How strong is an orbit?

[SkylabI]

So here's a question. If you could put the SpaceX Dragon on a non-orbital trajectory where it would re-enter but was somehow timed to fly past the ISS in a way the ISS could grapple it with it's RMS, what would happen?

 

[Hielor]

So here's a question. If you could put the SpaceX Dragon on a non-orbital trajectory where it would re-enter but was somehow timed to fly past the ISS in a way the ISS could grapple it with it's RMS, what would happen?

The RMS would be torn off, probably.

Relative velocity is what you'd be looking at.

 

[Mojave]

Yes, the velocity would overpower the RMS and rip it off the module because of the extreme difference of speed between the two vessels.

(This, however, is impossible to even try/do)

 

[Zatnikitelman]

If the Dragon was going slow enough to be captured by the SSRMS, it'd be in a similar orbit as the ISS, so either it'd not be reentering, or the ISS would be in your scenario.

 

[Mojave]

If the Dragon was going slow enough to be captured by the SSRMS, it'd be in a similar orbit as the ISS, so either it'd not be reentering, or the ISS would be in your scenario.

That's a good point. So that would mean that the dragon was coming in for docking in the first place?

 

[C3PO]

If ISS is in a circular 400 km orbit and Dragon was in a 60 x 400 km orbit, the DV at apogee would be 99.44 m/s.

We'll have to wait for the ninja version of SSRMS. :lol:

 

[SkylabI]

If the Dragon was going slow enough to be captured by the SSRMS, it'd be in a similar orbit as the ISS, so either it'd not be reentering, or the ISS would be in your scenario.

Well, I'm not familiar with all the maths, but conceivably couldn't the sub-orbital Dragon's apogee just happen to perfect match up with the ISS in a place to rendezvous? I was under the impression that you slowed down at the apogee in a ballistic orbit like that right before you started back down. I mean, isn't there a point where you'd hit a velocity of zero before accelerating back down?

I was just wondering if there was a hypothetical scenario where you could launch something up in a very steep arc and time the apogee where -- providing you could have an RMS arm that was strong enough -- kind of "snag" the object and take it along with you.

I wonder if it would be possible to try this in Orbiter and see what happened. Alas, I am not capable enough at these kinds of maneuvers.

 

[Hielor]

Well, I'm not familiar with all the maths, but conceivably couldn't the sub-orbital Dragon's apogee just happen to perfect match up with the ISS in a place to rendezvous? I was under the impression that you slowed down at the apogee in a ballistic orbit like that right before you started back down. I mean, isn't there a point where you'd hit a velocity of zero before accelerating back down?

I was just wondering if there was a hypothetical scenario where you could launch something up in a very steep arc and time the apogee where -- providing you could have an RMS arm that was strong enough -- kind of "snag" the object and take it along with you.

I wonder if it would be possible to try this in Orbiter and see what happened. Alas, I am not capable enough at these kinds of maneuvers.

Your vertical velocity will be 0 momentarily, yes, but your horizontal velocity most certainly won't be. The ISS will have a higher horizontal velocity (since it's in orbit and you're not), so your relative velocity will be nonzero.

 

[Ripley]

Well, I'm not familiar with all the maths...
...Alas, I am not capable enough at these kinds of maneuvers.

These 2 sentences (and your original question as well) basically sum it up.

First you have to try an ISS approach (and succeed in the subsequent docking).
Particurarly, you'll need to master the "orbit synchronization", then you'll understand the rest, and looking at your original question you'll probably facepalm and go "...doh!" (no offence intended).

Don't worry, we were all newbies one day.

 

[C3PO]

Well, I'm not familiar with all the maths, but conceivably couldn't the sub-orbital Dragon's apogee just happen to perfect match up with the ISS in a place to rendezvous? I was under the impression that you slowed down at the apogee in a ballistic orbit like that right before you started back down. I mean, isn't there a point where you'd hit a velocity of zero before accelerating back down?

Only if tangential velocity is zero. The problem is that when you hit apogee, you hit the lowest Vtan too. The ISS zooms by at ~100 m/s relative speed because it's in a circular orbit. If you speed up to match the speed, you raise the perigee out of the atmosphere.

 

[Jarvitä]

Well, I'm not familiar with all the maths, but conceivably couldn't the sub-orbital Dragon's apogee just happen to perfect match up with the ISS in a place to rendezvous? I was under the impression that you slowed down at the apogee in a ballistic orbit like that right before you started back down. I mean, isn't there a point where you'd hit a velocity of zero before accelerating back down?

I was just wondering if there was a hypothetical scenario where you could launch something up in a very steep arc and time the apogee where -- providing you could have an RMS arm that was strong enough -- kind of "snag" the object and take it along with you.

I wonder if it would be possible to try this in Orbiter and see what happened. Alas, I am not capable enough at these kinds of maneuvers.

If you launched the object straight up (90 degrees vertical) to intercept the ISS at apogee, its velocity at apogee would indeed be zero. But it wouldn't "snag", it would collide with the ISS at ~7.8 kilometres per second. Calculating the energies involved and picturing the consequences is left as en exercise to the reader.

 

[Urwumpe]

But it wouldn't "snag", it would collide with the ISS at ~7.8 kilometres per second. Calculating the energies involved and picturing the consequences is left as en exercise to the reader.

Simple rules of thumbs for the task:

A collision of 2400 m/s means the energy equivalent of each kg of spacecraft exploding like TNT.

Double velocity means four times the energy and the TNT equivalent.

 

[Jarvitä]

Simple rules of thumbs for the task:

A collision of 2400 m/s means the energy equivalent of each kg of spacecraft exploding like TNT.

Double velocity means four times the energy and the TNT equivalent.

Actually, I got 2892.75 m/s.

[MATH]\frac{1 kg\times \left(x \frac{m}{s}\right)^{2}}{2}=4.184MJ[/MATH]

 

[Urwumpe]

Well, then the rule was 2800 m/s - at least it was not that high compared to orbital velocities.

 

[Screamer7]

I think another factor to be dealt with is the relative mass of the two vehicles.
Let us assume for a moment we broke the law of orbital mechanics momentarily, like a stopwatch,and is able to dock.
The dragon mass is much smaller than that of the ISS.
When we start the "stopwatch" again, the ISS will absorb the relative difference in speed of about 400 to 500 DV
I by no means are a math guru, so I may be wrong.

 

[martins]

I think another factor to be dealt with is the relative mass of the two vehicles.
Let us assume for a moment we broke the law of orbital mechanics momentarily, like a stopwatch,and is able to dock.
The dragon mass is much smaller than that of the ISS.
When we start the "stopwatch" again, the ISS will absorb the relative difference in speed of about 400 to 500 DV
I by no means are a math guru, so I may be wrong.

You don't need to break physics. Just imagine you dock the capsule by lassoing it with a long elastic rubber band.
Either way - Conservation of momentum. The docked assembly will have a different orbit. Restoring the orbit will require the same amount of energy as synchronising the capsule with the ISS in the first place. You can't cheat physics.

 

[Loru]

And to sum it up read first "Go play in space"

You'll catch basics of orbital mechanics with it.

After that check this tutorials: http://www.orbiter-forum.com/tutorials.php