Orbital maneuver by applying a thrust in the radial direction

In summary: What does this mean? Are you saying that the position of the perigee will change due to the radial acceleration?If radial thrust is turned on, then the increased velocity will cause the mechanical energy of the spacecraft to increase. However, from the equation of motion, it should not cause a change in the velocity. What is causing the discrepancy?
  • #36
I just made some simulations.

In case of a thrusted orbit (red) the radius of perigee and apogee remain constant but they are rotating around the center:

thrusted orbit.gif


With too much thrust the spacecraft escapes:
thrusted escape.gif


As I already expected the thrusted transfer (red) results in a new orbit (blue) with perigee below and apogee above the original circular orbit (green):

thrusted transfer.gif
 
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  • #37
DrStupid said:
As I already expected the thrusted transfer (red) results in a new orbit (blue) with perigee below and apogee above the original circular orbit (green):

thrusted-transfer-gif.gif
So the shifting of the orbit and the crash on the planet happens for radial burn of a certain finite time. In this case the image below and its labeling is misleading

1611275515558-png.png
 
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  • #38
DrStupid said:
I just made some simulations.

In case of a thrusted orbit (red) the radius of perigee and apogee remain constant but they are rotating around the center:

View attachment 276654

With too much thrust the spacecraft escapes:
View attachment 276656

As I already expected the thrusted transfer (red) results in a new orbit (blue) with perigee below and apogee above the original circular orbit (green):

View attachment 276658
A.T. said:
Constant in the sense of distance from the center, but they still can precess?

A quick test simulation gives me the following. The radial thrust is 0.05 (yellow), 0.1 (blue), 0.13 (pink) of the gravity at the initial circular orbit (red). Note that none of the trusted orbits goes below the initial radius.View attachment 276651
These diagrams look neat. Thanks for spending time making them.
So the conclusion is that both mechanical energy and angular momentum are conserved when the spacecraft receives a radius impulse during a short time interval, right?
 
  • #39
Leo Liu said:
So the conclusion is that both mechanical energy and angular momentum are conserved
Please define mechanical energy. I do not see why these calculations imply that it is conserved before and after.
 
  • #40
jbriggs444 said:
The simulation results clearly fail to match the scenario as I understand it.
I thought the simulation (post 34) matched your description wonderfully. It's what I expected. Precession of the orbit at low continuous acceleration, and escape at higher (nowhere near 1g needed) acceleration.
 
  • #41
hutchphd said:
Please define mechanical energy. I do not see why these calculations imply that it is conserved.
Mechanical energy = potential energy + kinetic energy.

We have a perfectly good conservative force here, so a well defined potential.
 
  • #42
I mean before and after application of thrust (the transition)
 
  • #43
hutchphd said:
I mean before and after application of thrust (the transition)
Ahhh, that makes sense. So we need to take some care to splice our definitions of "potential energy" together so that the unthrusted central force field and the thrusted central force field have the same potential at transition.

And that is indeed clearly problematic. You can splice the one way with no problem. But if you splice back the other way, you'll find an offset from the original potential.
 
  • #44
jbriggs444 said:
Mechanical energy = potential energy + kinetic energy.

We have a perfectly good conservative force here, so a well defined potential.
Mechanical energy is not conserved in the scenario plotted. It is conserved between perigee and the following perigee, but energy is higher at apogee because the continuous force acting on the object is doing work (force over distance, where distance is the increased radius).
 
  • #45
The dynamics are exactly what I intuited (and you too) but the total energy is proportional the the major axis of the ellipse.
 
  • #46
Halc said:
Mechanical energy is not conserved in the scenario plotted. It is conserved between perigee and the following perigee, but energy is higher at apogee because the continuous force acting on the object is doing work (force over distance, where distance is the increased radius).
That depends on how one is accounting for potential energy. The thrust is a conservative force and has an associated potential. The mechanical energy at apogee and perigee are identical once this is accounted for.
 
  • #47
That is very tricky. I guess we need to define our system.
 
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  • #48
hutchphd said:
That is very tricky. I guess we need to define our system.
Yes, I think we are in agreement about the physics. Just the terminology is in dispute.
 
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  • #49
hutchphd said:
That is very tricky. I guess we need to define our system.

It's not that complicate if you just do the math. With constant acceleration the radial force is the negative gradient of a potential

##F_r = m \cdot a \cdot \frac{r}{{\left| r \right|}} = - \nabla \Phi##

Integration results in the potential

##\Phi = - m \cdot a \cdot \left| r \right|##

which contributes to the total energy. Such forces conserve total energy. You don't need to wonder where the field actually comes from. It is just a mathematical formalism.But you need to keep in mind that the force is only conservative as long as the field doesn't change. If you change it (e.g. by switching the thrust on and off) total energy is not conserved anymore. That's why orbital energy can be changed by switching between thrusted and free orbits.
 
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  • #50
hutchphd said:
Please define mechanical energy. I do not see why these calculations imply that it is conserved before and after.
$$\int_t^{t+\Delta t}\vec F_{impulse}\cdot d\vec r=0=\Delta E=\Delta U+\Delta T$$
Edit: typo
 
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  • #51
Leo Liu said:
$$\int_t^{t+\Delta t}\vec F_{impulse}\cdot d\vec r=0=\Delta E=\Delta U+\Delta T$$
Wait a minute. I am not following this. What physical situation does this refer to? I will assume that we are talking about a transient impulsive outward central force delivered at perigee.

[Edit: With the typo corrected we are now in agreement about integrating F dot dr]
Surely if we are computing a change in energy then we should be dotting F with dr rather than dv. For a given total impulse, the resulting integral should be largely independent of the time interval and should be non-zero in the limit as ##\Delta t## goes to zero despite ##d \vec{r}## tending to zero.

Heuristically, if you halve the time interval, you double the force and halve the resulting displacement. The work done is unchanged.
 
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  • #52
Halc said:
...and escape at higher (nowhere near 1g needed) acceleration.
In my simulations the thrust needed to escape is very close to 1/8 of the gravity on the initial circular orbit. I wonder if it is exactly 1/8 and if there is a nice proof for it.

0_01249_01251.png
 
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  • #53
A.T. said:
In my simulations the thrust needed to escape is very close to 1/8 of the gravity on the initial circular orbit. I wonder if it is exactly 1/8 and if there is a nice proof for it.

View attachment 276745
Can you tell us what the name of the software you use is?
 
  • #54
Leo Liu said:
Can you tell us what the name of the software you use is?
This is just simple Euler integration I wrote myself.
 
  • #55
A.T. said:
In my simulations the thrust needed to escape is very close to 1/8 of the gravity on the initial circular orbit. I wonder if it is exactly 1/8 and if there is a nice proof for it.
I suspect it is exactly 1/8. I've not done the proof.
If you accelerate at exactly 1/8g, it will probably approach but never reach a perfect circular orbit at the apogee radius, never returning to the low radius like your green line. This is an unstable equilibrium, so your simulation probably will not show this unless it has infinite precision and infinitely small time steps.

As for the software, I know how to write my own simulations, but the nice picture output uses a library set with which I'm not familiar. I can't make pretty pictures like that, only tables of numbers and such.
 
  • #56
Halc said:
I suspect it is exactly 1/8.
Me too. I tried different radii and gravity strengths, and it always starts escaping a this thrust ratio to initial gravity. I guess to show this, one would modify the effective potential:

https://en.wikipedia.org/wiki/Effective_potential

by the additional potential from the thrust described by @DrStupid:

DrStupid said:
It's not that complicate if you just do the math. With constant acceleration the radial force is the negative gradient of a potential

##F_r = m \cdot a \cdot \frac{r}{{\left| r \right|}} = - \nabla \Phi##

Integration results in the potential

##\Phi = - m \cdot a \cdot \left| r \right|##

An then find the value of ##a## that allows to move over the outer local maximum of the modified effective potential.
 
  • #57
DrStupid said:
##\dot r^2 = \dot r_0^2 + r_0^4 \cdot \dot \varphi _0^2 \left( {\frac{1}{{r_0^2 }} - \frac{1}{{r^2 }}} \right) + 2 \cdot a \cdot \left( {r - r_0 } \right) + 2 \cdot G \cdot M \cdot \left( {\frac{1}{r} - \frac{1}{{r_0 }}} \right)##

Solving for ##r## at ##\dot r^2 = 0## should result in perigee and apogee. It's just a quadratic equation but I'm too lazy to do that.
I am a pretty lazy SOB as well. But member @A.T. has come up with the conjecture that the critical thrust is ##\frac{g}{8}## where g is the gravitational acceleration in a pre-existing circular un-thrusted orbit.

The above equation looks like an excellent starting point. So I had a second look. It is not a quadratic. It is a cubic. As it should be. It should have three solutions.

Solution 1 is perigee for the bound, looping orbit.
Solution 2 is apogee for the bound, looping orbit.
Solution 3 is for an unbound orbit with the given radius as its perigee.

More formally, one can see that the equation is a cubic because it has an ##\frac{1}{r^2}##, a ##\frac{1}{r}## and a ##r## term. Multiply through by ##r^2## and you have the cubic.

If I were less lazy, I would do synthetic division to remove the solution for the radius at perigee (##r=r_0##), yielding a quadratic. That quadratic will have either zero, one or two solutions depending on the discriminant. The critical thrust we are after is the thrust value that makes the discriminant zero.
 
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  • #58
jbriggs444 said:
I am a pretty lazy SOB as well. But member @A.T. has come up with the conjecture that the critical thrust is ##\frac{g}{8}## where g is the gravitational acceleration in a pre-existing circular un-thrusted orbit.
Another interesting parameter to explore is the apogee for thrust just below the critical escape thrust. Purely visually it looks like twice the original circular orbit radius.

This would fit the critical thrust of ##\frac{g_0}{8}##: At twice the initial radius, the tangential velocity is halved (AM conservation), so the centripetal acceleration ##\frac{v^2}{r}## required for a circular orbit is ##\frac{g_0}{8}##. Gravity itself is ##\frac{g_0}{4}##, but with ##\frac{g_0}{8}## opposite thrust the acceleration becomes exactly the value for an circular orbit.

That thursted circular orbit at ##2r_0## is unstable. With less than ##\frac{g_0}{8}## thrust you drop lower and gravity grows again (if you got to ##2r_0## somehow). With more than ##\frac{g_0}{8}## thrust you go higher continuously as gravity decreases.
 
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  • #59
In the original circular orbit I have

##g_0 = \frac{{G \cdot M}}{{r_0^2 }}##

##\dot \varphi _0^2 = \frac{{G \cdot M}}{{r_0^3 }} = \frac{{g_0 }}{{r_0 }}##

##\dot r_0 = 0##

That means for the orbital energy of the thrusted orbit

##\frac{E}{m} = \frac{1}{2} \cdot \left( {\dot r^2 + \frac{{r_0^4 \cdot \dot \varphi _0^2 }}{{r^2 }}} \right) - \frac{{G \cdot M}}{r} - a \cdot r = \frac{1}{2} \cdot \left( {\dot r^2 + \frac{{g_0 \cdot r_0^3 }}{{r^2 }}} \right) - \frac{{g_0 \cdot r_0^2 }}{r} - a \cdot r##

resulting in the initial energy

##\frac{{E_0 }}{m} = - \left( {\frac{1}{2} \cdot g_0 + a} \right) \cdot r_0##

In apogee the radial velocity is zero

##\dot r_a = 0##

Thus, orbital energy in apogee is

##\frac{{E_a }}{m} = \frac{1}{2} \cdot \frac{{g_0 \cdot r_0^3 }}{{r_a^2 }} - \frac{{g_0 \cdot r_0^2 }}{{r_a }} - a \cdot r_a##

I am looking for the apogee with the maximum energy. That means

##\frac{d}{{dr_a }}\frac{{E_a }}{m} = - \frac{{g_0 \cdot r_0^3 }}{{r_a^3 }} + \frac{{g_0 \cdot r_0^2 }}{{r_a^2 }} - a = 0##

Now I would need to solve this cubic equation (@jbriggs444 already corrected me in this regard) for ##r_a##, calculate the corresponding energy and than solve ##a## for equal initial energy. But fortunately we already have an idea for ##a## that just needs to be checked:

##a = \frac{{g_0 }}{8}##

results in

##r_{\max } = 2 \cdot r_0##

and

##\frac{{E_{\max } }}{m} = \frac{{E_0 }}{m} = - \frac{5}{8}g_0 \cdot r_a ##

That means the empiric estimates for ##a## and ##r_{max}## are correct.
 
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  • #60
DrStupid said:
But fortunately we already have an idea for ##a## that just needs to be checked:

##a = \frac{{g_0 }}{8}##

results in

##r_{\max } = 2 \cdot r_0##

and

##\frac{{E_{\max } }}{m} = \frac{{E_0 }}{m} = - \frac{5}{8}g_0 \cdot r_a ##

That means the empiric estimates for ##a## and ##r_{max}## are correct.

Thanks for checking this. Another interesting distance is ##\sqrt{8}{r_0 }##, where the barely unbound orbit has its inflection point, as the thrust becomes greater than gravity and the ship visibly "pulls away" from the planet.
 
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  • #61
jbriggs444 said:
If I were less lazy,
You wouldn't know a short fellow called Tom Sawyer would you ?...
Anyhow I took what @DrStupid did and made a little graph which I find useful. First I will choose some natural units ##r_0,g_0, m=1## to write the effective potential$$V_{eff}(r)=\frac 1 {2r^2}-\frac 1 r -a(r-1)$$ where I have changed the zero for the "added" radial potential . The graph and a "blown up" version near the minimum show the potential for various values of a. I think it comports with all the analyses.
I need pictures.
 

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  • #62
hutchphd said:
You wouldn't know a short fellow called Tom Sawyer would you ?...
ROFL. Only a really smart person is good enough to do those boards.
 
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  • #63
hutchphd said:
Anyhow I took what @DrStupid did and made a little graph which I find useful. First I will choose some natural units ##r_0,g_0, m=1## to write the effective potential$$V_{eff}(r)=\frac 1 {2r^2}-\frac 1 r -a(r-1)$$ where I have changed the zero for the "added" radial potential . The graph and a "blown up" version near the minimum show the potential for various values of a. I think it comports with all the analyses.
I need pictures.
Yes, thank you! This is what I envisioned in post #56. I added some additional marks to your plot. For thrust higher than 0.125 the right maximum of the potential will drop below the value of the circular orbit (dotted red line) and can be overcome to escape.

eff_pot.png
 
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