Jump to content

orbital mechanic question


unlikethem
 Share

Recommended Posts

Let's put a stationary spacecraft some distance (A) from a planet. It will accelerate towards the planet in a straight line. Now lets imagine we have a tunnel through the planet (or maybe the planet is made from dark matter and interacts with our spacecraft only gravitationally). So I think the spacecraft will fly decelerating through the planet the same distance (A'), then stops and goes back and keeps repeating it.

Now lets forget it for a while and imagine our spacecraft in a high elliptical orbit around the same planet. And at apoapsis we burn retrograde, so our periapsis gets closer to the planet's center and we need to make a tunnel for it so it doesn't crash (or it is made of dark matter and just goes through the planet). Now at apoapsis we burn retrograde again and again until we completely stop and our ellipse will look like a straight line. So our spacecraft will go accelerating from apoapsis to the planet's center, then instantly turn around and start decelerating towards apoapsis again. Looks weird and for some reason not the way we had it in our first example.

So obviously there is a mistake somewhere, and I cannot find it. So I decided to ask for help here, on KSP forums, where I learned Orbital Mechanic in the first place.

PktG7HJ.png

Link to comment
Share on other sites

Really good question! Took me a moment of thinking to see where the glitch was. Good job.

The difference has to do with the Shell Theorem. An elliptical orbit only holds true as long as the entire mass of the primary remains inside the orbital path. In the right-hand case, the smaller elliptical orbit would begin to deviate from the elliptical path as soon as it entered the surface.

Link to comment
Share on other sites

@sevenperforce has it. Your example only holds for a point mass. The difference between a tight elliptical orbit with a periapsis of zero, and a vertical oscillation through the centre, is infinitesimal and both are allowed. Think of the zero-periapsis tight elliptical as having both vertical parts of example #1 coincide. If it seems weird, its because point masses are weird, they are not really possible (it is expected that even the singularity at the centre of a black hole has a finite size.) What you have found is a mathematical convergence where there is an infinitesimal area where a question has two solutions. Mathematics does not always represent physical reality, however, as the convergence relies on a physical impossibility.

**edit**

Oh and welcome to the KSP forums!

Edited by p1t1o
Link to comment
Share on other sites

49 minutes ago, unlikethem said:

Thank you people, I really could not sleep wondering what am I missing.

*Googles Shell Theorem - Oh God, heavy math! - Closes the tab in horror*

Shell theorem is simple: if you are inside a sphere, none of the mass at a higher altitude than you has any gravitational effect on you.

Proving the shell theorem is not very simple.

Link to comment
Share on other sites

1 hour ago, sevenperforce said:

Shell theorem is simple: if you are inside a sphere, none of the mass at a higher altitude than you has any gravitational effect on you.

Proving the shell theorem is not very simple.

Should indicate that gravity is weaker in an deep mine.

Note that this is not true in KSP, all bodies in KSP is an shell around an singularity with all the mass. If you manage to clip trough the body of the  Mun and don't blow up doing so you will accelerate towards that singularity.
This will glitch KSP physic a lot and because physic in only measured in 50 millisecond you are likely to exit the Mun at very high speed :)

 

Link to comment
Share on other sites

2 hours ago, magnemoe said:

This will glitch KSP physic a lot and because physic in only measured in 50 millisecond you are likely to exit the Mun at very high speed :)

 

Shouldn't you exit the body's surface with the same speed that you entered at? Your path accelerating towards the core should perfectly mirror your path moving away from it barring any other forces being applied to the craft (such as burning engines).

Link to comment
Share on other sites

1 hour ago, Jaelommiss said:

Shouldn't you exit the body's surface with the same speed that you entered at? Your path accelerating towards the core should perfectly mirror your path moving away from it barring any other forces being applied to the craft (such as burning engines).

Realistic yes, however because time is quantified in 50 ms intervals (KSP Planck time) and you are accelerated very rapidly the last meters before the singularity it will give you an very high speed so you will be many kilometers away from the singularity at next tick, here the gravity is far lower so you keep most of the last speed buff. 

Link to comment
Share on other sites

46 minutes ago, Cheif Operations Director said:

I think you may go back in forth in the first example and slowly loose energy until your in the center. That's just my two cents I'm not a physicist. 

So long as the only force acting is the gravity between the two bodies then no, it won't degrade.

Of course in the real world there'd be other things to worry about like the fact that space (especially near a massive body) isn't a pure vacuum, so it would.

Link to comment
Share on other sites

14 hours ago, sevenperforce said:

Shell theorem is simple: if you are inside a sphere, none of the mass at a higher altitude than you has any gravitational effect on you.

Proving the shell theorem is not very simple.

Proving the shell theorem using calculus is actually pretty simple. Proving it using algebra is insane.

Newton devised it using calculus, but because he was trying to keep calculus secret, he re-derived it using only algebra, which was much, much harder to do and to understand.

Link to comment
Share on other sites

11 hours ago, sevenperforce said:

It is.

It's actually a bit more complicated than that. 

1280px-EarthGravityPREM.svg.png 

Quote

Earth's gravity according to the Preliminary Reference Earth Model (PREM).[13] Two models for a spherically symmetric Earth are included for comparison. The dark green straight line is for a constant density equal to the Earth's average density. The light green curved line is for a density that decreases linearly from center to surface. The density at the center is the same as in the PREM, but the surface density is chosen so that the mass of the sphere equals the mass of the real Earth.

 

Link to comment
Share on other sites

Just now, sh1pman said:

It's actually a bit more complicated than that. 

1280px-EarthGravityPREM.svg.png 

 

Very good catch; I hadn't thought of that.

Explanation: as you descend into a deep mine, you have more mass above you, reducing the total amount of stuff attracting you, but you are closer to the bulk of the mass in the Earth's core. Gravity is inverse squared, so as long as the bulk of the core remains beneath you, you pick up more gravity by being closer to the core than you lose from being beneath portions of the crust or mantle.

Link to comment
Share on other sites

8 hours ago, 5thHorseman said:

So long as the only force acting is the gravity between the two bodies then no, it won't degrade.

Of course in the real world there'd be other things to worry about like the fact that space (especially near a massive body) isn't a pure vacuum, so it would.

That actually depends on the exact configuration. In some configurations no energy is lost but in others energy is lost as gravitational waves. 

Two objects orbiting each other will radiate, but generally this is such a small amount of gravitational wave emission that it doesn’t really matter.

Since two objects orbiting their common center of mass give off gravitational waves and lose energy, and the OP’s system is essentially a system of this type, then it stands to reason that the orbit would eventually decay. Of course that would take an immense amount of time, but nonetheless it would.

Link to comment
Share on other sites

3 hours ago, Bill Phil said:

That actually depends on the exact configuration. In some configurations no energy is lost but in others energy is lost as gravitational waves. 

Two objects orbiting each other will radiate, but generally this is such a small amount of gravitational wave emission that it doesn’t really matter.

Since two objects orbiting their common center of mass give off gravitational waves and lose energy, and the OP’s system is essentially a system of this type, then it stands to reason that the orbit would eventually decay. Of course that would take an immense amount of time, but nonetheless it would.

I was thinking about "frame dragging", is that effect at play in the creation of gravity waves?

Link to comment
Share on other sites

21 minutes ago, p1t1o said:

I was thinking about "frame dragging", is that effect at play in the creation of gravity waves?

That generally has to do with rotatin objects in general relativity, and is somewhat related, I suppose. I’m no expert but in this context I don’t think its relevant.

Link to comment
Share on other sites

8 minutes ago, Bill Phil said:

That generally has to do with rotatin objects in general relativity, and is somewhat related, I suppose. I’m no expert but in this context I don’t think its relevant.

I figure if a rotating mass can frame-drag the space around it, an object moving in a straight line would do so as well, ergo any mass passing any other mass, exhibits a "drag" on the other.

What it boils down to is that gravity is not a zero-sum effect, more energy is extracted from the object than is gained by the object on the infall.

Meaning that no orbit can be perpetual under any circumstances.

I think.

Link to comment
Share on other sites

1 hour ago, p1t1o said:
1 hour ago, Bill Phil said:

That generally has to do with rotatin objects in general relativity, and is somewhat related, I suppose. I’m no expert but in this context I don’t think its relevant.

I figure if a rotating mass can frame-drag the space around it, an object moving in a straight line would do so as well, ergo any mass passing any other mass, exhibits a "drag" on the other.

What it boils down to is that gravity is not a zero-sum effect, more energy is extracted from the object than is gained by the object on the infall.

Meaning that no orbit can be perpetual under any circumstances.

I think.

An object in an infinite oscillation through the center of a primary IS in an orbit, it's just a very weird orbit because it deviates from Kepler's laws thanks to the shell theorem. Two objects in circular orbit around each other induce frame-dragging; that is how two orbiting black holes will eventually collide. Thus, any orbit, even that weird oscillating orbit, involves frame-dragging and loses specific orbital energy over time.

That being said, the timeframe for this to cause significant decay is too large to do anything.

Link to comment
Share on other sites

i think you would need to treat the single mass as a pair of masses, like 2 smaller denser planets (but together being the same mass as the original planet) in a binary orbit. in that case your orbit would intersect one of the lagrange points between the two objects. that should throw things off. maybe. idk.

anyway thats a gross oversimplification. the division of mass would be along the orbiral trajectory, and the resulting planet sections would be non-spherical gravity sources and how close you actually pass the lagrange point between them would factor in. there is a reason we simplify down to point masses. you might be able to treat the planetary sections as point masses however part of one of those bodies would be concave and that might factor in as well.  you aren't really being pulled toward the planet but to all the particles with mass which the planet is made of so its a big n-body problem with n being pretty damn huge. 

Edited by Nuke
Link to comment
Share on other sites

On 3/13/2019 at 1:22 PM, sevenperforce said:

Thus, any orbit, even that weird oscillating orbit, involves frame-dragging and loses specific orbital energy over time.

Curious: the orbital energy that is lost, what happens to it?  Turned into another kind of energy?  I read the wiki on frame dragging but it's over (or is it orbiting) my head.

https://en.wikipedia.org/wiki/Frame-dragging 

Link to comment
Share on other sites

15 hours ago, justidutch said:

Curious: the orbital energy that is lost, what happens to it?  Turned into another kind of energy?  I read the wiki on frame dragging but it's over (or is it orbiting) my head.

https://en.wikipedia.org/wiki/Frame-dragging 

It is emitted as graviton radiation...energetic ripples in the fabric of spacetime that propagate outward forever.

12 hours ago, Xd the great said:

Drag from the few atoms of particles, solar pressure...

That happens, yes, but that's not related to frame-dragging.

It would be interesting to evaluate which has a greater effect on orbital decay: general-relativistic frame-dragging, or local drag. A Fermi estimation says local drag but I have no way of knowing without sitting down and doing the math.

Link to comment
Share on other sites

This thread is quite old. Please consider starting a new thread rather than reviving this one.

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.
Note: Your post will require moderator approval before it will be visible.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

 Share

×
×
  • Create New...