SpaceX Starship : First Flight : Starbase, TX : 20 April 2023 - DISCUSSION

[OTV Booster]

It can go half way around, 3/4 around, 99.99999% around.

Uh, actually, no. Look at the equation for an orbit: radius = p/(1 + ecc * cos(theta)) where radius is the distance from the center of the Earth, p is a constant (the semi-latus rectum of the orbit, but that's not important), ecc is the eccentricity, and theta is the angle between the cannon and the current position of the shell.

At 180°, the shell is at its apex (the perigee). But because cos(-theta) = cos(theta), the orbit is symmetrical. (As if I needed to prove that.) :-) As a result, if it does not touch the ground in the first 180°, it will not touch the ground at all until it reaches the launch point.

So you're saying that in the idealized world described it is impossible to launch tangent and hit anywhere on the back 180?


My amateur status is more real than my math skills but I do have some conceptual questions. On a flat earth, a ballistic trajectory is a parabola. On a spherical earth it is something else. In that idealized earth the minimum viable orbital cannon shot would be the special case of an eclipse called a circle. Between the short round that almost follows a parabola and the circular orbit shot is something else, possibly an ellipse, but that is unclear to me.


Would the short round trajectory (not a parabola) be symmetrical? If it's a segment of an ellipse it would be. My gut says it would be a non-conic shape, but I don't really know. If the trajectory is non-conic it would have no symmetry and it should be possible to hit the back 180. It looks like my question hinges on 'not a parabola' being, or not being, an ellipse.

[OTV Booster]

But because cos(-theta) = cos(theta), the orbit is symmetrical. (As if I needed to prove that.) :-) As a result, if it does not touch the ground in the first 180°, it will not touch the ground at all until it reaches the launch point.

Exactly, because orbit is symmetrical, if it touches the ground at some theta between 90-180°, then it must have travelled another theta above ground BEFORE the point of the cannon. So the angular distance between these two intersections with ground is 2theta, a value greater than 180°

There used to be a clear distinction between orbital & suborbital , a wide gulf between the two. A space launch vehicle was one or the other. But now SpaceX have, as they have with other industry terms like "flight proven", thrown themselves directly into the grey area and forced us to probe concepts that never had a clear boundary. This is what has lead us to strained definitions involving spherical cows in a vacuum.

I suspect this whole discussion is just a sign that we're all bored because not a lot is happening at the moment.

I think it's been explored often in ICBM circles, and is the original meaning of "suborbital".

The "Suborbital" that VG and BO/NS are doing is very very very suborbital.  It's what a sounding rocket does.  Calling it "suborbital" is charitable.  I'm saying it because on the grand scheme of Earth, you can just approximate it with a parabola, since the gravity vector doesn't have time to shift.  A true suborbital trajectory is elliptical.

... and, I like the energy definition of suborbital better, since there are orbital-energy trajectories that still intersect the surface of the Earth, and from a vehicle capability point of view, are still orbital.  (e.g. "suborbital around the moon", as folks here often point out) - all that's gone with an energy-based definition.

Hmmm. Maybe you just answered my 'not a parabola'/elliptical question.

[OTV Booster]

But because cos(-theta) = cos(theta), the orbit is symmetrical. (As if I needed to prove that.) :-) As a result, if it does not touch the ground in the first 180°, it will not touch the ground at all until it reaches the launch point.

Exactly, because orbit is symmetrical, if it touches the ground at some theta between 90-180°, then it must have travelled another theta above ground BEFORE the point of the cannon. So the angular distance between these two intersections with ground is 2theta, a value greater than 180°

There used to be a clear distinction between orbital & suborbital , a wide gulf between the two. A space launch vehicle was one or the other. But now SpaceX have, as they have with other industry terms like "flight proven", thrown themselves directly into the grey area and forced us to probe concepts that never had a clear boundary. This is what has lead us to strained definitions involving spherical cows in a vacuum.

I suspect this whole discussion is just a sign that we're all bored because not a lot is happening at the moment.

It seems to me that none of this discussion about sub-orbital versus orbital is because the participants
haven't read the SpaceX FCC application whose link is at the beginning of this thread.

In the "Flight Profile" section it says:

"The Starship Orbital test flight will originate from Starbase, TX. The Booster stage will separate
approximately 170 seconds into flight. The Booster will then perform a partial return and land in the
Gulf of Mexico approximately 20 miles from the shore. The Orbital Starship will continue on flying
between the Florida Straits. It will achieve orbit until performing a powered, targeted landing
approximately 100km (~62 miles) off the northwest coast of Kauai in a soft ocean landing."

I'm surprised that the Soviet FOBS launches of the 60s wasn't mentioned in the discussion.
The payloads achieved orbit and were de-orbited before completing a full orbit, Just like Gagarin's
Vostok flight.

Carl

You take all the fun out of spherical cows 

[meekGee]

But because cos(-theta) = cos(theta), the orbit is symmetrical. (As if I needed to prove that.) :-) As a result, if it does not touch the ground in the first 180°, it will not touch the ground at all until it reaches the launch point.

Exactly, because orbit is symmetrical, if it touches the ground at some theta between 90-180°, then it must have travelled another theta above ground BEFORE the point of the cannon. So the angular distance between these two intersections with ground is 2theta, a value greater than 180°

There used to be a clear distinction between orbital & suborbital , a wide gulf between the two. A space launch vehicle was one or the other. But now SpaceX have, as they have with other industry terms like "flight proven", thrown themselves directly into the grey area and forced us to probe concepts that never had a clear boundary. This is what has lead us to strained definitions involving spherical cows in a vacuum.

I suspect this whole discussion is just a sign that we're all bored because not a lot is happening at the moment.

It seems to me that none of this discussion about sub-orbital versus orbital is because the participants
haven't read the SpaceX FCC application whose link is at the beginning of this thread.

In the "Flight Profile" section it says:

"The Starship Orbital test flight will originate from Starbase, TX. The Booster stage will separate
approximately 170 seconds into flight. The Booster will then perform a partial return and land in the
Gulf of Mexico approximately 20 miles from the shore. The Orbital Starship will continue on flying
between the Florida Straits. It will achieve orbit until performing a powered, targeted landing
approximately 100km (~62 miles) off the northwest coast of Kauai in a soft ocean landing."

I'm surprised that the Soviet FOBS launches of the 60s wasn't mentioned in the discussion.
The payloads achieved orbit and were de-orbited before completing a full orbit, Just like Gagarin's
Vostok flight.

Carl

Yes, correct. 

The document does not describe a suborbital trajectory but an aborted orbital one.

The rest of the suborbital discussion is good, it just doesn't directly apply here. Probably.

[Greg Hullender]

So you're saying that in the idealized world described it is impossible to launch tangent and hit anywhere on the back 180?

Correct.

My amateur status is more real than my math skills but I do have some conceptual questions. On a flat earth, a ballistic trajectory is a parabola. On a spherical earth it is something else.

It's a perfect ellipse. (Edit: or hyperbola. Or circle. etc. But always a conic.) If we know the muzzle velocity and the angle of elevation of the gun, we can compute the exact orbital elements, but it is always an ellipse. When the angle is zero (as in your example), the shell cannot impact the "back side" of the planet unless it passes through it first. If you use a different elevation of the gun, though, it's not hard to hit the other side.

[Comga]

But because cos(-theta) = cos(theta), the orbit is symmetrical. (As if I needed to prove that.) :-) As a result, if it does not touch the ground in the first 180°, it will not touch the ground at all until it reaches the launch point.

Exactly, because orbit is symmetrical, if it touches the ground at some theta between 90-180°, then it must have travelled another theta above ground BEFORE the point of the cannon. So the angular distance between these two intersections with ground is 2theta, a value greater than 180°

There used to be a clear distinction between orbital & suborbital , a wide gulf between the two. A space launch vehicle was one or the other. But now SpaceX have, as they have with other industry terms like "flight proven", thrown themselves directly into the grey area and forced us to probe concepts that never had a clear boundary. This is what has lead us to strained definitions involving spherical cows in a vacuum.

I suspect this whole discussion is just a sign that we're all bored because not a lot is happening at the moment.

I think it's been explored often in ICBM circles, and is the original meaning of "suborbital".

The "Suborbital" that VG and BO/NS are doing is very very very suborbital.  It's what a sounding rocket does.  Calling it "suborbital" is charitable.  I'm saying it because on the grand scheme of Earth, you can just approximate it with a parabola, since the gravity vector doesn't have time to shift.  A true suborbital trajectory is elliptical.

... and, I like the energy definition of suborbital better, since there are orbital-energy trajectories that still intersect the surface of the Earth, and from a vehicle capability point of view, are still orbital.  (e.g. "suborbital around the moon", as folks here often point out) - all that's gone with an energy-based definition.

Bingo
My opinion has been stated previously that the reason to call SS2 and New Shephard launches "sub-orbital" is to associate them with orbital flights, even by exclusion.

And an energy criterion for "orbital" is  equivalent to a semi-major axis criterion.
If a trajectory is in an orbit with enough energy to go around the world, the semi-major axis of the path is greater than the radius of the Earth (plus some margin for the atmosphere.)
That means a free return trajectory around the Moon is clearly "orbital".
If Gargarin's Vostok, the FOBS, and Starship/Super Heavy Flight 1 need to perform retro-burns to bring them down either before or after they complete an orbit, they are "orbital"

[OneSpeed]

If Gargarin's Vostok, the FOBS, and Starship/Super Heavy Flight 1 need to perform retro-burns to bring them down either before or after they complete an orbit, they are "orbital"

The Starship/SuperHeavy first flight doesn't have to have a retro burn, because its ellipse can enter the sensible atmosphere at perigee. But that would still be a valid test of an orbital re-entry.

https://forum.nasaspaceflight.com/index.php?topic=47179.msg2240197#msg2240197

[meekGee]

But because cos(-theta) = cos(theta), the orbit is symmetrical. (As if I needed to prove that.) :-) As a result, if it does not touch the ground in the first 180°, it will not touch the ground at all until it reaches the launch point.

Exactly, because orbit is symmetrical, if it touches the ground at some theta between 90-180°, then it must have travelled another theta above ground BEFORE the point of the cannon. So the angular distance between these two intersections with ground is 2theta, a value greater than 180°

There used to be a clear distinction between orbital & suborbital , a wide gulf between the two. A space launch vehicle was one or the other. But now SpaceX have, as they have with other industry terms like "flight proven", thrown themselves directly into the grey area and forced us to probe concepts that never had a clear boundary. This is what has lead us to strained definitions involving spherical cows in a vacuum.

I suspect this whole discussion is just a sign that we're all bored because not a lot is happening at the moment.

I think it's been explored often in ICBM circles, and is the original meaning of "suborbital".

The "Suborbital" that VG and BO/NS are doing is very very very suborbital.  It's what a sounding rocket does.  Calling it "suborbital" is charitable.  I'm saying it because on the grand scheme of Earth, you can just approximate it with a parabola, since the gravity vector doesn't have time to shift.  A true suborbital trajectory is elliptical.

... and, I like the energy definition of suborbital better, since there are orbital-energy trajectories that still intersect the surface of the Earth, and from a vehicle capability point of view, are still orbital.  (e.g. "suborbital around the moon", as folks here often point out) - all that's gone with an energy-based definition.

Bingo
My opinion has been stated previously that the reason to call SS2 and New Shephard launches "sub-orbital" is to associate them with orbital flights, even by exclusion.

And an energy criterion for "orbital" is  equivalent to a semi-major axis criterion.
If a trajectory is in an orbit with enough energy to go around the world, the semi-major axis of the path is greater than the radius of the Earth (plus some margin for the atmosphere.)
That means a free return trajectory around the Moon is clearly "orbital".
If Gargarin's Vostok, the FOBS, and Starship/Super Heavy Flight 1 need to perform retro-burns to bring them down either before or after they complete an orbit, they are "orbital"

Thx...

Also, I just realized I said "in ICBM circles" on a trajectory discussion, and so I apologize. That door should remain closed.

And yes about association by exclusion, making "sub-" hint at "almost-"...   

Shrug.  It doesn't matter anymore.  But I still expect one of these clowns to "welcome to the club" when Starship carries its first crew.

[OTV Booster]

So you're saying that in the idealized world described it is impossible to launch tangent and hit anywhere on the back 180?

Correct.

My amateur status is more real than my math skills but I do have some conceptual questions. On a flat earth, a ballistic trajectory is a parabola. On a spherical earth it is something else.

It's a perfect ellipse. (Edit: or hyperbola. Or circle. etc. But always a conic.) If we know the muzzle velocity and the angle of elevation of the gun, we can compute the exact orbital elements, but it is always an ellipse. When the angle is zero (as in your example), the shell cannot impact the "back side" of the planet unless it passes through it first. If you use a different elevation of the gun, though, it's not hard to hit the other side.

You've made my day. You helped me learn something. Thanks.

[capoman]

All of this could be understand with a bit of time in Kerbal Space Program, even in the tuturials, but the real answer is whether starship will need to do a deorbit burn? If so, it was orbital, if it was going to decay on the first trip around the earth without a deorbit burn, it's not orbital.

[meekGee]

All of this could be understand with a bit of time in Kerbal Space Program, even in the tuturials, but the real answer is whether starship will need to do a deorbit burn? If so, it was orbital, if it was going to decay on the first trip around the earth without a deorbit burn, it's not orbital.

Not sure about that..

Since there's no such thing as a gravitational decay in a two body classical system, then you're talking about aerodynamic decay, and then no, decay during the first orbit doesn't mean you weren't orbital, not any more than if you actively de-orbited.

The best definition is whether at insertion you had enough energy to ballistically go around the earth.

[ppb]

All of this could be understand with a bit of time in Kerbal Space Program, even in the tuturials, but the real answer is whether starship will need to do a deorbit burn? If so, it was orbital, if it was going to decay on the first trip around the earth without a deorbit burn, it's not orbital.

Not sure about that..

Since there's no such thing as a gravitational decay in a two body classical system, then you're talking about aerodynamic decay, and then no, decay during the first orbit doesn't mean you weren't orbital, not any more than if you actively de-orbited.

The best definition is whether at insertion you had enough energy to ballistically go around the earth.

I agree the speed/energy definition of full orbital is the simplest.  At insertion above most of the atmosphere, flight path angle will be close to zero and inertial speed will be at least 7.8 km/s.

[meekGee]

All of this could be understand with a bit of time in Kerbal Space Program, even in the tuturials, but the real answer is whether starship will need to do a deorbit burn? If so, it was orbital, if it was going to decay on the first trip around the earth without a deorbit burn, it's not orbital.

Not sure about that..

Since there's no such thing as a gravitational decay in a two body classical system, then you're talking about aerodynamic decay, and then no, decay during the first orbit doesn't mean you weren't orbital, not any more than if you actively de-orbited.

The best definition is whether at insertion you had enough energy to ballistically go around the earth.

I agree the speed/energy definition of full orbital is the simplest.  At insertion above most of the atmosphere, flight path angle will be close to zero and inertial speed will be at least 7.8 km/s.

I think what I and several others were saying is that if you have a vehicle capable of such an orbit, then it should be considered orbital.

Or rather the other way - it should not be considered "suborbital".

In other words: "It's not your impact point that defines you"

[erictant]

All of this could be understand with a bit of time in Kerbal Space Program, even in the tuturials, but the real answer is whether starship will need to do a deorbit burn? If so, it was orbital, if it was going to decay on the first trip around the earth without a deorbit burn, it's not orbital.

Not sure about that..

Since there's no such thing as a gravitational decay in a two body classical system, then you're talking about aerodynamic decay, and then no, decay during the first orbit doesn't mean you weren't orbital, not any more than if you actively de-orbited.

I think we are really discussing the difference between the behaviour of Spherical Cows orbiting Billiard Balls and Real Rockets orbiting Planets.

The best definition is whether at insertion you had enough energy to ballistically go around the earth.

For the Cow, this will mean you stay in orbit forever. For the Rocket, you will immediately experience aerodynamic drag and effectively begin your deorbit manoeuvre. You are now in a region where you must balance drag and boost if you wish to maintain orbital velocity.

If you want to stay in orbit around your Planet you have to go higher to reduce the drag to a manageable level. The ISS is near the top of this region and is effectively deorbiting all the time and left to its own devices would re-enter. It is only kept there by regular orbit boosting burns.

SpaceX are deliberately exploiting the bottom edge of this region to allow for a controlled and predictable flight profile that does not require additional burns and all the complexities that entails. This allows them to maximise the chance of getting permits and also of being able to test high speed re-entry with a passive deorbit system.

So, yes, Starship will be orbital momentarily (English meaning) but will immediately start to deorbit so will not go all the way around.

[meekGee]

All of this could be understand with a bit of time in Kerbal Space Program, even in the tuturials, but the real answer is whether starship will need to do a deorbit burn? If so, it was orbital, if it was going to decay on the first trip around the earth without a deorbit burn, it's not orbital.

Not sure about that..

Since there's no such thing as a gravitational decay in a two body classical system, then you're talking about aerodynamic decay, and then no, decay during the first orbit doesn't mean you weren't orbital, not any more than if you actively de-orbited.

I think we are really discussing the difference between the behaviour of Spherical Cows orbiting Billiard Balls and Real Rockets orbiting Planets.

The best definition is whether at insertion you had enough energy to ballistically go around the earth.

For the Cow, this will mean you stay in orbit forever. For the Rocket, you will immediately experience aerodynamic drag and effectively begin your deorbit manoeuvre. You are now in a region where you must balance drag and boost if you wish to maintain orbital velocity.

If you want to stay in orbit around your Planet you have to go higher to reduce the drag to a manageable level. The ISS is near the top of this region and is effectively deorbiting all the time and left to its own devices would re-enter. It is only kept there by regular orbit boosting burns.

SpaceX are deliberately exploiting the bottom edge of this region to allow for a controlled and predictable flight profile that does not require additional burns and all the complexities that entails. This allows them to maximise the chance of getting permits and also of being able to test high speed re-entry with a passive deorbit system.

So, yes, Starship will be orbital momentarily (English meaning) but will immediately start to deorbit so will not go all the way around.

I don't think they're trying to naturally decay within one-half of an orbit - I think they specifically said they'll do a de-orbit burn.

They want to a) demonstrate orbital velocity and b) hit a specific landing zone.

Besides, natural decay within one-half of an orbit is really difficult and serves no useful purpose.

[Greg Hullender]

Personally, if I were drafting a definition of suborbital, I'd want to capture the idea that i's a short, fairly flat arc that still gets into space.

So something like "a ballistic trajectory from ground to ground that leaves the atmosphere, does not enter the Van Allen belts, and subtends less than 180°."

To generalize that to bodies other than Earth, you could use something like "a ground-to-ground partial orbit (50% or less) with apapsis between 1.01 and 2 planetary radii."

Whether it's really worth trying to exclude very high orbits and/or orbits that go (say) 3/4 of the way around is debatable, of course, but I do think it's important to exclude things like mortar shells and basketballs. It's also a question whether you want to include things that merely go straight up and come straight down.

[JaimeZX]

My opinion is not necessarily scholarly, but I would draw the distinction between Orbital and Sub-Orbital solely by imparted dV. If you don't get the 9km/s, you ain't getting an (Earth) orbit, no matter what. This is therefore sub-orbital.

If you impart the requisite 9km/s, you COULD fly to a stable orbit, but you can also impart this dV on a trajectory that will *not* result in an orbit. Hence "orbital class rocket" and the first SS/SH "orbital" flight that only goes 3/4 of the way around. It COULD go into a stable orbit, but they didn't point it in that direction... So still "orbital."

[DavP]

This has been posted in another thread

It shows  what appears to be a countdown to the orbital test launch and the engine configuration of the rockets.
Also some of the engines are marked in green and others in white.

[Skyrocket]

Also some of the engines are marked in green and others in white.

The green engines are gimballed while the white are fixed.

[RotoSequence]

Also some of the engines are marked in green and others in white.

The green engines are gimballed while the white are fixed.

Out of frame here is a Starship shot, showing two green center engines and one white, and we know that all three gimbal.