rendezvous with planets orbiting in the star system's orbital plane. And vice versa.

As a writer, I'm having trouble wrapping my head around the orbital dynamics involved. How hard would it be for a spacecraft to leave Earth's orbit, transit out of the orbital plane to a point "above" the star, and rendezvous with another craft that's stationary relative to that point? (I.e,. the rendezvous point is not orbital to the sun.)

If it helps the discussion, let's assume the distance to the point above the star is the same as the distance from the sun to Saturn.

I've been looking at the flight path of the Ulysses spacecraft, but as it orbited the sun and didn't have to "stop" relative to a point above the star, I'm not sure it's a good model for what would have to happen.

If it helps, the setting assumes that the spacecraft rarely accelerates/decelerates greater than 1 g for long periods for the comfort of its crew. The technology is vaguely like ours, but it assumes that the amount of fuel required is not a limiting factor.

• Why would they want to park there? They'll need to constantly fire their thrusters to keep from falling into the sun, and they're not on average any closer to (and are indeed usually further from) planets in the system than they would be by simply occupying a small orbit around the sun on the ecliptic plane. Nov 1, 2023 at 19:38
• @Scottoooooo You're somewhat misusing the word "nadir." Astronomically, the nadir is the direction from an arbitrary point above a surface to its gravitational center (said another way, it's the vector describing the force of gravity). In other words, you can travel along the sun's nadir, but the "nadir" isn't a place at which you can stop.
– JBH
Nov 1, 2023 at 20:04
• Sorry but there is no such thing as "the rendezvous point is not orbital to the sun.". You seem to be under the impression that the orbital plane of the planets is somehow special. The gravitational pull of the sun doesn't disappear at the solar system orbital axis. Nov 2, 2023 at 10:10
• @vsz You could simply orbit the star at 90 degrees to the planets, and you would at regular intervals occupy that point for that sweet cover shot... not that you'd ever see the planets from that distance... but you know, artistic license is a thing that needs no basis in reality. Nov 2, 2023 at 13:31
• In space, you don't "stop". If you want to go somewhere you can shut your engines off and just chill while maintaining distance from your destination, you either "orbit", "land" or "rendezvous".
– Wyck
Nov 2, 2023 at 17:33

A magic search term that might be of interest to you is statite. A statite balances the force of gravitational attraction to a body (such as a planet or star) with light pressure on a solar sail (though other mechanisms could be used instead, such as a magnetic sail or some other suitably science-fictional gadgetry). Such a spacecraft would not be orbiting and could remain "fixed" relative to that body, though it would probably have to trim its sails a bit from time to time if there were changes in light pressure or local gravity.

From recollection, the jump-ships in the Battletech universe assumed this position, using an energy-collecting solar sail to charge their FTL transport system.

The major issue you have is that transitioning from a solar equatorial-ish orbit (like the one that the Earth is in) to a solar-polar orbit is extremely expensive in terms of delta-V, and then once you reach your pole-sitting statite you need much more delta-V again to stop. If you have a super-efficient torch drive you might not have to worry about such a thing, of course.

If it helps, the setting assumes that the spacecraft rarely accelerates/decelerates greater than 1 gee for long periods for the comfort of its crew. The technology is vaguely like ours

Accelerating at 1G for extended periods of time requires rockets of astonishing power and capability, and would be far beyond what humans are capable of for a considerable time. Be careful about going down the hard scifi route, for real world physics is tough to beat!

Now you've given a few more hints about what you're thinking about, I can add the following observations:

• Solar flux is negligible outside of the inner solar system. Even at Marslike distances from the sun it is too low, and by the time you get out to Saturnlike distances you'll barely notice it... a perfectly absorbing sail would experience ~50 nanonewtons of thrust per square metre. Solar collectors and solar sails are both largely pointless out here
• Gravity apparently obeys the same inverse-square laws as light, so at Saturnlike distances a Sunlike star only exerts about 6.5 microgees. You can acheive station-keeping thrusts against this using something an ion drive... the Dawn space probe acheived a greater acceleration than this when fully loaded, and had enough fuel to run its rocket for years. Unlike a solar statite, your rocket-based things would need periodic refuelling.
• A brachistochrone transit crossing 8 AU (eg. from an Earthlike distance to a Saturnlike distance from a star) at 1G continuous acceleration takes about 8 days and requires a Delta-V of ~7000 km/s. Even to fly a single AU at that acceleration is ~2400 km/s. That's a lot. Dawn had 11 km/s. You can't do this without serious nuclear rocketry, but if you can do it then you can consider the laws of orbital mechanics more as polite requests. For example, injecting yourself into a polar heliocentric orbit from an earthlike heliocentric orbit requires ~42 km/s delta-V... an outrageous amount for a modern-day or near-future spacecraft, but a rounding error for your monstrous things. They can just point where they want to do and blast and not worry too much. "Parking" by the non-orbital target and station-keeping to oppose the local gravity gradient would be trivial.

Because I thought it was an interesting discovery, even if it is obvious in retrospect, there's an equation you can derive from the force imparted by sunlight on a perfectly absorbing surface, and the force caused by the gravity of the sun an equation that tells you how large a sail would have to be in order to make a statite: $$C_{ad} = { L_\odot \over {4\pi c G M_\odot} }$$

The critical areal density of a solar sail, $$C_{ad}$$, in kilograms per square metre, is the ratio of the total mass of your statite (including the sail) to the surface area of the sail pointing directly at the Sun. $$L_\odot$$ is the luminosity of the Sun, in watts, $$M_\odot$$ is the mass of the Sun, $$c$$ is the speed of light, and $$G$$ is the gravitational constant.

This ratio is independant of the distance of the sail from the Sun, because both light pressure and gravitational attraction obey the inverse square law, and so cancel each other out nicely. If the areal density of your statite exceeds this critical ratio, it will fall back towards the Sun and a firey demise. If it is lower than the critical ratio, it will be blown out into deep space, though possibly very slowly. In the latter case at least you can do station keeping by pointing your sail ever so slightly away from the sun to reduce sunlight pressure.

For the Sun, the critical ratio is ~0.765g/m2. Clearly, statites of any significant mass will need colossal sails of some exceedingly light, low density material. Using a solar-powered ion drive is probably a much more sensible way to hold yourself up, even if you do need periodic refuelling.

• It is, in fact, a Battletech related writing question. Haha. Those are all things I was thinking, but I wanted to make sure I wasn't missing something in the physics or orbital dynamics that made it more feasible somehow. Thank you for your thoughtful answer! One of the bottlenecks of my own research is my limited vocabulary for the questions I'm asking. Nov 1, 2023 at 19:59
• if you have a statite, couldn't the trimming of the sail alone let you move around pretty much anywhere in the solar system? (close/away from the star is tricky to balance, but any sideways on a sphere of equal distance should be fairly easy, no? (something like piloting a kite) Nov 1, 2023 at 22:06
• @njzk2 you can, in fact, go more or less anywhere with the aid of a solar sail... you just need patience. Moving further away and staying there would be awkward, but you could fall in towards the star where radiation pressure is stronger and inject yourself into a suitably eccentric orbit that would carry you a long way out. Tricky to get out of that orbit again, though. Nov 1, 2023 at 22:52
• FWIW, in the BattleTech universe, the sails the JumpShips use (lots of CamelCasing in BT) only serve to collect energy from the sun to charge the jump drive. Any role of the sails in keeping stationary is not addressed, and per canon, JumpShips have "a weak station-keeping drive" for that purpose. Nov 2, 2023 at 6:26
• @Blueriver poor phrasing on my part, more or less... it was mostly intended to mean "you need even more on top of what you've just used up". The equatorial-to-polar transition might need as much as ~42km/s (though a heliocentric polar injection from a low polar orbit about Earth could be cheaper) though braking to a relative halt would only take ~30km/s. I think about some rewording. Nov 5, 2023 at 18:19

It's a hugely expensive thing to do, in terms of orbital mechanics.

The main issue is that simply being on (or in a bound orbit around) a planet that already orbits near the ecliptic means you already have a lot of energy from that planet's orbit. If you're going to another planet that orbits near the ecliptic (in the same direction), then most of that energy is "reusable" and thus a good thing.

For example, Earth orbits the Sun at about 30 km/s. Mars is a little less because it's further out, around 25 km/s. A simple Hohmann transfer from Earth orbit to Mars orbit only needs about 3 km/s change in velocity; this is an order of magnitude less than the orbital velocities that Earth and Mars are travelling at. So if you were "starting from scratch" with no useful orbital velocity in the ecliptic, it's much harder to reach Mars than it would be from near the orbit of just about any body orbiting the Sun, since you would need to provide enough energy to get up to the speed that Mars orbits at.

Travelling to or from a point really far from the ecliptic (like directly above or below the Sun) is also really difficult. If I'm orbiting near Earth and I want to get to the far side of the solar system, then I don't really need any energy; Earth's orbit will simply take me there if I wait 6 months. And we've seen that it takes relatively low amounts of change in velocity to travel interplanetary distances (eventually), by re-using the energy we already had form the starting orbit. But if I want to get to a point directly above the Sun and a similar distance as Earth's orbital distance, then "coasting" there would require me to be on an orbit with the same speed as Earth's orbit, but rotated 90°, so that I'm travelling "up" where Earth is travelling "sideways". Earth's orbital velocity is never pointing in a direction where I can re-use any of its energy for that, so I would have to provide all 30 km/s of it myself. Having done that, I would also still have the 30 km/s "sideways" velocity from Earth's orbit that is not useful, so I would have to provide enough energy myself to cancel out that velocity on top of the energy required for the "vertical" orbital velocity. (And if you want to "stop" at a position directly above the Sun, you then need to pay a similar magnitude energy cost again to zero out your orbital velocity as you pass overhead, not to mention whatever you're doing to staion-keep once you're there. I'm not saying that "thrust Earth-retrograde 30 km/s, then thrust perpendicular to the ecliptic 30 km/s, then coast to above the Sun and thrust retrograde 30 km/s" would actually be the most energy-efficient way to do this, but it gives you a ballpark idea of the magnitudes involved)

Nobody would plan these sorts of courses unless they have so much energy available that these astronomical energy costs are a minor issue to them, or unless there is some very compelling reason to go there. Given the existence of an ecliptic plane where most interesting things are orbiting near that plane (in the same direction), it's just much cheaper to travel in orbits on that plane.

• The OP's stated spacecraft capabilities imply delta-V values of thousands of km/s or more. Injection into polar or retrograde heliocentric orbits would be trivial. Nov 2, 2023 at 9:05
• @StarfishPrime The capabilities of the OP's ships are up to them. I was just trying to convey an intuitive sense of just how much more costly this would be than more "normal" space journeys. The OP can decide how that affects their worldbuilding.
– Ben
Nov 2, 2023 at 9:37

Pointless if you are using Newtonian physics.

Let's start with a minor frame challenge explaining why you should never under any circumstances see this done in any science-based setting. Newtonian physics makes stopping over the sun a pointless waste of energy no matter how advanced your civilization becomes. Even if you need to be able to see the whole solar system at once, you could orbit the sun and apply a much smaller Z-axis thrust to hold your ship "high" enough to be able to see planets on the far side of the sun.

Mercury's orbit is 47 million km from the sun, and the sun has a coronal radius of about 7-14 million km and a diameter of 0.7 million km. So, if you orbit the sun at a the edge of its corona, and thrust your way up to about 0.8 million km above the planetary plane, you will be able to see Mercury from the far side of the sun allowing you to see all of the planets in the solar system without having to spend fuel on fully stopping and while using about 10-20 times less fuel to hold yourself above the planetary plane.

Even when you look at Starfish Prime's answer, orbital mechanics allow one to offset some of the needed efficiency for solar wind based lift. So, instead of having to design an insanely light ship/sail, it makes more sense to design a sail big enough to capture the ideal amount of energy for cost of manufacture and maintenance, and then derive the remaining needed energy from orbital momentum.

Whether the thrust comes from rockets, the sun itself, or some theoretical antimatter drive etc. does not matter, because it will always be way cheaper to orbit.

The only solution that makes since is if you use a reactionless drive

The only way to make a this a reasonable consideration is if you use a reactionless drive. This is any theoretical engine that moves by non-newtonian principles like wormholes, improbability drives, parallel dimensions, etc. Because these drives move without imparting any momentum, when you turn it off, you instantly return to a stationary position relative to your original reference frame. If this is the case, coming to a dead stop over the sun could be a trick for using the star's gravity to slow you down so you can match reference frames after an interstellar jump. If you need to stay there, you could just jump from one side of the star to the other, yo-yo-ing above the poles.

How hard is it to use a reactionless drive in this manner? ... no one knows. The whole point of reactionless drives is that they don't obey the known laws of physics; so, until one is invented, there is no hard science answer to this question. But the good thing as a writer when dealing with them is that no one can contradict the physics of a jump drive. So, as long as you stay internally consistent, it's as hard or easy as you want it to be.

You could also perhaps have an Alcubierre meta-material drive. A material that imparts a constant forward acceleration by warping spacetime. This would be a sort of "thruster" than once made, can never be turned off; so, by parking over the sun or a planet, at just the right distance, you could use the gravity well to cancel out your acceleration in which case, the answer is that it takes no energy at all... you'd just have to maneuver a bit inside the corrona at a range of 3.68 million kilometers to offset your 1G of acceleration.