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OmniAtlas

Here is a dumb physics question for you smart people

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You're in a futuristic vehicle flying at a given speed. 

 

You need to dock with the mothership through its trap door on the underside. 

 

You match the speed of the flying mothership so you are under the trap door. 

 

Trap door opens and you use your vertical thrusts to enter into the hanger bay. 

 

Trap door closes. 

 

At what point of the sequence does your horizontal speed become zero? 

 

When is your speed relative to the inside of the mothership? When the trap door closes or when you enter inside the ship? 


Soarbywire - Avionics Engineering

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You match the speed of the flying mothership so you are under the trap door.

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Whenever you are at the same speed, regardless of distance, you have the same relative speed to the inside of the ship.


David Zambrano, CFII, CPL, IGI

I know there's a lot of money in aviation because I put it there. 

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I suppose in this example you'd always give your speed relative to the mothership, giving it relative to anything else is useless to you in your calculations... So even approaching the ship, you'd say you were moving at 5m/s or whatever, but always relative to the ship, not 500m/s relative to earth back in the distance.

 

Regards,

Ró.


Rónán O Cadhain.

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Whenever you are at the same speed, regardless of distance, you have the same relative speed to the inside of the ship.

 

How about you 'hover' inside the ship and then the trap door closes. You are still maintaining your speed relative to the mothership's speed. Will you slam into the end of the ship? 


Soarbywire - Avionics Engineering

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How about you 'hover' inside the ship and then the trap door closes. You are still maintaining your speed relative to the mothership's speed. Will you slam into the end of the ship? 

No you won't slam as your speed all along would have been relative to the ship, not just when the trap door closes or you enter the ship. I understand why you could find that confusing though, it does require quite the bit of logical thinking through...

 

Regards,

Ró.


Rónán O Cadhain.

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I imagine your indicated airspeed would reduce to zero the moment the pitot tube

no longer received the intake of air from the earth's atmosphere.

 

ANAMIV

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You really need to specify, if you are still in the earths atmosphere (RIP ! ), or in outer space.
 

( Much easier in Outer Space ! )

It's all relative ...

"God does not care about our mathematical difficulties. He integrates empirically."

Albert Einstein ( 14 March 1879 – 18 April 1955)

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Lets say the mothership is travelling at 100 knots and your vehicle enters the hanger at a hundred knots.

 

When the trap door closes you are still going at a hundred knots but now relative to the inside of the ship which is at zero knots. So wont you have to cut your throttles and speed to zero the minute the trap door closes to 'land' inside?


Soarbywire - Avionics Engineering

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That is like jumping in an aircraft and flying to the back at 500 mph because you are not touching the aircraft anymore.

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Lets say the mothership is travelling at 100 knots and your vehicle enters the hanger at a hundred knots.

 

When the trap door closes you are still going at a hundred knots but now relative to the inside of the ship which is at zero knots. So wont you have to cut your throttles and speed to zero the minute the trap door closes to 'land' inside?

Okay, but you're not getting the relativity bit here. The Mothership is travelling 100kts, but relative to what?

 

So let's just say it's relative to earth, and your ship is also travelling 100kts relative to earth, so when they dock, both will still be going 100nts relative to earth, but your ship will be going 0kts relative to the mothership. Making any sense?

 

That is like jumping in an aircraft and flying to the back at 500 mph because you are not touching the aircraft anymore.

Perfect example.

 

Regards,

Ró.


Rónán O Cadhain.

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You really need to specify, if you are still in the earths atmosphere (RIP ! ), or in outer

 

"God does not care about our mathematical difficulties. He integrates empirically."

 

Albert Einstein ( 14 March 1879 – 18 April 1955)

Earth!

That is like jumping in an aircraft and flying to the back at 500 mph because you are not touching the aircraft anymore.

So in this example when the trap door is open but you are inside the mothership, are you still relative to earth or the ship?


Soarbywire - Avionics Engineering

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Actually, I have "sim" experience with this. Anybody here a fan of the space-sim "Orbiter"?

 

Well, it turns out that if you're, say, 5 meters below the mothership and your relative speed is 0 m/s then that (obviously) going to have (basically) identical orbital parameters. If you thrust "up" into the mothership you've changed your orbital parameters. You're generally going to continue in that direction until you reach your new apoasis (highest point in orbit) and begin the other part of the ellipse and start heading back downhill towards your periapsis. If you're in a perfectly circular orbit to begin with (which really never stays circular due to orbital perturbations/solar wind/etc), then by thrusting in any direction you will change your orbital into an ellipse by some degree. It will not change it very much, but it is without a doubt changing the shape of your orbit.

 

If (for example's sake) you entered in infinitely large room, then you would not always just drift towards the ceiling. At some point you'd start drifting down and around the inside of the ship. Unless you thrust "down" and null your relative velocity again, you will hit the ceiling.

 

This is basically why astronauts strap themselves down to sleep. Over time they will drift around the cabin. Which is not good since the walls/ceilings are generally filled with switches and buttons.

 

EDIT: Docking is a different story. Even during the docking process the ships are moving towards/away from each other. It isn't until they make contact, connect and attach to each other that they have zero relative velocity. If you even watch videos of spacecraft docking (older Apollo videos of the CSM docking to the LM show this the best) you can see a definate "rocking" of the spacecraft as they connect and the docking ship gets "grabbed" by the larger, more massive target vehicle.

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As you entered the hangar bay, you would no longer be subject to the resistance (and aerodynamic lift) imparted by a mass of air moving towards you at your initial speed. The air in the hangar bay would be moving at zero horizontal velocity relative to your ship, so you would lose all of that lift!!!

 

EDIT: This is assuming that you are in the atmosphere. If you are in space, then you would not have any forward thrust at all once you are directly underneath the "trap door", otherwise you would accelerate beyond it. All you would need to do is use a small amount of "upward" thrust to get into the hangar bay, and then a small amount of "downward" thrust to land inside the mothership once the "trap door" is closed.


Christopher Low

UK2000 Beta Tester

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Okay, but you're not getting the relativity bit here. The Mothership is travelling 100kts, but relative to what?

 

So let's just say it's relative to earth, and your ship is also travelling 100kts relative to earth, so when they dock, both will still be going 100nts relative to earth, but your ship will be going 0kts relative to the mothership. Making any sense?

 

 

Yes that makes sense. So if the trap door is open, and you are inside the ship hovering, you are still relative to earth and going 100 knots?

 

The minute the trap door closes you become relative to the inside of the ship and you should cease all speed?


Soarbywire - Avionics Engineering

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