8
$\begingroup$

I'm sure most worldbuilders will have heard the aphorism that in space "Everyone sees Everything", or that "There Ain't No Stealth In Space". I'd like to propose a possible solution to this.

The stealth ship would be a drone, probably used as a missile bus, and used for short to medium term missions (weeks to months). Firstly, it would be shaped like a double cone with an inclination of 0.5 degrees, the angular size of the sun at 1AU, and would be continuously pointed towards the sun to minimize solar heating. Secondly, the hull would be coated in VantaBlack to minimize the probability of detection via radar systems or reflection of sunlight. Thirdly, to avoid the emission of infrared radiation, the hull of the ship would double as a storage container for liquid hydrogen, which evaporates at a temperature of 20.28 K. This would ensure that the exterior temperature of the ship is kept at this temperature or lower until the hydrogen has completely melted. The evaporating hydrogen would be used as fuel for cold gas thrusters, which would be positioned around the center of mass to avoid turning the ship and exposing more of its hull to sunlight. This would give the ship maneuvering capability.

The ship would be deployed in a very high earth orbit, in roughly the same orbit as the moon but opposite to it to mitigate risks of discovery by moon-based craft. The sensors attempting to detect the craft are in LEO or MEO, so detection distance is about 360,000 km (roughly lunar perigee). The ship's dimensions, chosen completely arbitrary, will be a 10m central diameter, which, if my math is correct, gives me a length of 1146m or so, and a volume of around 30,000 m^3.

So the question is: Would this be a viable method of stealth in space, and, if so, how long could stealth be maintained for before the heat sinks run out, assuming the ship carries a few missiles and a guidance computer as payload? (Bonus marks for anyone who wants to calculate the delta-v of such a ship)

$\endgroup$

This question asks for hard science. All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Answers that do not satisfy this requirement might be removed. See the tag description for more information.

  • $\begingroup$ What sort of detection distances are you concerned with? Is this vessel intended to operate in orbit around a planet, or be interplanetary? Detection capabilities change dramatically as the distances change. $\endgroup$ – Cort Ammon Jan 16 at 2:21
  • $\begingroup$ Ah, yeah, forgot to add that, one second. $\endgroup$ – Gryphon - Reinstate Monica Jan 16 at 2:22
  • $\begingroup$ What are the dimensions of the ship? Also, how are the missile bays, propulsion systems, and any necessary sensors being masked? VantaBlack can't work for all of them. $\endgroup$ – Bewilderer Jan 16 at 3:05
  • $\begingroup$ @Bewilderer as stated, the propulsion system is a simple cold gas thruster fueled by the melting hydrogen, so it can be masked with VantaBlack. The missiles would be stored inside the ship, which would break apart if the missiles were to be fired, given that its entire purpose is to be a missile bus. Finally, sensors are unnecessary given that the bus is a drone that will simply be commanded, probably by encrypted radio (which merely requires a radio receiver of the sort in my alarm clock). $\endgroup$ – Gryphon - Reinstate Monica Jan 16 at 3:09
  • 1
    $\begingroup$ I reccomend Matter Beam's series on Stealth in Space: toughsf.blogspot.com/2018/04/… $\endgroup$ – Mranderson Jan 16 at 9:05
10
$\begingroup$

No, it is probably terrible stealth.

A simple google search yields good results, such as this site with a picture of the reflectance.

We're talking about 2% reflectance, and increasing, for radio waves of wavelength of 25 microns and greater. A stupid crude simple radar could detect that (if close enough). If we are talking about big distances, possibly a powerful enough radar could do the trick.

See also this answer.


To deal with the comments: What if the ship is far away?

Assume a radar that shoots an EM wave with a power $P_0$. Assume our radar has a maximum detecting distance of $x$. This is equivalent of saying that, our radar can can detect a receiving EM wave with minimum power of: $$ P_1 \approx P_0\frac{1}{x^2}\frac{1}{x^2} = P_0\frac{1}{x^4} $$

This takes in consideration a hull of reflectance $100\%$. If, our reflectance is, instead, a number $R$, then, the new minimum detecting distance $r$ is: $$ P_1 \approx P_0\frac{1}{r^2}\frac{R}{r^2} = P_0\frac{R}{r^4} $$

Because $P_1 = P_1$, we can relate $r$, $x$ and $R$: $$ P_1 = P_0\frac{R}{r^4} = P_0\frac{1}{x^4} \quad\implies\quad \frac{r}{x} = R^{\frac{1}{4}} $$

For $R=2\%$ we have $r/x \approx 37.6\%$. Thus, the maximum detection distance would now be $37\%$ the initial one. So, if we acquire a radar that can detect 958,000 Km maximum, it can also detect such vantablack coated ship at 360,000 Km.

If we already have a radar that can detect 360,000 Km maximum, to adapt and detect the vantablack coated ship, it would be just a matter of tripling the power (sometimes not hard to do), or tripling the sensitivity (perhaps by increasing the antenna dish by a bit less than two times the initial size).


Reflectivity: enter image description here

$\endgroup$
6
$\begingroup$

Subterfuge

If you want stealth, hide behind a comet. It will be seen if someone is looking, but won't raise any warning bells - there are a lot of space rocks out there. If you could engineer the approach to be close, but far enough off that no preventive responses occur, all the better.

Better yet, a comet will release a gas trail, this will mask any cold-propulsive residue, and naturally act as interference for any radar or other electro-magnetic detection system. At worst such systems will recognise the ship, but will likely be dismissed as a false positive due to inconsistent observations at that and other frequencies.

Earth/Moon System

Asteroids are known to approach the Earth/Moon system, aim your asteroid to pass behind the moon some distance away from the Earth so as to not draw Preventative responses.

There is no longer any need for subterfuge. At these distances any detection facility will identify your spacecraft as at least anomalous when detected. Due to how close it is, there is a reasonable chance that such news will be passed around various observatories quickly. Soon after confirmation, it will likely be brought to the attention of those who can respond.

While deception is useless now, that is no reason to announce your presence. Leave the cover of the asteroid while it is occluded by the moon.

Target Earth

Use the lunar gravity and hot propulsion to slingshot the vessel to ramming speeds. The moon is 384,400 km away, the Juno spacecraft had a max speed at 58km/s which could travel that within 2 hours. As the key is to reach the target prior to any interception by a counter measure the faster the better:

  • 6406km/s takes 1 minute.
  • 3203km/s takes 2 minutes.
  • 1602km/s takes 4 minutes.
  • 801km/s takes 8 minutes.

Kinetic Bombardment

At these speeds the space ship does not need any fancy high-grade explosive. Though if you want that go ahead. Simply fill it with one or more tungsten metal cylinder and make a Kinetic Missile.

Mach 25 is about 8.5km/s so this spaceship will need to slow down enough that it simply does not explode in the atmosphere. As the ship itself can be a delivery mechanism, one method would be to fire smaller propelled rockets in the counter trajectory. The delta-v budget to counter 6000km/s is roughly the budget needed to launch a rocket from earth to the moon. (Irony abounds).

The tech to build such a rocket has existed since the early 1970's with a Saturn V. Obviously the rocket will need to be resized as not all the energy needs to bleed off, and the payload is about twice as massive as the lunar module+lander at ~16000kg vs a kinetic missile at ~33110kg (given a 6.1m Height 0.3m radius cylinder ~1.72m^3 of tungsten). With a slower lunar-earth approach the rocket can be significantly less powerful. Although this does jeopardise the success of such a vehicle attacking.

The kinetic missile literature indicates that a strike from orbit could occur anywhere on the surface within about 12-15minutes. As the spaceship is already moving at speed and presuming that suitable course corrections had already occurred, the optimal firing area will be close and aligned. Additionally as the ship is already moving at speed, there is no time required for acceleration (other than the burn to slow for atmospheric entry), this would drop the target time down to atmospheric entry time at top-speed + time to travel from moon.

The fastest known meteriod traveled at about 72km/s through the atmosphere exploding a little too soon at 100km above the ground. Most asteroids average ~20km/s in atmosphere. The Atmosphere at its thickest is somewhere between 7km and 20km high, although the top of the atmosphere stands as high as 100km. Presuming the later, a tungsten rod could travel that distance in 5 seconds, but the atmosphere really is in the way. It would take about 2-4 minutes with deceleration due atmospheric compression to reach the surface.

Thus we have a window of 5minutes to 2hours for an attack once the ship becomes detectable after leaving the occluding lunar shadow.

Counter Measures and Response

It takes time to identify, verify, inform, and respond.

Some military installations with dedicated missile defense systems, if active, and linked to a hot-trigger for orbital strikes, might, be able to identify, and respond to this weapon successfully if undetected till atmospheric-entry. But due to the amount of energy in the system (20km/s is no joke), the chances are that the best that could happen is an aerial detonation, which will produce concussive shock waves that will damage and potentially demolish buildings depending on how high the detonation occurred in the atmosphere, and the relative distance to those structures.

If the Earth had orbital defense systems, it might be possible to apply the force of MegaWatt/GigaWatt lasers/masers to overheat your spaceship on its approach to Earth. It would need to overwhelm the Cyrogenic hydrogen cooling system installed in your ship. But again as stealth is no longer a requirement, this could be more formidable if the surface of the ship had been replaced by a highly reflective material. Thus most of the heat energy generated by the lasers/masers would be reflected, reducing the effectiveness of such a system. Obviously speed is critical here, an approach taking handful of minutes will have to deal with less heat than an approach taking two hours.

An ICBM usually carries large explosives and is a rocket capable of sub-orbital flight. That is it can be launched from the surface and travel to space, but does not gain sufficient speed to enter orbit, and will (usually on purpose) crash back to earth. It takes 3-5 minutes for such a craft to boost to max. velocity, and stays up to 25minutes in space. If a similar rocket were constructed with the intention of intercepting solar missiles (such as the space ship) it would need somewhere between 10 and 25 minutes to launch and position itself for intercept.

If suitably targeted such an anti-solar missile could deflect and or sufficiently damage the space-ship so as to nullify its capacity for damage. Such a missile will however be vulnerable at its zenith (low velocity or capacity for maneuvering) its effective zone of control would be some 10-15 minutes prior to intercept. This leaves several counter-counter measures available.

  1. use a counter missile specifically to hit the anti-solar missile.
  2. alter course with a significant burn (you would need to do so anyway to reduce speed) after the missile had reached a critical point in the trajectory.
  3. confuse the anti-solar missiles guidance systems with something approximating chafe, or other disruptive mechanism.

Arguably the slowest aspect of any counter-measure are the humans. At best 1minute from identification to response, presuming an alert, active, and responsible group. The less coherent the group, such as relying on civilian/researchers, cross-site verification, bureaucratic involvement will slow such response times down. 15 minutes might be super fast in the worst of these scenarios. Conversely a fully automated system would be able to respond within seconds (the moon is about 1.3 light-seconds away). This of course presupposes that the counter-measure systems automatic or biological are looking for and have perceived a possible issue.

Constraint Solving

  1. Use subterfuge to close distance.
  2. Once at distances where detection is definitive reduce transit time.
  3. Forgo cloaking on attack run - ditch the cloaking layer in favour of a highly-reflective armour. If hoping for the best, ditch the cloaking on the attack run only after observing counter measures.
  4. Ensure sufficient coolant is available to prevent over-heating on the attack run.
  5. Attempt to keep lunar-earth approach time low.
    • reducing the likely-hood of response.
    • reducing the likely-hood of successful intercept by anti-solar missile
    • minimising the time available to laser/maser defenses.
  6. Have anti-solar missile counter-measures.
  7. Use the deceleration process to avoid anti-solar missiles.
  8. Do not target a defensive location with counter-measures without employing a multiple missile strategy to saturate and overwhelm those counter measures.

In short, cloaking would actually be a handicap. Yes black paint, an advanced cooling system using hydrogen and rearward radiation would minimise detection chances particular at the periphery, but will fail quickly under scrutiny. The best bet is subterfuge, then a fast attack run.

$\endgroup$
1
$\begingroup$

Im going to have to make some assumptions here, based on my experience with weapons.

First, since you are using Vanta black, I assume most all (98%) of the solar energy would be absorbed but the hull material.

To accommodate the missiles, each launcher should be ~ 2m diameter by about 10 meters long. This doesn't include much of the support systems that would need to go in to launching them, but you can find space in the rest of the ship. to have a a 0.5 degree inclination of the hull, my rough calculations would put the ship at 470 m long. this would accommodate 4 missiles

$sin(0.5) = 2m/H => h= 2m/sin(0.5)$ which is to the end of the launcher, + a minimum 10 m and the other slope of the ship.

Now the surface area. We only need to consider the cross-sectional area facing the sun assuming the cone is strictly orientated-toward the sun, the surface area increases if its not. Fortunately that cross-section is the area of a circle.

Cross-sectional Radius: $r = sin(0.5) \cdot 470 = ~4.1 \ m$

Cross-sectional Area = $πr^2 = 4.1 \cdot 4.1 = ~16.82 \ m^2$

The total energy absorbed at 1 AU (around earth) is ~3600 $kJ/m^2$.

Absorbed heat (at 1AU) $= 16.82 \cdot 3600 = ~60559 \ kJ$

The specific heat of vaporization of hydrogen is 0.449 kJ/mol

Hydrogen evaporated $= 60559 / 0.449 = ~134876 \ mols$

A mol of Hydrogen atoms is about 2.01588 grams.

Hydrogen evaporated = ~134876 mols = ~271.893 kilograms

So rough estimates, using the information I have found is you would need 272kg of hydrogen to keep the hull cooled. Could not find a good description, but I think that is per hour.

That's a lot of hydrogen you would then have to figure out what to do with. Your exhausting of relatively warm hydrogen in those quantities would be detectable.

Sorry about not including pretty graphs and pictures. Hope someone with better info can edit my answer to better suit your needs.

$\endgroup$
  • $\begingroup$ Actually, thinking about it more, I think the surface area in this scenario would be πr^2 of the ship itself relative the sun, so significantly less surface area absorbing the suns energy. I may need help to full answer the question. $\endgroup$ – Sonvar Jan 16 at 5:40
1
$\begingroup$

Outside of the good answers already provided, any stealth spacecraft will be revealed through occultation.

In orbit, it will pass in front of stars, planets and potentially the moon (depending on the parameters of its orbit). You might arrange an orbital path so it will not or minimally pass in front of objects in space, but this will only be true of a particular viewpoint or position on Earth. Someone looking from a different position will potential see the spacecraft pass in front of the moon or other celestial object. Indeed, with enough observation, it will become not only possible to see the object, but also work out the orbital trajectory.

This becomes even more important when you consider there are a lot of satellites in high orbit looking down on the Earth. A cold object passing into the field of view will be very conspicuous agains the bright background of the Earth from that POV. This illustration of a fictional Stealth spacecraft from Tough SF will illustrate the point perfectly.

enter image description here

Suddenly, everything becomes clear

This also does not take ordinary wear and tear into account. A spaceship orbiting the Earth is going to be "sandblasted" by micrometer particles, energetic radiation from the sun and high energy particles from the Solar Wind. This isn't going to destroy the spacecraft, of course, but the outer surface will be under constant attack, and a fine surface like Vantablack will suffer degradation almost immediately. Over a prolonged period of time, it is entirely possible that the coating will erode or be damaged and gradually become less able to absorb or reflect incoming radiation (solar light, radar waves, infrared etc.) If the spacecraft intersects the orbit of a discarded bolt or other small piece that has been shed from a spacecraft over the decades of the Space Age, then the surface will suffer an impact crater due to a high velocity impact, and potentially the "dewar flask" holding the liquid hydrogen will also be breached, dumping the coolant and revealing he spacecraft.

enter image description here

Effect of a hypervelocity impact. This will ruin your day

So while it may be possible to provide passive stealth for a spacecraft for a short period of time, this is not going to be a long term solution. The other issue which will have to be kept in mind is getting into a particular orbit is going to take energy (from the rocket launch and manoeuvring thrusters), which is highly visible and easily tracked. An observer who watched the launch has the potential to understand the orbital parameters just from observing that, and will have an understanding of where to start searching, especially if the launch vehicle gets to orbit, and then nothing shows up on radar or visual scanning. The observer will be very interested in knowing what just happened, and will look very hard for the spacecraft, especially if they have orbiting vehicles of their own. Colliding with an uncharted or unregistered space vehicle could have catastrophic effects, especially if a critical military satellite happens to be the one destroyed.

$\endgroup$
  • $\begingroup$ That latter issue is the most important. Sure, potentially you could have a dark, cooled, stealthy satellite located somewhere for a short period of time, but to get it there in a reasonable amount of time requires a very obvious non-stealthy method. To use an analogy, it would be like a submarine running at top speed and then going full silent at a dead stop. Sure, passive sonar would have a hard time finding it, but they'd have a good idea where to start looking with active sensors. $\endgroup$ – Keith Morrison Jan 16 at 19:17
  • $\begingroup$ If the spacecraft is in a Moon-distance orbit, you can keep it from occluding the Sun for at least a year: it's simply never passing through an orbital node at a time when that node is pointing at the Sun. Radar searches or hydrogen boiloff will become a limiting factor long before then. $\endgroup$ – Mark Jan 17 at 5:14
1
$\begingroup$

MIT has worked-out a solution for that, and it's still in the theoretical phase. The idea is that the ship's shell is coated with layers of gradually-increasing negative refractive index. Light hitting the spaceship is refracted and diverted away from the actual shell's surface. It would "go around" the ship and leave from the other side.

The system has few drawbacks and is still in the planning phase. Some of the problems are:

  • the shape of the ship. The incident (hitting) angle affects the refraction angle. The ideal shape is the sphere.

  • Breaches necessary for entry/exit may cause visible reflections

  • it is possibly made of metamaterial (composite materials) and micrometeorites may compromise structural integrity and therefore stealth quality.

  • different wavelengths may refract differently and material will act like a prism (this one is my own speculation, and the engineers may have worked it out).

There is partial success here and here.

P.S. it is important to note that unlike earthbound travel (land, sea or air) eye contact is far less important. Speeds and distances are too high for human reflexes. Most detections are usually by radar, detecting radio signals and and infra-red emissions. By the time you see a spaceship, it will be hitting you and shatter your ship with a big show of firework.

$\endgroup$
1
$\begingroup$

Assuming a 10 meter central diameter and pointed at the sun, the cross section can be simplified as a 10 meter diameter circle, with an area of 78.5 m^2. That's what's going to absorb solar radiation.

Solar radiance at 1 AU is about 1361 W/m^2. Given vantablack at 99.96% absorption, that's 106,799 Watts going into your vehicle, ignoring reflected light from the Earth and the Moon. Since a watt is 1 Joule/s, that's 106,799 joules per second of energy being absorbed.

To simplify things again, we'll just use the heat capacity of hydrogen, ignoring enthalpy of fusion and vaporization. Liquid Hydrogen has a heat capacity of 9.41 J/g⋅K at 20 Kelvin. Given the amount of energy being absorbed (106,799 J/s), the vehicle is absorbing enough energy to raise the temperature of 11.349 kilograms of hydrogen 1 degree per second.

In an hour, 40,836 kg. In a day, 98 tonnes.

Liquid hydrogen at 20 K has a density of 70.8 kg/m^3. So, in a day, that 98 tonnes that's gone up 1 degree and become hydrogen gas (and has to be vented, as per your premise) has a volume of about 13,849 cubic meters.

Short answer: your system isn't going to work for very long at all.

I will note, however, that you cheated a bit in your premise. You limited sensor positions, which isn't a reasonable premise: if someone is advanced enough and has a sufficient space presence so that you need to hide, why are you limiting where they have sensors?

$\endgroup$
  • $\begingroup$ I am curious. Where did you find this data (heat capacity, density) for hydrogen at 20K? $\endgroup$ – Physicist137 Jan 18 at 0:26
  • $\begingroup$ Nevermind. I found it on page 28 and 29. $\endgroup$ – Physicist137 Jan 18 at 1:06
-2
$\begingroup$

It sounds reasonable. You could pick it out by seeing it occlude stars behind it or the sun if you were at the right angle but you would have to be looking hard (or have an automated sky searcher).

Black ship is going to get hot.

There are 3 conventional ways to dump heat - conduction, convection and radiation. The first 2 don't work in space and that leaves the third. Two factors affect how much heat an object can radiate away: surface area and heat absorption. Your shape does not maximize surface area and you have maximized heat absorption with the Vantablack. There is a reason space stuff is always so shiny.

You have the cold hydrogen but you are going to boil it fast. Maybe if you made this ship a ramifying curling wonder of tiny twists you could keep it black but increase surface area enough to compensate?

I like the transparent ship better, but your enemies will see you in the shower.


Math!

https://www.engineeringtoolbox.com/hydrogen-d_1419.html

Liquid H2: 71 kg/m^3

Specific heat = 3.42 cal/gC

Volume from OP = 30,000 m^ 3 x 71 = 2,130,000 kg liquid H2

Liquid H2 will be assumed to be at 5K. It turns into a gas (boils) at 21 K

From http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/spht.html

Heat added is 487986612480 joules or 487986612 kJ

Solar constant is 1.36 kJ/second/m2

Area of black cone ship (considered as a rectangle 1146 x 5m) = 5730 m2

1.36 x 5730 = 7792 kJ/second radiation energy delivered to ship

487986612 kj required to boil / 7792 = 62626 seconds = 2087 minutes = 34 hours to boil the hydrogen.

$\endgroup$

This question asks for hard science. All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Answers that do not satisfy this requirement might be removed. See the tag description for more information.

  • 1
    $\begingroup$ The whole point of not radiating is to prevent detection via infrared radiation. We want to radiate as little heat as possible and instead put it into the hydrogen, which has a very high specific heat capacity. Additionally, I do not believe this answer adheres to the requirements of the hard-science tag. It also does not answer the question, which asks about the length of time the heat-sink would last and the viability of the stealth method. As a result, I'm going to have to downvote this answer. I will retract my downvote if the issues are rectified with an edit. $\endgroup$ – Gryphon - Reinstate Monica Jan 16 at 2:13
  • $\begingroup$ physics.stackexchange.com/questions/289168/… - everyone inside will bake. $\endgroup$ – Separatrix Jan 16 at 8:29
  • $\begingroup$ @Gryphon - how can one calculate how long the heat sink will last with none of the relevant variables? Like how much hydrogen, how much radiation and how big a ship? Granted: it is a (temporary) solution to where you put accumulated heat. The heat sink will last an amount of time depending on how big it is and how much incoming radiation there is. That is as hard as it gets without numbers. $\endgroup$ – Willk Jan 16 at 13:09
  • 1
    $\begingroup$ Ship size is stated in the question. It is stated to be orbiting the earth, so you can assume solar radiation at a distance of 1AU for incoming radiation. Does that answer your question? $\endgroup$ – Gryphon - Reinstate Monica Jan 16 at 14:47
  • 1
    $\begingroup$ "it would be shaped like a double cone with an inclination of 0.5 degrees, the angular size of the sun at 1AU, and would be continuously pointed towards the sun to minimize solar heating." and "a 10m central diameter, which, if my math is correct, gives me a length of 1146m or so, and a volume of around 30,000 m^3." It's two cones stuck end to end with a diameter at the center (e.g., the base of each cone) of 10 meters and the sides of the cones have a 0 degree slope. Subtract a realistic volume for the missiles and you have your hydrogen volume. $\endgroup$ – Gryphon - Reinstate Monica Jan 16 at 16:14

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.