Given a billion dollars, build the most visible orbiting satellite possible

Question

As per the title, given

• A billion ($10^9$) dollars.
• Present day technology.
• A year of time (including making projects and testing), (preferred, not compulsory)
• No restriction about maximal nor minimal size of the object nor its orbiting altitude.

Build and put into orbit a satellite that will have the greatest possible apparent magnitude when seen from the surface of earth.

I am interested both in details about the satellite and estimates of its apparent magnitude.

There is no minimum shining time

The device may be visible for any amount of time, you should maximize the average apparent magnitude. For example, if X works only half of the day but the apparent magnitude is more than double than Y that works all day, X is better.

There is no minimum area to be visible from

It does not matter how concentrated the light is, but if it can be seen from just one 'point' it should be adjustable to fire at other locations.

Monument to the future humans

I intend this light in the sky to be a sign for future humans (or whatever intelligent life will develop) that past space faring lifeforms existed in their planet. So it should last millions of year (I thought all orbits lasted forever, hence the lateness of adding this requirement)

Answer 1

The first problem we have to solve is power: where does the light energy come from? Our options are the same as for every other spacecraft:

• Nuclear thermal: Although this option provides a lot of power, there are two strikes against it: nuclear reactors are very heavy, and they are currently not ready for spaceflight. You'd be looking at a decade-long development program, which does not meet requirements.

• Radioisotope thermoelectric generators (RTG): These are a common choice for deep-space probes since they provide long-term power with high reliability. New Horizons and the Curiosity rover both use plutonium RTGs. However, they are not efficient (typical powers are around a hundred watts), and they are very expensive. (NASA is paying the US Department of Energy over a hundred million dollars to restart production at the level of about a kilogram per year.)

• Fuel cells: These can provide high power by consuming a fuel and oxidizer. These consumables limit the lifespan of fuel cells, so they do not meet requirements. (I'm making the assumption that you want the spacecraft to remain visible for at least a month.)

• Solar power: Solar photovoltaic is an OK option. However, we can also use the incoming sunlight directly by reflecting it at the Earth. This eliminates the need to collect, process, and re-emit all of the light power, replacing that equipment with a simple reflective surface.

The idea of beaming power from a 'generator' satellite in high orbit to a 'lightbulb' satellite in low orbit is a good one which draws on technically plausible concepts. However there are a couple reasons that I'm avoiding it in my design:

• Microwave power transmission is nowhere near mature, and would require a development period of many years.
• There is a lot of inefficiency. Even assuming very high efficiency solar panels (30%), LED lights (40%), power transmission (90%) and conversion (95%), the overall efficiency is only about 10% (and the transmission efficiency is likely to be far lower), compared to around 95% for a reflector.
• Solar panels are far heavier per area than reflectors (on the order of 100 watts per kilogram). Add to that the cost of launching that weight into high orbit and the project quickly becomes infeasible.

Choosing the passive reflector, our initial design looks something like a solar sail.

The next problem is the structure of the spacecraft. Unfortunately the reflector can't be tensioned by centrifugal force like a solar sail, because the rotation would interfere with the pointing (since the reflector is not an isotropic radiator, steerability is a requirement). I envision a folding truss structure like SMAP's antenna (Note that SMAP's budget was around 900 million dollars). The structure doesn't need to be as "dense" since the flatness of the reflector is not as critical.

Now we should determine the reflector size. Assuming that the reflector is very close to flat, it will appear (to an observer within the reflected beam of sunlight) to have the same surface brightness as the Sun. (To put it another way, it looks like a window showing a small part of the Sun.) Thus, the total apparent brightness is equal to the apparent brightness of the Sun times the reflector's apparent size relative to the Sun. To get some rough numbers, I'll assume the reflector has $90\%$ efficiency. The apparent magnitude of the reflector is:

$$ m_\text{sc} = m_\text{Sun} - 5\log_{10}\left(\frac{d/r}{32'}\right) $$

The quantity in the logarithm is the angular size of the spacecraft (it's diameter $d$ divided by distance to the observer $r$) divided by the angular size of the Sun (in minutes of arc). The apparent magnitude of the Sun is $-26.74$. Putting this into a plot, we get:

We can see that this rough estimate has good agreement with the magnitude of Iridium flares, caused by reflective antennas 1–2 meters in size.

Assuming again that the reflector is close to flat, the width of the reflected beam will be around 30-40 arcminutes. The diameter of the beam at the surface will be about one percent of the reflector's altitude (from a 4 km spot in low orbit to a 350 km spot in geosynchronous orbit).

However, we need to maximize not the peak magnitude, but the average magnitude. This is affected both by visibility of the spacecraft from the ground and shadowing of the spacecraft by the Earth.

I performed some simulations to determine the optimal altitude. I took into account four factors:

• Visibility: If the spacecraft is below the horizon, it is not visible to the observer and its relative brightness is $0$.
• Viewing angle: Imagine extending two lines from the spacecraft, one to the Sun and one to the observer. Call the angle between these lines $\theta$. If $\theta=0$, the Sun is directly behind the observer as they look at the spacecraft and the relative brightness is $1$. If $\theta=\pi/2$, the reflector must be turned at 45 degrees and its apparent size (and apparent brightness) is only $1/\sqrt{2}\approx 71\%$. If $\theta=\pi$ the spacecraft is directly in between the Sun and the observer; in this configuration the spacecraft can't reflect any light towards the observer and its relative brightness is $0$. The relative brightness is $\cos(\theta/2)$.
• Distance: The apparent brightness of the spacecraft is proportional to the inverse square of the distance to the observer.
• Shadowing: I assumed that the Earth's shadow is perfectly sharp, so the spacecraft is either fully illuminated (brightness $1$) or fully shadowed (brightness $0$).

Surprisingly, the simulations indicate that you want the reflector to be in as low an orbit as possible. The decrease in brightness with altitude dominates the increased visibility. However, our orbit can't be too low, since with a large, lightweight structure drag becomes an issue.

Let's spec a 1000 kilometer orbit. At this altitude, even a 100-meter reflector would drop by only about ten kilometers per year. The inclination should be a little higher than the latitude of the northernmost (or southernmost) location you want it to be visible from. The reflector will cast a spot about ten kilometers wide on the Earth's surface.

At 1000 kilometers, we could make the spacecraft as bright as the full moon with a 20-meter reflector. This size is easily within the realm of plausibility; JWST's sunshield is close to this size at 18 meters long.

Answer 2

A metallized mylar balloon or light sail can be amazingly brilliant via reflected sunlight, although it would only be visible when the sun is illuminating it (so from twilight to dawn). If you are satisfied with that, then the builder will happily take the money, pay SpaceX $50 million for the launch and a few million more for the balloon/lightsail and pocket the change.

In terms of "how" brilliant it could be, Nazi scientists (who apparently had nothing better to do with their time) worked out the parameters of an orbital mirror made of sodium which was designed to focus sunlight on hapless targets on the ground (much like a child uses a magnifying glass to burn ants).

Some details can be found here:

http://www.worldatwar.net/chandelle/v1/v1n1/ww2space.htm

https://en.wikipedia.org/wiki/Sun_gun

Later during World War II, a group of German scientists at the German Army Artillery proving grounds at Hillersleben began to expand on Oberth's idea of creating a superweapon that could utilize the sun's energy. This so-called "sun gun" would be part of a space station 8,200 kilometres (5,100 mi) above Earth. The scientists calculated that a huge reflector, made of metallic sodium and with an area of 9 square kilometres (3.5 sq mi), could produce enough focused heat to make an ocean boil or burn a city.1

Now to make this visible all the time, use something like the Sonnengewehr in high orbit as a power generator, and beam the energy to a satellite in LEO which has IMAX projection lamps on board. A single IMAX lamp on Earth pointed into space is supposedly visible on the Moon:

The lamphouse on top of the IMAX projector utilizes two 15,000-watt liquid-cooled, short-arc xenon lamps. The lamps weigh 10 pounds each, and are nearly two feet in length. Costing more than $6,000 each, the lamps have a life expectancy of only about 1,200 hours of operation and are replaced 4 times per year. Because of the extreme high-pressure xenon gas inside the quartz glass envelope of the lamp, projectionists must wear ballistic safety gear when changing out a lamp. If dropped, the xenon lamp would explode with the destructive force of a hand grenade.

The average luminance of one of these xenon lamps is approximately 1.6 billion candles per square yard--about equal to that of the Sun as viewed from the Earth's surface! The lamp has a light output of approximately 600,000 lumens. NASA uses this same type of lamp at the Kennedy Space Center to illuminate the Space Shuttle at night on the launching pad.

During normal operation, the clear quartz glass envelope of the lamp has a surface temperature of about 1,300 degrees. To prevent the lamp from overheating, it has coolant "jackets" that allow cool distilled water to be pumped around the electrodes at the flow rate of 8 gallons per minute and a pressure rate of 100 psi. In addition, an exhaust fan removes 1,200 cubic feet of air per minute from the lamphouse. The xenon lamps operate at 37.5 volts DC, and 400 amperes of current.

So if we take 50 million/launch from SpaceX (X2), use 100 million to build a solar power station in high orbit with a microwave energy beam then we have 800 million left to purchase IMAX bulbs and build the satellite. (Lets put aside 100 million for the satellite body). So we can use 700 million to buy IMAX light bulbs, for a total of 166,666 bulbs. This will produce 99999600000 lumens of illumination until the bulbs start burning out.

I'm pretty sure you might need sunglasses to look at it from the Earth's surface. (Use SPF 100 sunscreen as well.....)

Answer 3

A square kilometer of metallized PET is available at retail prices for about \$4 million, and at a density of 1.4 grams per cubic centimeter, weighs about 71000 kilograms. SpaceX can put that into low orbit for around \$330 million.

For \$1 billion, you get 16 Falcon 9 launches, putting three square kilometers of highly-reflective material into orbit. At an altitude of around 200 kilometers, it will have an angular diameter of 34 minutes of arc, slightly larger than that of the Moon, but it will be much brighter, with an albedo close to 1 (in contrast to the Moon's graphite-dark 0.13). With careful angling of the surface, it will be visible day or night, only disappearing when it enters the Earth's shadow; a 10% reduction in size will let you put it in polar orbit, making it visible at all times.

Overall brightness will be around -14 (compare the full Moon at -12), flaring to around -27 (the same as the Sun) when the angle is right for it to focus sunlight somewhere.

Answer 4

A nuclear shaped charge aimed at specific point should do the trick.

Nuke shaped charges are an idea invented in the 1960s to improve the efficiency of projects like Project Orion. Instead of a spherical blast produced by conventional nukes, where much of the energy from the explosion would uselessly radiate away, nuclear shaped charges focused the energy into a relatively small area, namely the pusher plate of a Project Orion spacecraft.

The brightness from a blast like this is very short, a few seconds at the most. In order to meet the "fire at different locations" requirement, the spacecraft will require multiple charges be deployed into orbit.

How to spend the billion

• Get some geosynchronous launch services from SpaceX for a few tens of millions of dollars.
• Build a launch vehicle. Total production costs are likely to be below $100M.
  • The vehicle will basically be a axial bomb launcher that can dispense and detonate as many shaped charges as possible within the available space and weight constraints.
• Make fusion powered nuclear shaped charges. This is the highest risk portion of the project because no one (to the knowledge of Google) has ever made one. The math is well understood but the actual manufacturing techniques are not. Testing such devices in the current regulatory/treaty environment will be tricky at best (though this can be easily handwaved away).
  • Fissile materials can be appropriated from existing nuclear stockpiles (US or Russian) so the arduous and tedious process of making plutonium or fissile uranium can be avoided in whole or in part.
  • The costs of producing these weapons may be very cheap, <\$1M or more expensive at \$20M. These costs will determine how many you can put into orbit, obviously.
  • The cost of the explosives may be considerably less than the cost of the aiming mechanism attached to the charge. Careful attention will need to be paid here. It is possible that the explosion angle and the distance from Earth may make it so that each charge is able to illuminate an entire hemisphere.

Discussion

Conventional explosives don't pack the punch required to make a really bright new "star". Fission charges may provide the power necessary but if going fission, go fusion and pick up another two to three orders of magnitude in power output without too much more weight. The most powerful US nuclear weapon was the B-41 with a yield-to-weight ratio of 5.1 megatons per ton. Note that the Russians did make a 50 megaton fusion weapon at 27,000 kilograms (60,000 lb) for a yield-to-weight ratio of 1.6.

Project Rho reports that the minimum aiming angle for these shaped charges is 0.1 radians all the way up to 0.39 radians (6 degrees to 22 degrees). The angle depends on the material in front of the charge as well as the thickness of the plate.

The only other place to find energy yield greater than fusion weapons is to use anti-matter reactions, but those aren't available in the budget provided. Large solar energy concentrators may be available in the time and budget allocated, but no one has launched anything approximating a solar lens larger than 1 km².