Technical Liftoff Question

[esampson]

If you want to tinker about with orbital mechanics I can't recommend Kerbal Space Program highly enough. Spend a bit of time figuring this out and you, too, can become a miserable old grognard like me. You can contemplate launch trajectories and orbital mechanics - and then play 'let's crash the rocket' when you get bored with that.

It also does space planes so you can see what taking off into orbit looks like in a streamlined craft with lift.

I would actually counter that with Orbiter. Not only is it free but it more accurately models gravity (I've been told of people reaching stable orbit in Kerbal with nothing but scramjets which is not really possible).

 

[sudnadja]

If you want to tinker about with orbital mechanics I can't recommend Kerbal Space Program highly enough. Spend a bit of time figuring this out and you, too, can become a miserable old grognard like me and get to complain about the lack of verisimilitude. You can contemplate launch trajectories and orbital mechanics from having some idea about how they work - and when that gets boring you can play 'let's crash the rocket'.

It also does space planes so you can see what taking off into orbit looks like in a streamlined craft with lift.

I do play KSP - KSP is a patched conics system and not remotely realistic. Everything I have been posting here is a real nbody system, using RK45 or Velocity Verlot integration.

In another thread when I was posting about gravity drive launches and provided the simulation for that, for example, could not have been done with patched conics. Patched conics will only approximate the motion of objects in a system when they can be approximated as two bodies , in my case I was integrating the forces applied by the sun, venus and the moon while turning off the force for the earth.

I’m a little surprised that someone interested in orbital mechanics and a traveller player didn’t recognize that I’m already discussing a more accurate system than KSP for orbital motion - I’m leaving out an atmosphere model, though, but in all other respects my system is much more accurate than KSP, and similar to the system that JPl uses in horizons. I get the same results that JPL does, for example, when simulating future encounters between the Ulysses spacecraft and Jupiter when grabbing current solar system state vectors and integrating forward.

“Children of a Dead Earth” makes use of real nbody integrators, and might be interesting for you for a space game.

 

[sudnadja]

At 8000 seconds the 1G ship would be approximately 16km above the ground since it has no acceleration. At 16km it is still experiencing .995 G's. It would actually have a very small acceleration at this point of about 5 cm/s² so it would be gaining speed (and as it would have been experiencing small amounts of acceleration before this point it would actually be slightly higher than 16km, but not by much) but it is certainly not to the point where 'the gravity well is no longer a factor'.

At 8000 seconds the 1G ship would be at 12.049 radii. Of course there is acceleration, the earth’s gravity is continuously decreasing as the ship gains altitude.

This is a trivial differential equation:

x’’(t) = -G 5.9721986*10^24 * x(t)^-2 + 9.806, with x(0)==earths radius and as I mentioned x’(0) = 20. using kms (SI) units. G= gravitation constant and you see the mass of earth there, x(t) is position at time and x’(t) and x’’(t) are the first and second derivatives.

Acceleration, as I mentioned, becomes positive at 416 seconds in my system and continues to increase as gravity decreases.

 

[esampson]

At 8000 seconds the 1G ship would be at 12.049 radii. Of course there is acceleration, the earth’s gravity is continuously decreasing as the ship gains altitude.

This is a trivial differential equation:

x’’(t) = -G 5.9721986*10^24 * x(t)^-2 + 9.806, with x(0)==earths radius and as I mentioned x’(0) = 20. using kms (SI) units. G= gravitation constant and you see the mass of earth there, x(t) is position at time and x’(t) and x’’(t) are the first and second derivatives.

Acceleration, as I mentioned, becomes positive at 416 seconds in my system and continues to increase as gravity decreases.

Woof. You are correct. I just assumed that the minuscule small amounts of acceleration would not make that much of a difference but since they are over a fairly long period of time and they compound (as you continue to move further from the planet the acceleration increases) you would be at a point where the force of gravity from the planet would be less than 1% (I should have realized that 5 cm/s² over 8000 seconds would produce a velocity well in excess of the initial 20 m/s and so while it would be a completely inaccurate approximation of velocity it should have been enough to tell me not to discount the acceleration).

 

[aramis]

The gravitic drives used by things like air/rafts or just to assist with liftoff only work in close proximity to the mass whose gravity is being neutralized (which is why you can't put on a vacc suit and fly an air/raft to another planet). Thus, while you may have neutralized the interaction of gravity between the object and the planet you have not neutralized it between the object and the star the planet is currently orbiting. The object will now be in an independent orbit around the star with an initial vector that is very largely based on the planet's orbital velocity with a slight modification coming from the planet's rotation (if you were on the equator of Earth you would have an orbital vector of 65,700 mph from Earth's orbit modified by a pretty measly 1000 mph by Earth's rotation).

This means the initial vector will be really, really close to the planet's orbital vector which will put the ship in an almost identical orbit to the planet. I haven't run all the numbers but I suspect that you might drift out of the atmosphere as the two orbits diverge (the gravitational attraction between the star and the planet will be larger by an infinitesimal fraction than the attraction between the ship and the star due to the ship's lower mass and their orbital vector will be slightly different) but in the end the tiny, tiny differences in the gravitational forces and orbital vectors will result in an orbit that is only a tiny, tiny bit different from the planet's orbit. You aren't leaving the orbit of the star, much less anything else.

Addendum: I just realized that since the neutralizing effect only works fairly close to the planet you won't even drift that far away from the planet. As you get to whatever the limit of the drive is the planet will once again exert force on you, pulling you back into the planetary orbit around the star.

That's only true of the ones in TNE and T4.

CT (specifically, Striker) and MT gravitics are pure thrust generation

 

[sudnadja]

Woof. You are correct. I just assumed that the minuscule small amounts of acceleration would not make that much of a difference but since they are over a fairly long period of time and they compound (as you continue to move further from the planet the acceleration increases) you would be at a point where the force of gravity from the planet would be less than 1% (I should have realized that 5 cm/s² over 8000 seconds would produce a velocity well in excess of the initial 20 m/s and so while it would be a completely inaccurate approximation of velocity it should have been enough to tell me not to discount the acceleration).

Yes, unwary Travellers can often get caught off-guard by this sort of thing (it's happened to me frequently enough) and has taught me that you *always* should do the math.

As a side note, I highly recommend Hale's "Introduction to Space Flight" as well as Coddington's "An Introduction to Ordinary Differential Equations".

ODEs are *everywhere* in space flight and the universe, from population growth to modeling the behavior of interiors of stars (PDEs anyway) to (most importantly) orbital motion and motion of objects under continuous thrust with varying mass and gravitation fields. The classic Rocket Equation is an ODE and if you want to play Traveller with a somewhat higher degree of realism you can model a lot with a few simple equations.

 

[sudnadja]

Which software was this, because widening orbits make no real sense, at least not at that scale. When you have an object with a given orbital vector around a mass and no further energy put in there's really only two possible solutions, a parabolic orbit or a hyperbolic orbit (an impact is still a parabolic orbit. It is simply one in which periapsis is beneath the surface of the major body).

As for how the drive 'knows', it 'knows' the same way a jump drive 'knows' it is too close to a body. It's just an aspect of the physics. I assume the drive works off of any sufficiently massive body that is within range. The problem is that the range is really very short (I think the limit is 10D) and so you almost never get two objects of sufficient mass that are close enough to one another for a drive to be in range of both.

I do realize that Diameters is not really a good way to deal with gravity but it's how Traveller does it for the sake of convenience. If you really want to get nerdy we could assume that 10D is relative to Earth and so what really happens is the drive loses effectiveness when the force of gravity is less than .01 G's.

I didn't catch the question the first time, the software is Mathematica. I set up a system of ordinary differential equations as the equations of motion and either use Mathematica's RK integrator or a self written velocity verlet integrator.

This appeared in the news today, which talks about a near-earth companion asteroid that is not gravitationally bound to the earth - is in an independent solar orbit around the sun.

As a further comment to some of the other comments in this thread that didn't seem to understand that some of my plots were earth-centric, I'll use this an an example too:

I grabbed the October 14 2017 state vector for the asteroid referenced in that article, inserted it into the simulation, and got this:



Replotting the same simulation centered on the Earth gives this (the plot runs for 2 years):



(red = the moon, there is a green dot in the center of the second plot representing the earth, black is venus which makes a brief appearance, and blue is the companion asteroid. Note the interesting "orbit" that the companion appears to make as if it were somehow bound to earth, even though it is not. That was the same situation as the spiral that the traveller spacecraft appeared to make.

To answer the other aspect of this, why should the orbit of the spacecraft change at all, of course an object that is on the surface of the earth (at the equator) which has the angular momentum of any object at the surface of the earth will have a different amount of energy than if it were only in solar orbit.

Think of a spacecraft doing a close slingshot flyby of jupiter - if you suddenly turn off jupiter's gravity at the spacecraft's closest approach it will end up in a different solar orbit than if you left the gravity on the whole time. Exact same thing.