I Orbital maneuver by applying a thrust in the radial direction

[jbriggs444]

$$\int_t^{t+\Delta t}\vec F_{impulse}\cdot d\vec r=0=\Delta E=\Delta U+\Delta T$$

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.

 

[A.T.]

...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.

 

[Leo Liu]

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?

 

[A.T.]

Can you tell us what the name of the software you use is?

This is just simple Euler integration I wrote myself.

 

[Halc]

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.

 

[A.T.]

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:

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.

 

[jbriggs444]

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

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.

 

[A.T.]

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.

 

[DrStupid]

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.

 

[A.T.]

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.

 

[hutchphd]

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.

 

[jbriggs444]

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.

 

[A.T.]

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.