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Orbital Maneuver Question


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30 minutes ago, Red Iron Crown said:

assuming no fuel spent... ...Imagine you enter the SoI exactly retrograde and exit exactly prograde...

 

Cant resist...considering that trajectory the only thing I imagine is

Explosion.jpg

 

49 minutes ago, Red Iron Crown said:

 Escape velocity and excess escape velocity are not relevant aside from affecting how closely your trajectory can approach that ideal. 

and how close you can approach the ideal don't limit how much you can change the speed in practice? (I’m not trying to disagree, I really don't know)

 

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1 hour ago, Red Iron Crown said:

The theoretical limit is twice the planet's orbital velocity, assuming no fuel spent. You will enter and leave the SoI with the same speed relative to that body, but your speed relative to the grandparent body can be changed by no more than twice the orbital speed of the slingshot target. Imagine you enter the SoI exactly retrograde and exit exactly prograde, your speed relative to the grandparent body will have increased by exactly twice the orbital velocity of the slingshot target. Escape velocity and excess escape velocity are not relevant aside from affecting how closely your trajectory can approach that ideal. 

I think some terms need to be clarified here. Suppose we have a central star S, planet P where you want to do gravity assist, and a ship X, the thing that get doubled and applied to X's velocity in frame S is the velocity of X relative to P, not P's orbital velocity around S, nor X's around S. Because in a slingshot situation, X is usually on an escape trajectory relative to P, that's probably why people call it "escape velocity". It's not the periapsis escape velocity, though - it's the velocity upon entry of SoI.

Your term "planet's orbital velocity", my understanding is it means P's orbital velocity around S. This has nothing to do with slingshot - if you have a huge planet far far away from the Sun, its orbital velocity is tiny, but it can still give a tremendous gravity assist.

(And just in case people are picky, which does happen in this community - I do understand 2x is a theoretical unachievable upper bound)

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2 hours ago, Spricigo said:

and how close you can approach the ideal don't limit how much you can change the speed in practice? (I’m not trying to disagree, I really don't know)

The ideal is an upper limit, and that limit is twice the slingshotted body's orbital velocity around its parent. Heavier, denser bodies permit you to more closely approach that limit, so in practice that matters. Does not affect the upper bound, though.

1 hour ago, FancyMouse said:

I think some terms need to be clarified here. Suppose we have a central star S, planet P where you want to do gravity assist, and a ship X, the thing that get doubled and applied to X's velocity in frame S is the velocity of X relative to P, not P's orbital velocity around S, nor X's around S. Because in a slingshot situation, X is usually on an escape trajectory relative to P, that's probably why people call it "escape velocity". It's not the periapsis escape velocity, though - it's the velocity upon entry of SoI.

It's P's orbital velocity that sets the upper bound of the change in velocity due to the slingshot.

1 hour ago, FancyMouse said:

Your term "planet's orbital velocity", my understanding is it means P's orbital velocity around S. This has nothing to do with slingshot - if you have a huge planet far far away from the Sun, its orbital velocity is tiny, but it can still give a tremendous gravity assist.

No, it can't. At absolute most it is twice the orbital velocity of P, full stop. It can look like a tremendous assist because out there where orbital velocities are small a small change in v has a large effect on the trajectory.

 

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36 minutes ago, Red Iron Crown said:

No, it can't. At absolute most it is twice the orbital velocity of P, full stop. It can look like a tremendous assist because out there where orbital velocities are small a small change in v has a large effect on the trajectory.

I have to disagree. I've never mentioned how trajectory looks like, all I've been saying is velocity change. Suppose planet's orbital velocity is v0, and your velocity is v (relative to planet P, upon entry of SoI of P). Assuming you enter retrograde and exit prograde. You're changing your Sun velocity from v0-v to v0+v, which is 2*v difference. This has nothing to do what v0 is. Absolute amount of change (we talk about change because it's the amount we get from a slingshot) is 2*v, not 2*v0.

Another way of proving this is, suppose I use Jool for gravity assist. Let's look at the trajectory from entering Jool's SoI until leaving. This particular slingshot should give a fixed velocity change on your Sun velocity. Now, assume we move the whole Jool system (including your slingshot trajectory) way out so that Jool's orbital velocity becomes 1m/s. Now do I get a different velocity change because of this "upper bound" of 2m/s?

----

EDIT: okay you could still say it's an upper bound because planet's orbital velocity is usually much higher than your relative velocity from the planet (it would mean you either enter from a Sun fly-by trajectory or Sun retrograde orbit). But that would mean the upper bound is dumb - Jool's orbital velocity is still around 4km/s and it doesn't help much saying slingshot gives at most 8km/s, does it?

Edited by FancyMouse
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...and now In reaserching this subject instead of playing KSP. Thanks you, stand-up guys!*

@Red Iron Crown is correct about twice planets orbital velocity, I found this explanation:

Quote

The net effect is almost as if we "bounced" off the front of the planet. From the planet's perspective we approached at the speed U+v, and therefore we will also recede at the speed U+v relative to the planet, but the planet is still moving at (virtually) the speed U, so we will be moving at speed 2U+v. This is just like a very small billiard ball bouncing off a very large one.

Where v its the velocity of the ship and U velocity of the planet in the Star frame of reference.

 

6 hours ago, FancyMouse said:

I have to disagree. I've never mentioned how trajectory looks like, all I've been saying is velocity change. Suppose planet's orbital velocity is v0, and your velocity is v (relative to planet P, upon entry of SoI of P). Assuming you enter retrograde and exit prograde. You're changing your Sun velocity from v0-v to v0+v, which is 2*v difference. This has nothing to do what v0 is. Absolute amount of change (we talk about change because it's the amount we get from a slingshot) is 2*v, not 2*v0.

ok, if v is the ship's velocity relative to the planet, U is the planet's orbital velocity. We can find the magnitude of ship's velocity relative to the star using the rule of cosines

|v+U|²=|v|²+|U|²+2|v||U|cos α   where α  is the angle between v and U, notice that -1<cos α <1

|v|²+|U|²-2|v||U| <|v+U|²<|v|²+|U|²+2|v||U|

(|v| -|U|)x(|v| -|U|) <|v+U|²  <(|v| +|U|)x(|v| +|U|) 

|v| -|U| <|v+U|  <|v| +|U|

 

*dont worry I love you all and only used this word for theatrical purposes. (but yes, 4h reasearching this instead of playing KSP)

 

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2 minutes ago, Spricigo said:

correct about twice planets orbital velocity

Read my EDIT section above. You could still argue it's correct in common cases, but it is a not-quite-useful bound since it's too large. And in extreme cases it will still be incorrect (e.g. a hypothetical Jool at 1m/s orbital velocity and you enter from a Sun flyby or retrograde orbit and Jool completely turns you around).

Bouncing argument in general is valid, but you need to use the correct velocity/frame of reference, which you didn't. (I'm just rephrasing my same argument again, in your bouncing setup) If you have object A travelling at v, hitting another object B travelling at u (both in inertia frame), and assume B is huge such that its own velocity change can be neglected. Then object A in B's frame has velocity v-u. A totally elastic bouncing changes the sign, so after bouncing, A has velocity u-v in B's frame. Transform back to inertia frame, it is 2u-v. In particular this is not 2u+v. Total velocity change is thus (2u-v)-v=2(u-v). So it's 2(u-v), i.e. twice the velocity of A relative to B is the change, not 2u.

In general, when there's an angle, it's really (u-v)sqrt(2-2cos(theta)). It's easiest to do all the calculation in B's frame.

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Actually, you know what - I don't know where your quoted reference come from, but if it starts from u and v in different directions (meaning my v is really its -v), then what we said both agree, meaning that after bouncing, the velocity of the object is indeed 2u+v. However, the v here has an opposite direction from the object's initial v, so the velocity change is really 2u+2v. This now agrees with my result with v sign-flipped.

Edited by FancyMouse
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OK, so I was wrong.

I was conflating speed (a scalar) with velocity (a vector). It is true that the vessels speed can only be increased by a theoretical maximum of twice the orbital speed. But velocity means direction, too, and the bending of the trajectory can have a significant effect on that even if the speed didn't change at all. To my thinking, the theoretical maximum change in velocity should be twice the speed at entry to the SoI plus twice the orbital speed.

Apologies for the misinformation, and thanks for teaching me something today.

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3 hours ago, FancyMouse said:

Bouncing argument in general is valid, but you need to use the correct velocity/frame of reference, which you didn't. 

Please,  show me why the frame of of reference I choose is incorrect. I'm pretty sure both the planet's and the star's are valid inertial frames of reference. The laws of (classical) physics are invariant for inertial frames of reference. 

3 hours ago, FancyMouse said:

Actually, you know what - I don't know where your quoted reference come from, but if it starts from u and v in different directions (meaning my v is really its - v

Actually your -v is his "approach at the speed v+U"  and your v is his "recede at the speed v+U". You use signal while he use words. 

 

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13 minutes ago, Spricigo said:

Actually your -v is his "approach at the speed v+U"  and your v is his "recede at the speed v+U". You use signal while he use words. 

Yeah but then the incorrect part is, the original velocity is really -v, so if you change to 2u+v, your speed change is 2u+v-(-v)=2u+2v, not 2u+v-v=2u. Considering that u and v are in opposite directions (which is still my assumption - I haven't read your source yet), 2u+2v is actually much, much smaller than each of u and v themselves.

That's why it's better to use velocity in the same direction - reduces confusion.

 

20 minutes ago, Red Iron Crown said:

twice the speed at entry to the SoI plus twice the orbital speed.

Eh... still not right if I'm reading correctly - there should be only the first part, no second part of orbital speed. i.e. if Jool's orbiting at 4km/s, you enter Jool at 3.5km/s (relative to Sun), then your upper bound of gravity assist velocity change is 1km/s, not 7 or 8 or 9 km/s.

And of course, all of these are ignoring bending - bending will introduce the extra sqrt(2-2*cos(theta)) factor, which we intentionally ignore for calculating upper bound.

Edited by FancyMouse
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6 minutes ago, FancyMouse said:

Eh... still not right if I'm reading correctly - there should be only the first part, no second part of orbital speed. i.e. if Jool's orbiting at 4km/s, you enter Jool at 3.5km/s (relative to Sun), then your upper bound of gravity assist velocity change is 1km/s, not 7 or 8 or 9 km/s.

You still steal some of the planet's momentum. The bouncing of a ball off a moving truck analogy works well for this, or this diagram:

300px-Gravitational_slingshot.svg.png

Note that the 2U term only becomes apparent after exiting the SoI of the red body, it's relative to that body's parent. Total velocity change relative to that parent is 2U+2v.

Edited by Red Iron Crown
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6 minutes ago, Red Iron Crown said:

You still steal some of the planet's momentum. The bouncing of a ball off a moving truck analogy works well for this, or this diagram:

300px-Gravitational_slingshot.svg.png

Note that the 2U term only becomes apparent after exiting the SoI of the red body, it's relative to that body's parent.

Yeah, but the stolen part is swallowed by the huge mass of the planet (m/M) that's why we ignore it.

Yeah this is exactly what I said the sign flipping above. If you calculate carefully, the "velocity difference" you get is actually 2U+2v, where U+v is exactly the relative velocity between the ship and red ball. No standalone v and U involved in here.

Edited by FancyMouse
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13 minutes ago, Red Iron Crown said:

Total velocity change relative to that parent is 2U+2v.

Yup agree - but see, both u and v are velocity relative to background. U+v, OTOH, is exactly the relative velocity between ship and the red ball (which one relative to which depends on the sign/direction, but the scalar speed is the same). Using my previous 4km/s and 3.5km/s example here, U=4km/s, v=-3.5km/s (we're not going retrograde Sun orbit anyway, so we need a minus sign if we define v as in the diagram), so U+v=0.5km/s, whose double, 1km/s, is our desired upper bound.

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16 hours ago, FancyMouse said:

(I'm just rephrasing my same argument again, in your bouncing setup) If you have object A travelling at v, hitting another object B travelling at u (both in inertia frame), and assume B is huge such that its own velocity change can be neglected. Then object A in B's frame has velocity v-u. A totally elastic bouncing changes the sign, so after bouncing, A has velocity u-v in B's frame. Transform back to inertia frame, it is 2u-v. In particular this is not 2u+v. Total velocity change is thus (2u-v)-v=2(u-v). So it's 2(u-v), i.e. twice the velocity of A relative to B is the change, not 2u.

 

 

If you have object A travelling at v, hitting another object B travelling at -u (both in inertia frame), and assume B is huge such that its own velocity change can be neglected. Then object A in B's frame has velocity v+u. A totally elastic bouncing changes the sign, so after bouncing, A has velocity -u-v in B's frame. Transform back to inertia frame(both frames are inertial), it is -2u-v. In particular this is not -(2u+v). Total velocity change is thus -(2u-v)-v=2(-u-v)  =2v-2u=2u-2v (not a mistake) 

 

Maths...

 

Edited by Spricigo
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17 minutes ago, Abastro said:

I think you got a mistake here: (-2u-v)-v=-2(u+v).

V=0   :wink:

Edit: that is why I said "maths..."  the same calculation "proved" opposite arguments because no one noticed that either u or v need to be 0. A 180° turn it's impossible with gravity slingshot only. 

Edited by Spricigo
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23 minutes ago, Spricigo said:

V=0   :wink:

Edit: that is why I said "maths..."  the same calculation "proved" opposite arguments because no one noticed that either u or v need to be 0. A 180° turn it's impossible with gravity slingshot only. 

You could always define an initial direction, and use a negative scalar value to flip the direction. That's not a big deal - but if you mixed the directions/signs up somewhere during the calculation. it will lead to errors like 2u+2v versus 2u.

Thinking physically it's actually quite clear. In planet's frame, you only know ship's velocity relative to planet. All the velocity change you get can only be a function of that. So if you transform the velocity change back, it should not change, so in particular planet's orbital velocity should not be relevant to the velocity change. This kind of error would suddenly become obvious if you think this way.

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1 hour ago, Spricigo said:

Edit: that is why I said "maths..."  the same calculation "proved" opposite arguments because no one noticed that either u or v need to be 0. A 180° turn it's impossible with gravity slingshot only. 

Well, the velocity difference is always -2(u+v) in vector when planet vel = -u & ship vel = v.

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3 hours ago, Abastro said:

Well, the velocity difference is always -2(u+v) in vector when planet vel = -u & ship vel = v.

OK,  take u=1 and v=1, u+v=2

Now u=-1 and v=1, u+v=0

Now u=1 and v=-1,  u+v=0

Now u=-1 and v=-1,  u+v=-2

We called orbital velocity -u and ship velocity for convenience but the equations need to works for any valid combination of  orbital velocity and ship velocity. And it need to keep working if we do any valid transformation also. 

 

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6 minutes ago, Spricigo said:

OK,  take u=1 and v=1, u+v=2

Now u=-1 and v=1, u+v=0

Now u=1 and v=-1,  u+v=0

Now u=-1 and v=-1,  u+v=-2

That's scalar... scalar does not contain direction while vector does.

7 minutes ago, Spricigo said:

We called orbital velocity -u and ship velocity for convenience but the equations need to works for any valid combination of  orbital velocity and ship velocity. And it need to keep working if we do any valid transformation also.

Yes it is. Also, if it works on some system, it works on every system since reflection/rotation/translation are linear.

If there were contradiction like this, nobody will be using vector arithmetics.

(Also similarly one can deduce a=b for any number a,b - the secret is dividing by zero which is packed up with formula, it is hard to realize that it's actually zero)

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15 minutes ago, Abastro said:

That's scalar... scalar does not contain direction while vector does.

Let me rephrase : multiply u and v for those scalars. A vector multiplies by a scalar is vector.

 

15 minutes ago, Abastro said:

If there were contradiction like this, nobody will be using vector arithmetics.

I'm not sure what contradictions are you referring to.  Please explain.

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5 hours ago, Spricigo said:

the same calculation "proved" opposite arguments

This is contradiction.

17 minutes ago, Spricigo said:

Let me rephrase : multiply u and v for those scalars. A vector multiplies by a scalar is vector.

What do you mean? Multiply u and v? Why?

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