# How complicated is Rocket Science?

Bzzzt! No, when there is no movement, there is zero relative velocity.

If there was no acceleration you would be weightless. Since you aren't weightless when you hang suspended, or stand "still", you must NOT have zero acceleration. Therefore you must be accelerating, therefore gravity accelerates mass. G is a constant, g is only a constant locally.

When you "eliminate F = mg", what are you actually doing? Where does it "go"?

But first I'll wait for your proof that springs and weights are no use to an observer in orbit around the earth, in a windowless capsule. You claim that the observer could not tell any difference between orbital and free velocity, using simple accelerated masses at the end of springs. I would like you to demonstrate that this is true...

The fundamental group.

Assume G is a universal set of "accelerations", let H be a subset of "velocities and accelerations" in G.
If h is the generator of accelerations and m is a mass, then hm = mh means h is symmetrical.
Also hmh', where h' is the inverse (acceleration) of h equals m. Mass is a density operator that generates 'self-inductance' L in M, the set of mass-densities. h is normal in H locally, since hh' = h'h =1. Is every h a member of H normal?

p.s. DH, that's a nice picture, but where is the windowless room with springs, masses and a record of their motions ?

Hmm. that person floating weightlessly, is in relative motion with the capsule. If they release any small objects in a small radius around themselves, after a time interval, t, the radius will change as the small objects accelerate in the same orbit as the person and the capsule.

The apparent divergence is due to something...

I strongly suggest you read up on the equivalence principle, noodler.

I already did. If you leave a small object "motionless" inside a space capsule (if you dare), go away and come back "later", it won't be where you "left" it.
Equivalence of acceleration in a gravitational field with and without "material gravity", is the principle of inertial equivalence.

inertia is something induced "in" mass when it accelerates. When a mass is at constant velocity, this is equivalent to being "multiplied by 1 mass" in a closed spacetime interval.

Induction in metals is also an equivalence relation, to induction of acceleration (self-induction) in mass.

Bzzzt! No, when there is no movement, there is zero relative velocity.
When there is zero relative velocity there is zero relative acceleration.

Let's look at what all that means more closely:
You leave an object with mass m1, "at rest" inside a comoving frame with mass m2.
After an interval of time the two objects are at different positions x1 and x2.
This happens because they "start at rest" in independent orbits - this won't happen if they both are at constant velocity, and moving through a field (of only) G.
If you want to know about the alteration of positions over time, you need a discrete or continuous time derivative of change in position over change in time dx1/dt1, dx2,/dt2,...
You want a map of velocities v and times t that relates the changes in orbit to g, m1 and m2.

DH said:
When there is zero relative velocity there is zero relative acceleration.
When there is a mass density there is acceleration, of the mass towards its center.

Replace a mass density with a small radius "ball of test particles". If they are in a "box of spacetime" and have a constant velocity they will accelerate toward each other and the radius of the ball will contract. Over time the mass density at the center will increase and the radius will shrink. You are standing on a larger ball, but the effect is the same. over time gravity compresses balls of particles together. Therefore gravity generates density of mass, a singularity of curvature.

So
if Einstein is correct, launching a rocket that reaches a constant velocity v > 0 and leaves the singularity behind, its inertia will be due to the fact it has density and is accelerating towards its own center; if its velocity changes it will accelerate "toward G", which is the field.

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Suppose that H is a "determinant" or has a subgroup which is, that '"determines" a maximum and minimum density, of mass. Well, matter is mass and charge, so you have to deal with charge density as well; then, spin density has an operator that "compresses" or induces accelerative forces in density of spin (see "Spin Waves in Gases", by someone whose name eludes me).

So if H is a subgroup and R is the set of relations between accelerations of mass, charge and spin densities, you want to know if H is normal in G. If you assume a universal set of "relators" E exists that maps densities to densities, in each case, you can construct a graph of velocities V, and edges E, in G, using the subgroup H, and the "full" set of relations R.

In each case, the densities have operators, in H. Forces ma or mg are in the relational map as "edge weights" in E.

Next, let N be the group of locally normal relations in G(V,E), with local symmetry (gg' = g'g - see phases of the moon, over periods of t-i-m-e ).
N is a local subgroup of H, which accounts for local ellipsoids of revolution, with "theories" T, and "closed forms" S. For example, a combustion engine is a closed form S, of densities in M. S represents the "machine space", and the density operator, in combustion is "heat transfer" dH.

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You are posting nonsense, noodler.

What is the definition of acceleration?

What's yours?
Acceleration is a change in velocity. Velocity is a change in position. In both cases you need a d/dt whatsit.
What is nonsensical to you? You understand a constant velocity is a closed form, so you can say its a node in a graph? To change velocity, you need to get there, with an edge-constructor in the set of edges E. If the E are universal, maybe you can.

Please also point out why, in that picture you posted, the people and the container they're in, both in orbit, aren't in different orbits, and why the orbits won't diverge?

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OK, so if velocity is identically zero, what is the acceleration?

The acceleration of what?
I'll assume you mean "of both bodies, initially at rest", so they are at the same node of G.
Because they orbit independently and because orbits are elliptical, their different accelerations during orbit, toward g means they will "build" independent orbital positions.
You can tell you're in a windowless orbiting container with the small radius of particles experiment - if they don't diverge you aren't orbiting a gravitationally dense body (in a closed elliptic form S),

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Please also point out why, in that picture you posted, the people and the container they're in, both in orbit, aren't in different orbits, and why the orbits won't diverge?
Please do read up on the equivalence principle, particularly on the concept of local experiments.

Even though Einstein was a theoretician, he was also a realist. He tried to bring his thought experiments back to the realm of what is practically measurable.

DH said:
Please do read up on the equivalence principle, particularly on the concept of local experiments.

Even though Einstein was a theoretician, he was also a realist. He tried to bring his thought experiments back to the realm of what is practically measurable.
I'm sorry, I can't see an explanation of why two bodies that start at relative rest, both in orbit around the earth either do or don't diverge over time. That's diverge positionally because of elliptical motion.

Please explain this time why you haven't provided an explanation.

OK, the OP "why is building a rocket complicated" to paraphrase.
The result you want is one rocket either at a highest point (suborbital) or at a v that "cancels" g, and so F, with an opposing F' = -kx; the x is a distance (h) and -k is the constant [of the] propulsive force you need to build in so the rocket gains momentum -mv.

That's an integral of "F", for the interval x=0 to x=h which is equivalent to a differential equation in terms of h(t) and weighted mass, with the critical points when v=0 at t and t'; t' is when the integral reverses, and you know mv going up will equal -mv coming down - you have a two-sided transform from the ma to the mv domain and back.

Then you see that this motion is exactly like what a weight at the end of a spring does, when it oscillates x = -x each time (if you factor the damping), and the velocity reaches zero at the extremes, which are just like the critical height and the launch points.

So you see, launching a rocket is equivalent to building a big spring, and compressing it so it will launch the rocket when it springs back (a reactive force), or by using a spring to pull the rocket up (sounds complicated).

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Wrong.

This is not a simple one dimensional problem. It is a six dimensional problem (position and orientation), plus time, plus the fact that any rocket that is going to go into orbit comprises multiple stages. The thrust is not constant, and if it was, the acceleration would not be constant. Any vehicle that is large enough to launch a reasonable sized payload into orbit cannot be launched at the optimal launch angle. It instead must be launched vertically. Ignoring stresses on the vehicle, the optimal trajectory is one in which the angle of attack is identically zero. However, vehicle stresses, particularly at max Q, constrain the trajectory design.

The real vehicle never follows the designed trajectory. Thrusters don't produce the amount of thrust you think they should produce, nor is the thrust vector pointed in the direction you think it should be pointed. The vehicle's mass properties are never quite what you think they are. Worse, those mass properties change with time because the vehicle is consuming fuel, and those changes are not measured (they are not even measurable).

The reason "rocket science" is hard is because getting a rocket into space (as opposed to launching a toy) requires many distinct advanced skills. The real rocket scientists, the people who design the rocket engines, don't know much about trajectory design. The rocket scientists and trajectory designers don't know a lot about computing the atmospheric stresses on the vehicles. These people don't know how to make the rocket dock with another vehicle. None of them really know how to design the avionics or write the flight software. Electrical engineers are needed to design the electrical power systems. Making the vehicle have the right thermal properties in space is a black art. This is just a start.

But launching all the stages, at t=0 and v=0 of any rocket is essentially equivalent to compressing a big spring. You store reactive force in a spring to launch an object. You transform this into reactive force stored in fuel. The problem then includes the weight of this stored energy and the fact it changes.

If you do launch a plastic rocket with compressed air and water as the reactants, this is also in the same domain as a big heavy multiple stage job. Unless of course plastic and metal react to gravity differently (haven't seen any evidence, myself).

But launching all the stages, at t=0 and v=0 of any rocket is essentially equivalent to compressing a big spring.
Nonsense. Or rather, that is akin to saying that the rollout of the Boeing 787 is essentially equivalent to folding a paper airplane.

This is not a one dimensional problem, noodler. Since this is so simple, please design an optimal trajectory that will bring a vehicle from a vertical launch to a circular orbit at 160 km altitude (nominal insertion altitude). At engine cutoff the vehicle is to be at the desired altitude, is flying horizontally, and is to have circular orbit velocity.

Folding a paper airplane is akin to doing an experiment in design, so is launching a water powered plastic rocket.
But tell you what, I'll figure out a rough trajectory for a "rocket" of some kind and post it alongside your proof that two bodies in uniform relative motion, which are however in elliptic orbits and accelerating towards g, remain at the same distance from each other.