Section 4.4: Phase 2B - Industrial Locations for Space (page 2)< Space Transport and Engineering Methods
The Hypervelocity Launcher is one of the rocket replacement alternatives. It is based on the Particle Bed Heated Gas Gun from Section 2.2. In order to replace a large part of conventional rocket propulsion, high accelerations are used. This makes it suitable for bulk cargo, but not for people or complex equipment. Gunpowder-based Guns have a long history. Light-Gas Guns use lower molecular weight gases instead of the explosive products of gunpowder to reach higher velocities. They have been used for high-speed research since about the mid-20th century. Use of such guns for launch to orbit has been proposed several times in the past, but has not yet been put into practice. Therefore they would require R&D work to go from existing research gun experience to versions capable of orbital launch. We refer to this alternative as a Hypersonic Launcher because the gun muzzle velocity is greater than Mach 5, and therefore the projectile flies through the atmosphere at Hypersonic Speeds. We choose the particle-bed version because it is relatively simple and has good performance, although there are a number of other gas gun designs.
After establishing an Advanced Manufacturing capability, the next step is a low cost launcher, mainly for bulk materials. Examples of bulk materials are fuel, water, oxygen, structural components, and even frozen food. If there is enough cost advantage in this launcher, you can purposely design parts for the higher acceleration. For example, electronics can survive high g's if they are mounted properly, but most commercial units are designed for lowest assembly cost, not high g's. In general the acceleration of this type of launcher goes down with size. For example, compared to the US M777 field artillery, it may have 5 times higher muzzle velocity, but 150 times longer barrel, thus the acceleration is lower. In the limit of the largest practical gun the acceleration would be low enough for humans (60 m/s^2 or 6 g's), but this initial launcher is at the other end of the size scale, and thus relatively high g-level.
Conventional rockets are used in parallel with this launcher for the balance of delicate cargo and humans which cannot withstand the high acceleration. The particular type of launcher selected is 17 Particle Bed Heated Gas Gun, which is within current technology, has the lowest development cost to start delivering cargo, and a considerable cost advantage over current rockets. If conventional launch costs get low enough, though, this step may be eliminated.
The largest known hypersonic light gas gun was the Lawrence Livermore SHARP Gun in the early 1990's, which reached 3 km/s with a 5 kg projectile, for a kinetic energy of 22.5 MJ. The largest hypervelocity gun launcher was it's namesake predecessor, the High Altitude Research Project (HARP) in the 1960's. That gun was made by welding two 16 inch battleship guns in series, and was able to fire 250 kg at 2300 m/s using a guncotton charge. This had a kinetic energy of 660 MJ, and could have put measurable payload into orbit given suitable propulsion on the projectile. The largest known gun of any kind by energy was the Gustav type German railway siege guns of World War II. It fired a 4800 kg projectile at 820 m/s, for an energy of 1.6 GJ.
For development purposes, it is not good to make too large a jump in scale if you are going beyond past experience. Since the SHARP gun was the largest of that type ever built, we will assume a prototype of about 5 times the energy at 100 MJ (12.5 kg projectile at 4 km/s). This would be followed by a gun large enough to deliver a useful payload to orbit, which we will assume is 10 kg to a 250 km altitude circular orbit. Beyond that, larger guns would be sized by expected cargo traffic. Detailed analysis may change these numbers, but we at least need a starting point to design to. Please note all the following calculations are only preliminary to show how the scaling is performed. For more detailed work, refer to a manual such as Interior Ballistics of Guns, which accounts for more of the real world factors, or do a computer simulation.
The location of the prototype gun is not critical. It's main purpose is to reduce the unknowns before designing the orbital gun. For convenience we will assume the White Sands Missile Range as the landing point of the projectiles, since it is large, empty, and designated for missile testing. The gun will then be placed on a mountain slope a suitable distance away, avoiding any population centers under the projectile path. Any other combination of launch and landing points meeting similar conditions will work. We will further assume intermediate physical dimensions and characteristics between the SHARP and Orbital guns, unless there is some technical reason to choose otherwise.
- Projectile Mass = 12.5 kg
- Muzzle Velocity = 4000 m/s
- Barrel Length = ~200 m - This is a geometric mean between the 50 m SHARP barrel, and estimated 800 m orbital gun. May be updated by later calculations.
- Barrel Elevation = 12 degrees - This is based on typical mountain slopes north of White Sands
- Projectile L/D = 8 - This is the length-to-diameter ratio based on a cylinder shape. The actual shape will be to minimize drag, so conical at the front. You want to minimize area to lower drag and barrel size, so a long and skinny projectile, but not so skinny that bending becomes an issue. 8 is a reasonable starting point before structural analysis.
- Projectile Density = 1 g/cm^3 - This is the density of water, a likely filler of test projectiles, and similar to the density of LOX/Kerosine, a likely fuel for an orbital gun projectile.
- Projectile Diameter = 12.5 cm. - Cylinder volume is pi*r^2*h, and we have assumed that h = 8D = 16r. Thus v = 16*pi*r^3, and we have v=12,500 cm^3 from the mass and density. Solving for r gives 6.29 cm, and we round the diameter of 2r to 12.5 cm. The projectile diameter is also the barrel diameter, if you add a small tolerance for a sliding fit.
- Projectile Acceleration = 40,000 m/s^2 (4,000 g's) - This compares to the peak acceleration of 640,000 m/s^2 of the SHARP gun and 60,000 m/s^2 for conventional artillery. Muzzle velocity is sqrt(2*a*d) where a is the average acceleration and d is the barrel length. Solving for a gives the quoted value. Because gun efficiency falls off at high velocity, we assume the *peak* acceleration is 25% higher than the average acceleration, thus 50,000 m/s^2.
- Peak Pressure = 51 MPa (7,400 psi) - Barrel area is 0.01223 square meters, and gas pressure has to produce a force of F = m * a = 12.5 kg * 50,000 m/s^2 = 625,000 Newtons (N). Pressure is Force/Area. Note this is much lower than the SHARP gun peak pressure of 400 MPa.
- Projectile Range = 54 km - If the barrel slope is 12 degrees, then the projectile is rising at 830 m/s as it leaves the barrel. The density of the Earth's atmosphere decreases with altitude, but the equivalent thickness at constant pressure, known as the scale height is about 7.5 km. At a 12 degree elevation, the total path through the atmosphere then is equivalent to 37.5 km. We can find drag on the projectile from the formula F(D)= 0.5* CD * rho * A * v^2. CD is about 0.15 for a conical hypersonic projectile. Rho is air density, which for a starting altitude of 2000 m is about 0.95 kg/m^3. Area and initial velocity of the projectile are given above.
The initial air drag is about 14,000 N, giving a negative acceleration of 1119 m/s^2. This is a significant fraction of the initial velocity per second, so the actual trajectory needs to be found by numerical integration (ie a spreadsheet) using small time intervals so the changes in velocity and drag per time interval, and thus the errors, are small. When this is done the projectile range is found to be 54 km, after a 63 second flight, and reaching a peak altitude of 7350 m above sea level. The impact velocity with no landing devices is about 400 m/s. The small size of the projectile and low gun elevation means air drag has a severe effect on it's path. This is not a problem for the prototype gun, since we are mainly testing the gun. A range of 54 km means the projectile can land within the White Sands Missile Range and not endanger the public.
The prototype gun was mainly concerned with demonstrating the components of the gun function properly. The orbital gun demonstrates a larger version, and additionally a functioning projectile that can deliver a small payload to orbit. The gun energy of 1.18 GJ is about 12 times larger than the prototype.
- Net Payload to Orbit = 10 kg to 250 km
- Barrel Elevation = 23 degrees - As noted below under Location, Nevada Cayambe is the best location, and we use the actual slope of the west side of the mountain between 4200 and 4600 m elevation. Above that altitude is a glacier, so we try to stay below that.
- Projectile L/D = 8 - Use the same value as for the prototype gun. An actual projectile design will be needed for for a better estimate.
- Projectile Density = 1 g/cc - Use same value as for the prototype gun.
- Barrel Length = 800 m - This is an initial guess at a reasonable number. To determine the real length requires a detailed enough design in a form you can vary barrel length, see how the changes affect the rest of the gun system, and then find the optimum value. This approach is called variation of parameters or system optimization, but we need a lot more design detail to attempt it. For now we pick a reasonable starting point.
- Muzzle Velocity = 4500 m/s - This is a reasonable guess based on past gun launcher work. It will also be subject to optimization later.
- Projectile Acceleration = 12,650 m/s^2 (1290 g's) - Found by the same method a for the prototype gun above. Again, this is the average number, so peak acceleration is estimated to be 15,835 m/s^2 (1615 g's). Note this is about three times lower than the prototype gun, mainly because of the longer barrel.
- Peak Pressure = 33.88 MPa ( 4910 psi ) - Found by the same method as the prototype gun. Note the peak pressure is lower by about 1/3 relative to the prototype gun, mainly because of the longer barrel.
- Projectile Mass = 122.5 kg - With a known muzzle velocity and slope, we can try various projectile masses (in the next several paragraphs) to find the one that gives us 10 kg of payload to orbit. The projectile mass can be divided into three main parts: payload, fuel, and empty vehicle. The latter includes all the components like guidance electronics, besides fuel tank and payload support. The on-board rocket is assumed to have an exhaust velocity of 3.3 km/s, typical of a good LOX/Kerosine engine in vacuum.
- Drag Loss = 1725 m/s - As a first approximation, assume that drag losses equal 1200 m/s, the rotation of the Earth at the equator is 465 m/s, and the sea level orbit velocity plus energy of 250 km altitude to be provided by the rocket + gun is 8065 m/s. If the net velocity after drag loss is 3,300 m/s, that leaves 4300 m/s for the rocket. From the rocket equation, the net mass after the rocket burn is 27.2% of the total. We make a first estimate of the vehicle empty hardware mass of 15% based on past rocket designs. Then the net payload is 27.2% - 15% = 12.2%, and the original mass is 10 kg/12.2% = 82.2 kg. Assume the projectile density is 1.0 g/cc. Then it's volume is 82.2 liters (0.0822 m^3), and can be approximated by a cylinder 23.5 cm in diameter and 188 cm long. Since the projectile structure is subjected to a known acceleration, we can estimate the structure mass from the load it sees.
A mass of 82.2 kg subjected to 15,835 m/s^2 peak acceleration requires a force of 1.302 MN. Graphite composite can be assumed to have a strength of 600 MN/m^2, with a density of 1.82 g/cc. For the given load we need 1.302/600 square meters of structure = 21.7 square cm. The forward parts of the structure only have to support what is ahead of that point, so we assume the structure averages 65% of the area over the length of the body. Thus the total structure will be 21.7 x 65% x 188 = 2651 cc, with a mass of 4.8 kg. This is only 5.87% of the total mass of the vehicle, so our 15% assumption for the total empty mass is reasonable.
To get a better estimate of the actual velocity for the rocket, we have to make a better drag loss estimate. Using a spreadsheet trajectory calculator, we find the projectile will fall to 2613 m/s after drag losses have fallen 99%, giving an estimated loss of 1900 m/s. Since this is higher than our original estimate, we recalculate the projectile mass and try again several time until we get a consistent answer. This is called converging to a solution. After 5 iterations we can estimate the final result is a projectile mass of 122.5 kg, and a drag loss of 1725 m/s.
- Projectile Diameter = 27 cm - this is found from the mass and formulas above. Projectile and barrel areas are both 0.05725 m2.
- Projectile Range = 615 km - This is the distance the projectile will travel if the rocket does not ignite. This is found by following the trajectory calculator until the altitude reaches ground level again. When choosing a launch site, be aware of the impact point of a failed ignition. In this case, the impact point would be near Chiribiquete National Park in Colombia, reached after a 4 minute flight, at a terminal velocity of 1500 m/s if the projectile reaches the ground intact, which is not assured with a full load of fuel and aerodynamic heating, If the rocket engine ignites and then stops before reaching orbit, the impact point will be on the Equator somewhere east of the ballistic range, with reduced amount of fuel.
Orbital Gun with 2 Stage ProjectileEdit
This is an alternate concept to deliver the same payload as the previous design. It uses a 2 stage rocket to see if it improves the overall size. The gun inherently has about 50% higher "exhaust velocity" than the rocket engines, so we divide the total velocity into 3.5 parts to equalize the "difficulty" for each stage. From the previous design, we have a total mission velocity, including drag losses of 9325 m/s, therefore the gun would do (1.5/3.5)* 9325 = about 4000 m/s. Each rocket stage then has to perform about 2660 m/s.
Using our previous assumptions of 15% empty hardware weight and 3.3 km/s exhaust velocity, we can calculate the stage masses as follows:
- Mass Ratios/stage = 2.241 - found from rocket equation. This implies final mass = 1/mass ratio = 44.63% of start mass.
- Stage 2 payload = 10 kg = (44.63% final mass - 15% empty weight) = 29.63% stage 2 start mass.
- Stage 2 start mass = 33.75 kg - From 10 kg / 29.63%.
- Stage 1 final mass = 44.63% start mass, Stage 1 empty weight = (15% x 57.2% fuel used) = 8.31% start mass, therefore stage 2 start mass = 36.32% stage 1 start mass.
- Stage 1 start mass = 93 kg = 33.75 kg / 36.32%. This is significantly smaller than the previous mass, and the gun velocity is assumed to be 500 m/s lower, so we recalculate drag losses using the ballistic calculator and converge to a solution. We end up with 1655 m/s drag loss, and can thus lower the gun velocity by 70 m/s to 3930 m/s.
- Average acceleration is found from a = v2/2d = (3930 m)2/ 2 * 800m = 9653 m/s2 ( 984 g's ). Peak acceleration is 1.25 a = 12,066 m/s2.
- Projectile diameter is found as above to be 24.5 cm, and barrel area is 0.0473 m2.
- Peak pressure is found from P = (mass x acceleration)/area = (93 kg x 12066 m/s2)/0.0473 = 23.72 MPa.
The combination of lower barrel area and lower pressure leads to a total barrel mass of 57.85% of the previous version. We assume other parts of the gun will scale along with the barrel. The projectile mass is 76% of the previous version. Whether the added complexity of a second stage outweighs the system size reduction is not known.
The operational gun is designed to deliver paying cargo to orbit. The actual size will be set by how much customer traffic is expected, but for discussion purposes we will assume a 1200 kg projectile launched at 5 km/s. The location is the same mountain as the previous gun. The kinetic energy of this gun, 15 GJ, would exceed the largest previous gun of any kind by nearly 10 times.
- Projectile Mass = 1200 kg - Set by assumption.
- Projectile L/D = 8 - As in previous sizes.
- Projectile Density = 1 g/cc - As in previous sizes.
- Muzzle Velocity = 5000 m/s - also set by assumption. This is towards the upper range for light gas guns.
- Barrel Length = 1600 m - This is set by the distance from the bottom of the mountain slope to the glacier line. For this size gun it is likely not worth the extra difficulty building through the ice layer.
- Barrel Elevation = 23 degrees as previous size.
- Projectile Dimensions = 57.5 cm diameter x 460 cm long - Calculated from mass and density as in previous sizes.
- Projectile Acceleration = 7810 m/s^2 ( 795 g's ) - Again found by same calculation as previous sizes. Allowing for 25% peak increase, this gives 9765 m/s^2 peak acceleration (just under 1000 g's). We want lower acceleration for larger projectiles both to keep the barrel pressure reasonable, and less load on the projectile structure with the larger mass.
- Peak Pressure = 45.1 MPa ( 6545 psi ) - This is slightly higher than the previous size. Note that in a large gun with a pressure drop as it fires, only the bottom end will see the peak pressure. The muzzle end can use a lower strength pipe. Using high strength steel for the barrel, the peak pressure requires a barrel wall thickness of around 6 cm, which is reasonable.
- Empty Vehicle = 180 kg - The projectile is sufficiently larger that we should check the empty weight rather than assume the previous percent fraction. The peak acceleration force is 11.7 MN, giving a structural area of 195 square cm at the base, and a total structure volume of 58,300 cc. This comes to 106 kg mass, so our assumption of 15% for total empty vehicle (180 kg) is still reasonable.
- Drag Loss = 1000 m/s - This is found by using the trajectory calculator to the point you are above 99% of the atmosphere ( 34.5 km above sea level )
- Payload = 180 kg - The velocity to reach orbit is 8065 m/s. Subtracting the rotation of the Earth ( 463 m/s ) and the remaining velocity after drag loss ( 4000 m/s ) gives 3602 m/s to be added by the projectile. The rocket equation gives a remaining weight of 360 kg. Subtracting the empty vehicle leaves 180 kg ( 400 lb ) of net payload.
Note that the projectile mass increased by about 10 times, while the payload increased 18 times over the previous size. This is due to higher muzzle velocity from the gun, and lower relative drag losses from the larger projectile.
This gun is towards the upper end of length that can be built on Cayembe's slope. The diameter is set to 1.2 meters. This assumes there is enough bulk cargo traffic to justify the larger gun size.
- Projectile L/D = 8 - As in previous sizes.
- Projectile Density = 1 g/cc - As in previous sizes.
- Muzzle Velocity = 5000 m/s - As in previous size.
- Barrel Length = 3200 m - This is the upper end of length that can be fit to the mountain slope. The bottom end is at 3960 m elevation, and the upper end is at 5300 m, giving a 1340 m rise. The mountain slope is not as constant at the ends because we are maximizing length, so some of the barrel will need to be supported above ground level. Also this extends above the glacier line, so the barrel will either have to be supported above the ice, or protected from ice movements.
- Barrel Elevation = 24.75 degrees - Found from simple trigonometry of arcsin(rise/barrel length).
- Projectile Dimensions = 1.2 m diameter x 9.6 m long - From design inputs above.
- Projectile Mass = 10,850 kg - from dimensions and density above.
- Projectile Acceleration = 3905 m/s^2 ( 400 g's ) average - Again found by same calculation as previous sizes. Allowing for 25% peak increase, this gives 4880 m/s^2 peak acceleration (just under 500 g's). By keeping the muzzle velocity the same, but doubling the barrel length, the accelerations are halved.
- Peak Pressure = 46.8 MPa ( 6790 psi ) - This is about the same as the previous size.
- Empty Vehicle = 1625 kg - We again check the empty weight rather than assume the previous percent fraction. The peak acceleration force is 52.95 MN, giving a structural area of 882.5 square cm at the base, and a total structure volume of 423.5 liters. This comes to 771 kg mass, so our assumption of 15% for total empty vehicle (1625 kg) is still reasonable.
- Drag Loss = 460 m/s total velocity - This is found by using the trajectory calculator to the point you are above 99% of the atmosphere ( 34.5 km above sea level ). Horizontal velocity component is 4,160 m/s at the peak of the ballistic arc, assuming the rocket stage does not fire, and altitude will be 194 km at this point.
- Payload = 180 kg - For this size gun we assume a Skyhook from a later step is available. Skyhook tip velocity relative to the Earth's center is 5074 m/s. Subtracting the rotation of the Earth ( 465 m/s ) and the remaining projectile horizontal velocity after drag loss ( 4160 m/s ) gives 449 m/s to be added by the projectile. The rocket equation gives a remaining weight of 9330 kg. Subtracting the empty vehicle leaves 7,705 kg ( 17,000 lb ) of net payload.
Note that the projectile mass increased by about 10 times, while the payload increased 43 times over the previous size. The additional 4 times gain beyond projectile size is due to the Skyhook. With a slight improvement in gun velocity or Skyhook velocity or orbit altitude, the projectile would not have to add any velocity, only do maneuvering to meet the Skyhook landing platform. In that case the net payload would rise another 20% to 9,225 kg.
With the scaling of the launcher and projectile completed, you can now do a preliminary design. Preliminary design stops just before you start the final drawings and calculations, and is in enough detail that you can do cost estimates, and tell if any parts require new research.
As noted above, the location for a sub-orbital prototype is not critical. For a full orbital gun, the best location on Earth is probably Nevada Cayambe, the 3rd highest mountain in Equador, and highest point in the world on the Equator at 0 deg N, 78 deg W, 5790 m above sea level. It has a nice slope (23 degrees elevation) pointing east just below the snow line, and you can position a gun so the terrain slope past the muzzle tends to decrease. This is obviously critical so projectiles don't hit the mountain itself. For now there is a glacier on top of this mountain, even though it sits on the Equator, so the uppermost 1200m of the mountain would be difficult to build on. Launching from 4600 m saves you from going through 46% the atmosphere, which significantly cuts drag losses and heating, and an equatorial space station as a destination will pass overhead every 90 minutes or so. Any launch site off the equator will be limited to one or two times a day to launch. Other places will work as a launch site, just somewhat less efficiently.
Because of it's rotation, the Earth is slightly fatter at the equator. So even though there are taller mountains measured from sea level, equatorial ones are "higher" in terms of reaching orbit, and get more benefit from the Earth's rotation. The latter can be found from the Equatorial radius (6378.1 km) plus launch altitude (4.6 km), divided by the time it takes to rotate once, which is called a sidereal day. This is 86,164 seconds, thus the rotation amounts to 465 m/s at this location, or 5.77% of the total velocity to reach orbit.
Other considerations for a launch site are distance to populated areas, since the gun is very loud, avoidance of avalanche zones, and optimum slope. To reach orbit you want the maximum kinetic plus altitude energy remaining after drag effects. Some barrel elevation will be optimum for this, and finding a mountain with a matching slope will minimize construction cost. The alternatives of tunneling or building a support structure for the barrel would be much more expensive. To find the optimum elevation for a given gun design, use the trajectory calculator and input different angles to find the best one.
The barrel for this type of launcher is basically a large high pressure pipe. The most similar industrial item is a natural gas pipeline. For cost reasons, the majority of the barrel strength will likely be high strength steel. Since the inside will be exposed to hot hydrogen gas for a short period, and some amount of wear from projectile friction, it may need a liner or coating. The need for that will be determined by materials and thermal analyses, which are a standard part of mechanical engineering. The gas and projectile do not produce great force on the barrel lengthwise, the reaction force to the gun firing will mostly be at the back end, which is called the breech in gun terminology. Even so, the barrel will likely be made of bolted sections, which allows removal for maintenance, mounting and alignment to the ground, and sliding fit sections for expansion from general weather changes and heating from firing.
The upper end of the barrel will likely have a flap system to keep most of the air out of the barrel. The flaps are pushed open by the remaining air piled up ahead of the projectile. A "silencer" type device may be needed at the muzzle, either to actually lower the acoustic levels, reduce the muzzle flash from the hot hydrogen burning when it meets air, or to capture the Hydrogen gas to be used again. Finally, a shaped nozzle or steam ejector at the muzzle may be needed to ease the change from high acceleration in the barrel to negative acceleration from air drag once outside it. These options should be analyzed and tested with the prototype gun.
The heat exchanger stores heat from a convenient source (probably gas burners or electric heating elements), then transfers it quickly to the Hydrogen gas when the gun fires. Aluminum Oxide particles, commonly used for sandpaper, are used as the storage medium. They can withstand high temperature, and in grain form can transfer heat quickly due to the large surface area.
This part of the gun is where projectiles are loaded, fueled if needed, and the main reaction force of the gun is passed into the ground.
The Hydrogen needed to fire the gun is stored at room temperature in ordinary high pressure tanks. The storage pressure in the tanks is higher than the operating pressure in the barrel so that the gas will flow correctly, but the tank volume will be smaller than the barrel because room temperature gas takes less volume than the same amount of hot gas.
Low Density Tunnel (Optional)Edit
- See also: Low Density Tunnel
In theory you could eliminate the drag and heating of climbing through the atmosphere by using a vacuum tunnel beyond the end of the barrel. This would be larger diameter than the barrel, to eliminate friction, and continue upward as high as drag savings and cost dictate. In practice, you can get 93% of the effect of a vacuum tunnel with a hydrogen tunnel. Drag is proportional to gas density, and hydrogen gas is 93% less dense than air. It also tends to float, so that simplifies holding the tunnel in the air above the end of your launch mountain. Also, hydrogen is your gun propellant, so it just mixes with the hydrogen in the tunnel, and it becomes fairly easy to pump back for the next launch.
You build a large enough tunnel so the hypersonic shock waves of the projectile won't destroy it. That also gives more lifting volume. How far you build the tunnel, or whether to use it at all comes down to cost. It can be added later to a basic gun to hold down initial cost. With such a tunnel in place, you can fire at higher velocities as well as lower the overall losses.
This example of the projectile design is for the operational gun with a 1200 kg projectile. The other sizes will be smaller and simpler versions will lower loads, so this is a "worst case" design challenge to solve.
Thermal and Re-entryEdit
The projectile needs to survive three periods of high heating: (1) Within the barrel from hot Hydrogen gas pushing it, (2) While flying up through the atmosphere at high speed, and (3) during re-entry so you can recover and use it again.
The first heating source comes from behind. Hypersonic guns often use a Sabot, a disposable structure to help fit the projectile to the barrel and provide a better seal for the gas pressure. Using a sabot in this design can protect the projectile from heating either by including insulation, or thermal inertia, which is simply that solid objects take time to heat up. If that time is longer than it takes to leave the barrel, then extra insulation is not needed. The alternative is to use the rocket nozzle at the back of the projectile, which already needs to withstand high heating. In that case it also needs to withstand the high acceleration force from the gas pressure.
The second and third heating sources come from the front. Flying up through the atmosphere at high velocity causes the higher peak heating rate, but re-entering from the even higher orbital velocity gives a higher total heating, although at a lower rate. The common method to deal with high heating rates is an ablative heat shield, which decomposes and generates a protective gas layer. To keep the projectile pointed in the right direction both flying up and during re-entry may require fins or control surfaces, which would require heat shielding on them.
Thermal protection is measured in terms of the amount of energy it has to dissipate. While flying up through the atmosphere, the projectile goes from 5 to 4 km/s via drag, thus losing 5.4 GJ of kinetic energy. During re-entry, when the projectile is empty, it dissipates 5.2 GJ. In the former, the heat pulse starts at the highest rate, decaying exponentially as the air pressure goes down with altitude with a time constant of about 4 seconds. In the latter, the heat pulse starts slowly as the projectile encounters the thin upper atmosphere at high velocity, reaches a peak at lower, thicker air density but still fairly high velocity, and then tapers off as velocity falls faster than pressure rises. The exact duration will have to be found by trajectory simulation, but is measured in minutes rather than seconds.
The projectile structure is designed to withstand 800 g's at launch, and therefore in theory should be able to handle a similar deceleration at landing. In reality, empty fuel tanks can buckle more easily than full ones (think of a full vs empty soda bottle for example), but the projectile is still a sturdy device. Terminal velocity is the velocity a falling object will reach where drag equals gravity, and so velocity stops changing. For our empty projectile, we can calculate terminal velocity assuming rear control surfaces add 50% to the total area and have a drag coefficient of 0.4. Since the projectile has an empty mass of 180 kg, gravity produces 1764 N downward force. The terminal velocity is then about 150 m/s. Assuming the empty vehicle can withstand 2000 m/s^2 (200 g's), it can reasonably sustain an impact velocity of around 20 m/s using crushable structures. Therefore it needs a device like a parachute or much larger control surfaces to create drag. Forcing the projectile to land sideways instead of nose-first will increase drag area and drag coefficient and lower terminal velocity.
If the launch site is on the Equator, then the landing point will also be on the Equator so long as the projectile propulsion is always aligned east-west. In this example, the landing point can be the relatively unpopulated areas about 80 km east of the Andean mountain chain, which makes return of the empty projectile to the launch site relatively easy. Then it is a matter of timing the re-entry burn of the on-board rocket to set the landing point. The size and weight of the empty projectile can be handled by a small truck.
We assume the cargo being delivered has the same average density as the rest of the vehicle, 1 g/cc. Allowing for the projectile structure, the cargo diameter is set to 50 cm. Given a mass of 180 kg, then the length is 92 cm. Depending how the structure needs to be designed, the cargo opening can either be a side hatch, or the end of the projectile can hinge open and the cargo exits that way. Two possibilities for delivering the cargo are (1) without any station or depot, and (2) with the assistance of a station or depot. On the first case, the projectile only needs to maneuver to the accuracy of the desired payload orbit. In the second case, it needs to maneuver close enough for the station or depot to rendezvous. The final docking can be done either by the projectile or station. It likely will be more efficient to put the rendezvous system once on the station, rather than launching it each time with each projectile. Given the size of the projectile, simply grasping it with a clamp around the middle may be the easiest way to dock.
If the cargo is liquid, an integral (built in) tank will probably save weight and deliver more mass. Then you will have two projectile designs, for liquid and dry cargo. The projectile can be designed with replaceable cargo modules for liquid, dense, or normal cargo. The weight savings for custom modules will have to be compared to the design and operations cost penalty for having multiple designs.