Orbital maneuver by applying a thrust in the radial direction

[Leo Liu]

The author of my textbook writes that a spacecraft 's "thrust in the radial direction at perigee changes the energy but not the angular momentum". Such a thrust increases the eccentricity of the elliptical orbit of the spacecraft because $\epsilon \equiv \sqrt{1+2EL^2/\mu C^2}$, where epsilon determines the eccentricity. When L is constant, an increase in E will give rise to a more prolate ellipse.

While I understand that the angular momentum does not change because the pseudo force or the added velocity is central, which implies $\vec\tau=\vec r\times\vec F=0$ or $\vec L_f=\vec r\times \vec p=\vec r_p\times m(\vec v_t+\vec v_r)=L_i+\vec 0$, the reason the energy increases when the thrust is turned on puzzles me. It is true that the thrust increases the velocity in the radial direction, and therefore increases the mechanical energy of the spacecraft , according to$E=1/2mv^2-C/r_p$ and $\Delta v_r=\text{the rocket equation}$. However, from
$\vec F_r\cdot d\vec r=0=m\vec v\cdot d\vec v$ we know that the pseudo force of the thrust, which is tangential to the trajectory, should not cause any change in the velocity.

I can't really see why my reasoning is wrong. I guess the contradiction is the result of viewing momentum change along the radial direction as force. Could you point out the reason(s)?

 

[.Scott]

Radial thrust at perigee (or apogee) will not change the horizontal component of the velocity.
But it will add a vertical component. So there will be an increase in velocity - and total energy.

In an extreme case, radial thrust can cause the craft to exceed escape velocity - an obvious increase in energy.

 

[kuruman]

Why do say that $\vec F_r\cdot d\vec r=0$? The two vectors (thrust and radial displacement) are non-zero and in the same direction. The same applies to $m\vec v\cdot d\vec v$.

 

[A.T.]

... the pseudo force or the added velocity is central ... we know that the pseudo force of the thrust,

Why do you introduce pseudo forces? Try using the inertial frame of reference.

 

[Leo Liu]

Why do say that $\vec F_r\cdot d\vec r=0$? The two vectors (thrust and radial displacement) are non-zero and in the same direction. The same applies to $m\vec v\cdot d\vec v$.

Hope this diagram can help you understand my reasoning. Here, $\vec r$ is the position vector of the spacecraft , and $d\vec r\parallel \hat\theta$.

 

[kuruman]

Hope this diagram can help you understand my reasoning. Here, $\vec r$ is the position vector of the spacecraft .

If the spacecraft receives an impulse $\vec F dt=F_r dt~\hat r$ in the radial direction, it will have an acceleration also in the radial, not tangential, direction whether it is moving or not. That's Newton's second law. Thus, the change in the position vector will have a radial as well as a tangential component, i.e. $d\vec r=dr~\hat r+r d\theta~\hat \theta.$

 

[DrStupid]

$\vec F_r\cdot d\vec r=0=m\vec v\cdot d\vec v$

That applies to a circular orbit only. If the spacecraft accelerates in radial direction the orbit is not circular anymore and the force results in work that changes the orbital energy.

 

[Halc]

Radial thrust at perigee (or apogee) will not change the horizontal component of the velocity.
But it will add a vertical component. So there will be an increase in velocity - and total energy.

I don't think is is carefully worded, and thus isn't true. Thrust perpendicular to the velocity vector cannot add energy. If very high thrust was put on an orbiting object, but always perpendicular to its motion, then it might do a loop-the-loop but its energy will be unaffected.

Most of the other posts speak more correctly of some finite impulse (a finite quantity of momentum) which does add energy because except at the first moment the motion changes and the unchanged thrust needed to transfer that momentum now has a forward component to it, adding to the energy.

Thrust perpendicular to motion will alter the orbit (make it more elliptical in the case in the OP) but as long as the thrust is maintained perpendicular to the current (not the initial) motion, no energy change in the orbit occurs.

 

[jbriggs444]

If you apply radial thrust at perigee then that thrust is no longer being applied at perigee.

 

[Leo Liu]

If you apply radial thrust at perigee then that thrust is no longer being applied at perigee.

What does this mean? Are you saying that the position of the perigee will change due to the radial acceleration?

 

[Leo Liu]

Thrust perpendicular to motion will alter the orbit (make it more elliptical in the case in the OP) but as long as the thrust is maintained perpendicular to the current (not the initial) motion, no energy change in the orbit occurs.

Can you please explain how it is possible to change energy and keep angular momentum constant at the same time if this is the case?

 

[.Scott]

When radial thrust is applied at apogee or perigee, the apogee or perigee immediately shifts. If you wanted to maintain thrust perpendicular to velocity, you would need to continuously shift the direction of thrust. The result would be a path that looped - maintaining constant energy bu continuously changing direction.

 

[jbriggs444]

What does this mean? Are you saying that the position of the perigee will change due to the radial acceleration?

Yes.

Of course as others have pointed out, once perigee has shifted the question arises: Is further acceleration the result of a central force (preserving angular momentum) or of a force that is perpendicular to velocity (preserving energy). The two no longer coincide.

 

[Halc]

If you apply radial thrust at perigee then that thrust is no longer being applied at perigee.

Perigee is but an instant, and thrust is force. Force times zero time means zero momentum transfer.

On the other hand, if an impulse (momentum) is applied at perigee (say our orbiting 8-ball is struck by the cue-ball), that will apply a nonzero impulse to it (and in theory, arbitrary high force for negligible time), which yes, will change the orbit, add energy, make the already eccentric orbit more eccentric, and move perigee to earlier in the orbit.

If the force applied over a short but finite time does change the orbit, then energy is added only if said force doesn't curve with the change of velocity of the thing being pushed. If it stays perpendicular, then no energy is added but he perigee still changes as you point out.

 

[jbriggs444]

move perigee to earlier in the orbit.

That remark jarred me for a moment. A little chuckle at human perceptions.

My mental image was an inward thrust. Your mental image was an outward thrust. Looking back at post #1, I do not see it being clearly stated either way. It's just "radial direction".

Not that it matters much. The effect of a momentary impulse is to increase the orbital eccentricity and deepen the point of perigee either way.

 

[Halc]

That remark jarred me for a moment. A little chuckle at human perceptions.

True. I took it to be outward, the OP doesn't say. I should have been clear on my assumptions.

Not that it matters much. The effect of a momentary impulse is to increase the orbital eccentricity and deepen the point of perigee either way.

Agree. Increase energy as well.

 

[Leo Liu]

On the other hand, if an impulse (momentum) is applied at perigee (say our orbiting 8-ball is struck by the cue-ball), that will apply a nonzero impulse to it (and in theory, arbitrary high force for negligible time), which yes, will change the orbit, add energy, make the already eccentric orbit more eccentric, and move perigee to earlier in the orbit.

Does this collision also alter the angular momentum? Thank you.

 

[jbriggs444]

Does this collision also alter the angular momentum? Thank you.

By assumption, the delivered impulse is radial (either inward or outward). It carries zero angular momentum.

If you try to do this with a cue ball (for instance, a cue ball launched on a radial upward trajectory striking the satellite), you will find that the cue ball's recoil velocity is zero [for equal mass ball and target and an elastic collision]. The cue ball ends up stationary and falls back to the ground. Assume Teflon so that no frictional forces arise.

 

[Leo Liu]

By assumption, the delivered impulse is radial (either inward or outward). It carries zero angular momentum.

Does this mean that the statement "thrust in the radial direction at perigee changes the energy but not the angular momentum " in the book is incorrect if we assume the spacecraft to be a rocket?

 

[jbriggs444]

Does this mean that the statement "thrust in the radial direction at perigee changes the energy but not the angular momentum " in the book is incorrect if we assume the spacecraft to be a rocket?

There is some ambiguity about what "radial direction" means.

Does it mean directed away from (or perhaps toward) the planet center?
Does it mean directed perpendicular to the satellite's path as it approaches original perigee? [which amounts to the same thing].
Does it mean directed perpendicular to the satellite's path as it departs from the position that was previously perigee?
Does it mean directed midway between the previous two possibilities -- toward the center of curvature of the limiting arc that would be formed if the next possibility were to be adopted and the force increased without bound while keeping the delivered impulse constant?
Does it mean continuously perpendicular to the satellite's path as the force is applied more than momentarily?

In my opinion, the author of the intends for the force to be "radial" in the first sense and definitely none of the last three. Given that interpretation, the author's conclusion is correct. Angular momentum is unaffected while energy is increased.

 

[Halc]

Does this mean that the statement "thrust in the radial direction at perigee changes the energy but not the angular momentum " in the book is incorrect if we assume the spacecraft to be a rocket?

Not to detract from all the ambiguities that jbriggs444 points out, but if it is a rocket, it is going to lose mass, and thus angular momentum as it thrusts sideways like that (either inward or outward). This is presuming the mass of the spent rocket exhaust is no longer counted towards the rocket's angular momentum.

The above posts (including the author of this book) assume an external force is acting on the orbiting object, with no change in mass. So no rocket.

 

[Leo Liu]

Does it mean directed away from (or perhaps toward) the planet center?

In this situation, I suppose the angular momentum of the spacecraft will change. As "Scrat" in post 3 pointed out, the infinitesimal displacement is not purely tangential. So it follows that $\Delta\vec L=(\vec r+\Delta\vec r)\times (m\vec v+m\Delta\vec v)-\vec r\times m\vec v$. Let's assume in this case the mass ejected from the spacecraft is negligible. Since $\Delta\vec r$ is not parallel to $\vec v$, the change in the angular momentum is not 0. I wonder why my reasoning went wrong.

Edit: $\text{I hate }\LaTeX$

 

[jbriggs444]

In this situation, I suppose the angular momentum of the spacecraft will change. As "Scrat" in post 3 pointed out, the infinitesimal displacement is not purely tangential. So it follows that $\Delta\vec L=\vec r'\times m(\vec r+\Delta\vec r)\times (m\vec v+m\Delta\vec v)-\vec r\times m\vec v$. Let's assume in this case the mass ejected from the spacecraft is negligible. Since $\Delta\vec r$ is not parallel to ###\vec v#, the change in the angular momentum is not 0. I wonder why my reasoning went wrong.

Rather than making the ejected mass negligible, I'd rather hypothesize an external force. But either approach will work. We are in the Newtonian model here, so we can have arbitrarily high exhaust velocities without reducing the mass of the craft.

My intuition says that any perceived discrepancy is going to involve a haphazard way of treating an impulsive burn as taking place over an extended time frame. For an impulsive burn, $\Delta \vec{r}$ is going to be zero and there is no problem. That is my recommendation..

But that just dodges the question you are asking.

If we have the burn taking place over an extended period, life gets complicated. We have to worry about the details. For instance, Coriolis rears its head. For an impulsive burn, Coriolis is a finite force applied over an infinitesimal interval. No effect. For an extended burn, Coriolis is a finite force applied over a finite interval. Finite effect. If we want to squash Coriolis then we could work in the inertial frame. That's all well and good, but if we are working in the inertial frame then we are not dealing with a constant force because "radial" is not a single direction.

Pick your poison. But note that angular momentum is not supposed to be conserved in the rotating frame.

 

[DrStupid]

So it follows that $\Delta\vec L=(\vec r+\Delta\vec r)\times (m\vec v+m\Delta\vec v)-\vec r\times m\vec v$.

Why $(\vec r+\Delta\vec r)$? Either the spacecraft receives a radial impulse, then you get

$\Delta\vec L=\vec r\times (m\vec v+m\Delta\vec v)-\vec r\times m\vec v = 0$

because $\Delta\vec r=0$ in this instant process, or there is a radial force acting on it, then you get

$\dot L = r \times F_r = 0$

PS: @jbriggs444 beat me to the impulse.

 

[Leo Liu]

Okay. It appears that my description of the situation is completely wrong, according to the following post on PSE:
https://physics.stackexchange.com/questions/70357/change-of-orbit-with-radial-impulse#:~:text=The radial velocity of a,cannot be a circular one.

The author of the post, who is a space engineer, made the statement that: "This is because the radial force does not bring about any changes to the angular momentum. It does not increase the total energy of the orbit, just its in-plane offset." if there a constant radial impulse applied on the spaceship. So the change in orbit will look like the image below.

Since the energy is constant, the length of the two axes of the ellipse remain the same. The impulse merely moves orbit without changing its shape.

So I think the statement the author made is not true, and the descriptions of the orbit I made were also incorrect.
Case closed. Thanks.

Edit: Situation would be quite different when the impulse takes place in an instant. IMO the energy of the spacecraft might change because the magnitude of the velocity vector increases. The detailed explanation can be found in the answer the link directs to.

 

[.Scott]

I doubt that that picture is correct. If I get a chance, I will attempt to to set up a simulation tomorrow.
If you with a circular orbit and apply a constant outward radial thrust (as described on that web page), what I would expect would be a roughly elliptical "orbit" with its "perigee" matching the original orbital altitude.

By "constant outward thrust", I mean a constant acceleration - so as fuel is burned, the force would decrease proportionally.

 

[DrStupid]

If you with a circular orbit and apply a constant outward radial thrust (as described on that web page), what I would expect would be a roughly elliptical "orbit" with its "perigee" matching the original orbital altitude.

Are you sure? A short radial burn would result in an elliptic orbit with perigee below and apogee above the original circular orbit. Old and new orbit would intersect at the position of the burn. An elliptic orbit with perigee at original orbital altitude can be reached with a short tangential burn. Thus, I would expect that continuous radial thrust would be somwhere in the middle, continuously decreasing perigee.

 

[jbriggs444]

Are you sure? A short radial burn would result in an elliptic orbit with perigee below and apogee above the original circular orbit. Old and new orbit would intersect at the position of the burn. An elliptic orbit with perigee at original orbital altitude can be reached with a short tangential burn. Thus, I would expect that continuous radial thrust would be somwhere in the middle, continuously decreasing perigee.

We would be looking at an orbit under a central force that does not vary as $\frac{1}{r^2}$. So an orbit that is neither required to be elliptical nor closed. However, it is still a conservative central force, so we would still have a conserved total mechanical energy and a conserved angular momentum. That places limits on both perigee and apogee. My expectation would be for a precessing pattern of some sort.

[Note that I agree that "constant thrust" is to be interpreted as constant acceleration. Either by arranging for an external force, rocket thrust decreasing in proportion to decreasing gross vehicle weight or infinitely efficient thrusters]

 

[A.T.]

We would be looking at an orbit under a central force that does not vary as $\frac{1}{r^2}$. So an orbit that is neither required to be elliptical nor closed.

For example if the constant radial thrust equals the initial gravity at the radius of the initial circular orbit. The trajectory will start out tangentially, and then bend away from the mass. It will never return.

 

[jbriggs444]

For example if the constant radial thrust equals the initial gravity at the radius of the initial circular orbit. The trajectory will start out tangentially, and then bend away from the mass. It will never return.

Yes, unbound orbits such as that are naturally not closed. In the case of bound orbits, a "closed" orbit is one that returns to its initial position with its initial velocity. An orbit that is subject to precession is not closed. [Unless the precession is a nice rational fraction of a complete circle]

Edit: Note that the potential energy function for this particular central force law is a bit unusual. It has a high point at a finite distance from the center of the planet and decreases without bound beyond that. It is natural to put the reference "zero point" at the place where peak potential energy is obtained. It would not be reasonable to put the reference point at infinity.

The fact that the reference point for potential energy is at a finite distance from the center has implications when one tries to determine whether an orbit is bound or unbound. For a customary reference point at infinity, it is enough to ask whether the total mechanical energy of an object is positive or negative. Positive or zero and it is unbound. Negative and it is bound.

With the reference point at a finite distance, this heuristic is no longer correct. Angular momentum figures in. An object with a positive total mechanical energy can still be in a bound orbit if it has non-zero angular momentum.

 

[.Scott]

After sleeping on it, the "orbit" pictured makes more sense. In an orbit, the satellite travels more slowly on the apogee side than on the perigee side. So what would happen on each "orbit" with our rocket is that the radial thrust would average more and more towards the apogee - thus propelling itself to ever higher apogees.

I will still do a sim.

 

[jbriggs444]

After sleeping on it, the "orbit" pictured makes more sense. In an orbit, the satellite travels more slowly on the apogee side than on the perigee side. So what would happen on each "orbit" with our rocket is that the radial thrust would average more and more towards the apogee - thus propelling itself to ever higher apogees.

I will still do a sim.

Given conservation of angular momentum and energy, it is immediately clear that neither apogee nor perigee can vary at all while the radial thrust is maintained. Yes, they will differ from the apogee and perigee of the original, un-thrusted orbit. But they must be constant from one loop to the next in the new, thrusted orbit.

Let me walk through that argument a bit more slowly.

At perigee, conservation of angular momentum dictates the tangential component of the velocity as a function of radius. At perigee, conservation of energy dictates the magnitude of the velocity as a function of radius. At perigee, the two must be equal.

That means that we can solve an equation for the radius at perigee. Same for apogee. Same equation. For a bound, non-circular orbit, it will yield [at least] two solutions.

One might look at the situation a bit differently, still without actually solving the equation. Consider the total mechanical energy of the satellite split up into three parts. One part is potential energy. Another part is the kinetic energy corresponding to the "tangential" component of velocity. That is, the component at right angles to the central direction. The third part is the kinetic energy corresponding to the radial component of velocity. Potential energy is a simple function of radius. Given conservation of angular momentum, tangential kinetic energy is also a simple function of radius. Once we factor in conservation of energy, radial kinetic energy must also be a function of radius.

For some radii, radial kinetic energy may be negative. That means that the radius is a forbidden zone. The satellite can never get there. Apogee and perigee are at the boundaries of forbidden zones.

An orbit is bound if it exists within a zone that is surrounded by a forbidden zone.

It might help to visualize a saucer with a series of concentric ridges and a marble rolling around in the trough between a pair of ridges. Or rolling within a central depression.

 

[DrStupid]

The energy of the thrusted orbit should be

$\frac{E}{m} = {\textstyle{1 \over 2}}\left( {\dot r^2 + r^2 \cdot \dot \varphi ^2 } \right) - \frac{{G \cdot M}}{r} - a \cdot r$

The last term is the contribution of the constant radial acceleration. That means switching the engine on or off decreases or increases the orbital energy by $a \cdot r$. The total change of orbital energy is the work done during thrusted flight:

$\Delta E = m \cdot a \cdot \Delta r$

Conservation of angular momentum should result in

$r^2 \cdot \dot \varphi = r_0^2 \cdot \dot \varphi _0$

That gives the orbital energy as a function of $r$

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

and the radial velocity as a function of $r$

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

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.

 

[A.T.]

Given conservation of angular momentum and energy, it is immediately clear that neither apogee nor perigee can vary at all while the radial thrust is maintained. Yes, they will differ from the apogee and perigee of the original, un-thrusted orbit. But they must be constant from one loop to the next in the new, thrusted orbit.

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.

 

[jbriggs444]

Constant in the sense of distance from the center, but they still can precess?

Yes.