What Speed is Needed to Launch a 1500 kg Cargo Container to 160 km Altitude?

[Zybex]

TL;DR Summary

Cargo to space launch problem

Let's say we have an egg-shaped (for better drag) 1500 kg cargo container and a acceleration system that accelerates that container and launches it at sea level to reach 160 km on its own without any engine. What speed should be applied to this 1500 kg container at sea level, so it reaches 160 km altitude and have its speed slowed down to 7.8 km/s? It seems unreal to implement without keeping the cargo accelerating like it's done with rockets nowadays. And, understandably, we can't apply a sudden speed to the cargo at sea level, we would need a runway to make it gradual so the container doesn't get damaged. But, let's say we don't care about damage and just want to know the speed that should be applied to it instantly to launch it into orbit.

 

[berkeman]

Welcome to PF.

Have you already seen the SPINLAUNCH system?

https://www.spinlaunch.com/orbital#p2

 

[Zybex]

Thank you! Yes, and that's why my question is what the initial speed should be. I know it is impossible, just a theory. But I would like to know the initial speed that isn't going to be applied all the way, but only at launch time.

 

[berkeman]

You probably want it to be more aerodynamic than egg shaped. I think the website gives some numbers, but I haven't looked through it for a while.

 

[Zybex]

Just wanted to see the calculation of this theory with a 1500 kg egg-shaped container without trying to improve it or make it look real.

 

[berkeman]

Just wanted to see the calculation of this theory with a 1500 kg egg-shaped container without trying to improve it or make it look real.

You're going to make me reread that website to find the numbers, aren't you...

 

[berkeman]

You're going to make me reread that website to find the numbers, aren't you...

The Orbital Accelerator spins up to approximately 5,000 mph prior to releasing the launch vehicle. To date, we’ve conducted tests over 6x the speed of sound.

https://www.spinlaunch.com/faq#p2

 

[Klystron]

The HARP Project provides an alternative launch device without rockets, particularly when ignoring condition of your cargo.

I have also read about hybrid launch systems that lift a cargo and rocket engine above most of the atmosphere where the rocket propulsion system then fires. IOW your 'first stage' consists of some novel propulsion method for the heavy lifting against Earth gravity followed by conventional rocketry for reaching a destination or precise orbit.

 

[russ_watters]

Let's say we have an egg-shaped (for better drag) 1500 kg cargo container and a acceleration system that accelerates that container and launches it at sea level to reach 160 km on its own without any engine. What speed should be applied to this 1500 kg container at sea level, so it reaches 160 km altitude and have its speed slowed down to 7.8 km/s? It seems unreal to implement without keeping the cargo accelerating like it's done with rockets nowadays.

The problem is under-specified and exceptionally difficult to solve. Also fortunately, there's not much difference between orbital velocity at 160km and orbital velocity at Earth's surface. So....8? 9? 10? The major complications I can think of (beyond converting kinetic to potential energy) are:

1. Atmospheric drag is going to be a big problem, though:
2. Depending on the trajectory it might not reduce the velocity much. But:
3. If you launch it straight up you'll be nowhere close to an orbital trajectory. Too flat and you'll spend a lot more time in the atmosphere sublimating. And:
4. Neither option results in an orbit...at least not one that doesn't still intersect the surface of the Earth. How much of an orbit-shaping burn do you still need when you get there? And once you do:
5. Did the payload survive the ride?

The challenge of even building a hypersonic trebuchet is not trivial, but I suspect the rest is much worse.

 

[ElfredaCyania]

You are essentially asking how the drag force should be applied to decelerate an elliptical orbit (you may think it’s a parabola but it's an oval with a focal at the centre of the Earth) into a circular orbit, this process involves a decrease in eccentricity. This is impossible because deceleration in the direction of the velocity (retrograde) near the apoapsis can only lower the periapsis, thus increasing eccentricity, this involves complex orbital mechanics, but to put it simple:

If you know the apoapsis and periapsis, you can find the eccentricity, using R_ap=a(1+e), R_pe=a(1−e), where a is the semimajor axis, a=(R_ap+R_pe)/2
As the periapsis decreases, e=(R_ap-R_pe)/(R_ap+R_pe) increases.

I think the only possible initial speed for any circular orbit is 7.9km/s horizontal, so it would reach orbit the moment it is fired, and then crash.

 

[Zybex]

I just wanted to know the initial speed required to launch 1500 kg into 160 km orbit and that cargo should still have velocity of 7.8 km/s after reaching 160 km altitude. Does anyone know the proper formula?

This is what AI gives me:

To throw a 1 kg object into a 160 km orbit from sea level, we need to consider the total energy required to reach that orbit. This includes both the kinetic energy to reach orbital velocity and the potential energy to reach the altitude of 160 km. However, achieving orbit also means the object must attain the required horizontal velocity for a stable orbit, not just reach the altitude.

$# Steps to Determine the Initial Speed: 1. **Gravitational Potential Energy at 160 km Altitude:** The gravitational potential energy (\( U \)) at an altitude \( h \) is given by: \[ U = mgh \] where \( m \) is the mass (1 kg), \( g \) is the acceleration due to gravity (9.81 m/s² at sea level), and \( h \) is the altitude (160,000 m). 2. **Orbital Velocity at 160 km Altitude:** The orbital velocity (\( v_{orb} \)) at altitude \( h \) is given by: \[ v_{orb} = \sqrt{\frac{GM}{R+h}} \] where \( G \) is the gravitational constant (\(6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2}\)), \( M \) is the mass of the Earth (\(5.972 \times 10^{24} \, \text{kg}\)), and \( R \) is the Earth's radius (\(6.371 \times 10^6 \, \text{m}\)). 3. **Total Energy Required:** The total energy required (\( E_{total} \)) to reach orbit includes both kinetic and potential energy: \[ E_{total} = \frac{1}{2}mv_{orb}^2 + mgh \]$# Calculations:

1. **Gravitational Potential Energy:**
\[
U = mgh = 1 \times 9.81 \times 160,000 = 1,569,600 \, \text{J}
\]

2. **Orbital Velocity:**
\[
v_{orb} = \sqrt{\frac{6.67430 \times 10^{-11} \times 5.972 \times 10^{24}}{6.371 \times 10^6 + 160,000}} \approx 7,784 \, \text{m/s}
\]

3. **Total Kinetic Energy at Orbital Velocity:**
\[
E_{kinetic} = \frac{1}{2}mv_{orb}^2 = \frac{1}{2} \times 1 \times (7,784)^2 \approx 30,300,000 \, \text{J}
\]

4. **Total Energy Required:**
\[
E_{total} = E_{kinetic} + U = 30,300,000 + 1,569,600 \approx 31,869,600 \, \text{J}
\]

To convert this total energy to the initial speed required at sea level, assuming no losses due to air drag or other factors (which in reality would be significant):

### Initial Speed Calculation:

1. **Initial Speed at Sea Level:**
\[
\frac{1}{2}mv_{initial}^2 = E_{total}
\]
\[
v_{initial} = \sqrt{\frac{2E_{total}}{m}} = \sqrt{\frac{2 \times 31,869,600}{1}} \approx 7,989 \, \text{m/s}
\]

This initial speed of approximately 7,989 m/s would be required in a theoretical, drag-free environment. However, considering atmospheric drag and other real-world factors, the actual required speed would be significantly higher. Real-world launches typically involve reaching much higher speeds initially to account for these losses and to achieve the desired orbit.

 

[ElfredaCyania]

the initial speed required to launch 1500 kg into 160 km orbit and that cargo should still have velocity of 7.8 km/s after reaching 160 km altitude.

You need to at least specify the direction of the velocity for any meaningful calculations. And even with that, the initial speed at sea level would be so high that most fluid dynamics formulas won't be applicable.

 

[Zybex]

You need to at least specify the direction of the velocity for any meaningful calculations. And even with that, the initial speed at sea level would be so high that most fluid dynamics formulas won't be applicable.

The direction of the velocity is perpendicular to Earth's surface. And by initial speed I mean a sudden thrust, after which there's no thrust at all.

 

[Zybex]

https://www.spinlaunch.com/faq#p2

SpinLaunch is just another scam project that exists only because investors and sponsors have no idea about physics. Just like Mars One, SpinLaunch will cease to exist pretty soon without launching a single satellite, and sponsors will just lose all invested capital.

 

[Tom.G]

Well, here is a close-enough approximation from
https://www.google.com/search?hl=en&q=earth+escape+velocity+m/s

Spoiler

11,176 m/s

Cheers,
Tom

 

[russ_watters]

The direction of the velocity is perpendicular to Earth's surface.

That's a terrible idea. It'll go straight up and then straight down.

 

[mfb]

SpinLaunch is just another scam project that exists only because investors and sponsors have no idea about physics. Just like Mars One, SpinLaunch will cease to exist pretty soon without launching a single satellite, and sponsors will just lose all invested capital.

The physics is sound, despite some people dismissing the idea because they don't understand it.
It's unclear if they can make the launcher cheap and reliable enough and find enough customers to make it economically viable, but that's an engineering and marketing problem not a physics problem.

Anyway. If we completely ignore drag then a simple energy conservation approach tells you the required launch velocity in order to move at 7.8 km/s at 160 km altitude. If you launch vertically, there is nothing special about that velocity however. You are not in an orbit, and there is nothing that could suddenly change your direction of motion in a useful way. The vertical velocity makes it harder to enter a circular orbit at that altitude.

 

[Drakkith]

The direction of the velocity is perpendicular to Earth's surface. And by initial speed I mean a sudden thrust, after which there's no thrust at all.

You MUST have some sort of propulsion onboard the cargo container if you want to get into an actual orbit. Launching an object from the surface of the Earth without some way to alters its velocity once it reaches space will always result in it coming back down to Earth somewhere. It doesn't matter what the trajectory is or how fast you launch it (as long as it is under escape velocity) the object will always come back down to the surface.

 

[Tom.G]

@Zybex, lets try this thought experiment. Drawing a sketch to work from will make this easier to follow.

a) There is a circular race track 2000 meters in diameter.
b) You are at its center and wish to go around the track in 0.82 seconds to make a new world record. (circumference) / (orbital speed)
c) To do this your speed around the track must be at least 7.6km/s. (orbital speed)
d) No problem you say, I can easily do 11.2km/s traveling the straight line from the center of the track to the track itself. (escape velocity, your post #13))
e) You take off from the center on a straight line at 11,186m/s and gradually slow down to 7.6km/s at the track circumference. (escape, orbital velocities)

So all is good you say. But wait! You are now traveling 7.6km/s AWAY from the circular track because you have no way to change direction.

You have to do two things to get around the track in world-record time:
1) Accelerate 7.6km/s onto the track (at a right angle to the direction you are traveling)
2) Stop your outward motion

Your posts do not include these last steps, 1) and 2).

Hope this helps!

references:
https://en.wikipedia.org/wiki/Orbital_speed

Low Earth orbit speed = 7.7 to 6.9km/s​

https://www.google.com/search?hl=en&q=escape+velocity+of+earth+m/s

escape velocity = 11,186m/s​

Cheers,
Tom
(edited to correct slipped decimal points. thanks to @Drakkith for pointing this out! {that's what I get for posting after midnite})

 

[Drakkith]

Low Earth orbit speed = 7.7 to 6.9m/s

I think you mean km/s.

 

[mfb]

7-8 m/s is an orbital velocity close to the surface of Phobos, Mars' larger moon.
People on Earth can sprint faster than that, but trying to run on Phobos would be really awkward as every "step" becomes a huge leap that has you tumbling around in a way that's hard to control (without rocket engines).

 

[Zybex]

Okay, let's simplify the problem.
What speed is required to throw a 1500 kg container from the sea level up into air to reach 160 km altitude and then fall down? I just want to see some numbers without getting too deep into efficiency, orbit, etc. Think about a 1500 kg container being accelerated below sea level and at the sea level the thrust is cut off, so the container continues to fly up and reached 160 km, then falls down. What speed has to be reached at sea level, so the container reaches 160 km?

 

[Drakkith]

What speed is required to throw a 1500 kg container from the sea level up into air to reach 160 km altitude and then fall down?

Roughly 1750 m/s, but that ignores drag from the atmosphere and some other things that are minor and aren't worth talking about if you just want a rough estimate.

 

[ElfredaCyania]

1500 kg container from the sea level up into air to reach 160 km altitude

Like I said before your container would be going absolutely insanely fast and I doubt that any man-made object has achieved controlled flight at such speed near the ground (particle accelerators don't count). But we do have things that move at that speed near the ground which is asteroids.

Assuming your container is 100% iron and it's a rigid sphere with a diameter of 0.7m. It would have a drag coefficient (Cd) of 0.9 for most of its journey. Earth has a scaled height of 8400m, and air density is 1.29, g is 9.81 near Earth, assuming no other factor changes (temperature, gravity etc.). Using the drag equation F=0.5Cdρv², a quick Matlab run gives 5933 m/s initial speed, it takes about 175sec to reach 160km.

Btw the energy it lost to heat alone is 91%, enough to melt the thing like 20 times. This is also why some think the infamous nuclear manhole never made it to outer space. edit: And it's also why we use rockets and planes to leave the dense atmosphere because being fast on the ground sucks.

(edit: You should consider something better for such speed, like a cone or a sabot.)
(edit 2:if you are still thinking about having 7800mps at 160km, the answer is 27879.37 m/s)

https://www.researchgate.net/public...ical_Property_Measurements_on_Iron_and_Steels

 

[Astronuc]

Just wanted to see the calculation of this theory with a 1500 kg egg-shaped container without trying to improve it or make it look real.

Egg shape is not an appropriate form. Instead, one would want a sharp nose relative to the shock cone.

One can learn from the Convair B58 design. See the following video starting around 4:15, and particularly about the fuselage at 4:45 and following.




One can also learn from the SR-71 and X-15 programs, and various hypersonic programs.


Electromagnetic launch systems have been under consideration for decades.
Electromagnetic Railgun Launchers: Direct Launch Feasibility
https://arc.aiaa.org/doi/abs/10.2514/3.51156?journalCode=aiaaj

https://ntrs.nasa.gov/api/citations/19850012948/downloads/19850012948.pdf

https://iopscience.iop.org/article/10.1088/1742-6596/927/1/012027/pdf

https://nnss.gov/mission/sdrd/fy-20...gnetic-launcher-of-high-velocity-projectiles/


Aerodynamic (aeroresistive) heating is a big deal, as well as aerohydrodynamic loading. There are materials that address those issues.

In the case of hypersonic aircraft, there were/are materials and cooling concepts, e.g., using the cryogenic fuel as coolant prior to injection into the engine.

I had a chance in graduate school to visit a facility where railguns were developed and tested. I was more interest in the power generation systems, but I did get to see some railguns.

The U.S. electromagnetic launcher can fire a projectile at speeds of more than 7,200 km/h (4,500 mph), which approximately corresponds to Mach 6

In grad school, we looked into EM launch concepts, and different scenarios, like building the launch system at an elevated altitude, e.g., near the top of a mountain, in order to mitigate some of the atmospheric hydrodynamics, to flooding the launch tube with He or Ne (too costly, but could be partially mitigated with a recirculating system), to including a rocket stage, . . . .

 

[Frabjous]

In grad school, we looked into EM launch concepts, and different scenarios, like building the launch system at an elevated altitude, e.g., near the top of a mountain, in order to mitigate some of the atmospheric hydrodynamics, to flooding the launch tube with He or Ne (too costly, but could be partially mitigated with a recirculating system), to including a rocket stage, . . . .

I saw a DARPA proposal back in grad school to build an EM gun up the side of the volcano on Ascension Island.

 

[Vanadium 50]

Egg shape is not an appropriate form.

I dunno - whatever gets launched will surely be scrambled.

 

[Astronuc]

Egg = Exceptionally great g-force(s)

 

[renormalize]

Egg = Exceptionally great g-force(s)

Eggceptionally

 

[Vanadium 50]

My fault. I shouldn't have egged people on.

 

[ElfredaCyania]

Now I started to wonder if it's possible, at least in theory, to fire something directly from the ground into Earth-Moon L4 or L5, it's probably the simplest solution for single-shot-to-orbit.

 

[Drakkith]

Now I started to wonder if it's possible, at least in theory, to fire something directly from the ground into Earth-Moon L4 or L5, it's probably the simplest solution for single-shot-to-orbit.

Good luck getting through the atmosphere at 10+ km/s...

 

[Tom.G]

Good luck getting through the atmosphere at 10+ km/s...

Yup, Mach 30 would be a, uhmm, uhh, shall we say, a definite challenge?

[EDIT] See thread https://www.physicsforums.com/posts/7091509
The USA military will be trying to hit Mach 7 in early 2025, just to see what happens. Kind of a slow start for your Mach 30.

 

[Flyboy]

Now I started to wonder if it's possible, at least in theory, to fire something directly from the ground into Earth-Moon L4 or L5, it's probably the simplest solution for single-shot-to-orbit.

From the surface of the Earth, not recommended. As others have mentioned, the drag and compressive heating loads are, frankly, absurd.

Now, from the surface of the Moon? Now that’s a whole different critter, and has been proposed several times over the years for mid to late colonization periods.

 

[sophiecentaur]

The direction of the velocity is perpendicular to Earth's surface. And by initial speed I mean a sudden thrust, after which there's no thrust at all.

SpinLaunch is just another scam project that exists only because investors and sponsors have no idea about physics. Just like Mars One, SpinLaunch will cease to exist pretty soon without launching a single satellite, and sponsors will just lose all invested capital.

From what you write in the first quote (nonsense) , I can't see how you can possibly feel qualified to make the comments in the second quote. If you want to be taken seriously, you should at least have some reference to justify your dismissal of Spin launch.
But I wouldn't invest my money in spinlaunch.