Is 3g's in outerspace more speed gain than 3g's at launch?

[askold]

I'm confused.<br /><br />Let's take a shutttle launch - you hear that they "experience 3 g's" during the launch. Does this mean they're accelerating (gaining) 64 f/s per second or 96 f/s per second relative to the surface of the Earth that they're leaving behind?<br /><br />I think it's 64 f/s per second beacuse they're experienceing 1g just sitting on the launch pad.<br /><br />Now, let's say they're half way to Mars and they fire their rockets until they feel 3 g's. Now they're gaining 96 f/s per second - right?

 

[Saiph]

well, you caught the minor nuance of incorporating the earth's gravity into your scheme. So I guess the question ends up being about convention.<br /><br />As for wether or not the figure they use is the total acceleration felt by the astronauts, or the total acceleration applied to the astronauts (i.e. incorporating gravity or not) I'm not sure.<br /><br />I suspect it is not. I would guess that the term means they are accelerating upwards at 3g's, after you remove 1g due to earth, and would feel 4gs of acceleration/force. <div class="Discussion_UserSignature"> <p align="center"><font color="#c0c0c0"><br /></font></p><p align="center"><font color="#999999"><em><font size="1">--------</font></em></font><font color="#999999"><em><font size="1">--------</font></em></font><font color="#999999"><em><font size="1">----</font></em></font><font color="#666699">SaiphMOD@gmail.com </font><font color="#999999"><em><font size="1">-------------------</font></em></font></p><p><font color="#999999"><em><font size="1">"This is my Timey Wimey Detector.  Goes "bing" when there's stuff.  It also fries eggs at 30 paces, wether you want it to or not actually.  I've learned to stay away from hens: It's not pretty when they blow" -- </font></em></font><font size="1" color="#999999">The Tenth Doctor, "Blink"</font></p> </div>

 

[igorsboss]

<font color="yellow">they "experience 3 g's" during the launch. Does this mean they're accelerating (gaining) 64 f/s per second or 96 f/s per second relative to the surface of the Earth that they're leaving behind? <br /><br />I think it's 64 f/s per second beacuse they're experienceing 1g just sitting on the launch pad.<br /><br />Now, let's say they're half way to Mars and they fire their rockets until they feel 3 g's. Now they're gaining 96 f/s per second - right</font><br /><br />I think you are correct. I hope I can straighten out your confusion.<br /><br /><i>The difference is the observer, not the rocket.</i> In both cases, 3g's means 96f/s per second relative to an <i>inertial</i> frame of reference. However, if you are sitting in a chair on Earth, you are not in an inertial frame of reference! You are in an <i>accelerated</i> frame of reference, with an acceleration of exactly 1 g.<br /><br />To prove this, I cite the fundamental principle of Einstein's theory of General relativity: that Gravity and acceleration are indistinguishable.<br /><br />Your chair is being accelerated by Gravity in exactly the same way as an identical chair <i>would be</i> accelerated if it were subjected to 1g of acceleration relative to an inertial frame of reference.<br /><br />So, an observer in an inertial frame would see your chair accelerating at 1g, and the rocket accelerating at 3g's. This is why you see the rocket accelerating at 2g's, while the astronauts feel 3g's.<br /><br />I've pretended Earth is not rotating, and that gravity doesn't diminish with distance. The full answer would involve stuff like the inverse square law of gravity, the corriolis effect, and possibly General relativity...<br /><br />(An inertial frame of reference is one in which the observer experiences zero acceleration, resembling frictionless free-fall.)

 

[henryhallam]

This is right, and it's the reason behind "gravity losses".<br /><br />Taken to the extreme, if you have a rocket capable of producing less than 1g of acceleration (initially) then you just sit there wasting propellant and not going anywhere. On the other hand if you burned the same amount of rocket fuel in an engine with the same specific impulse, but a huge amount of thrust so you get a very high acceleration, then you can change your velocity by a great deal.<br />So rockets are most efficient at high accelerations, and they tend to be designed to accelerate as quickly as possible, although a tradeoff must be made with the increased weight of a strong structure that is able to take more "g's" (and of course the payload must be able to deal with it).

 

[CalliArcale]

I believe the G force for a rocket is described relative to a payload inside the rocket. This is because it's the most significant frame of reference with respect to survivability. Even nonliving payloads care about the G-forces; you build the payload to tolerate a certain amount of G forces, and so you definitely care whether or not that is exceeded during the ascent.<br /><br />Rockets are actually <i>not</i> neccesarily designed to accelerate as quickly as possible. Usually they go for a balance between fuel efficiency and acceleration, but on some vehicle (especially manned ones) there is an effort to keep it down to a more comfortable rate. 3 Gs is quite common for manned vehicles; unmanned ones sometimes accelerate faster. <div class="Discussion_UserSignature"> <p> </p><p><font color="#666699"><em>"People assume that time is a strict progression of cause to effect, but actually from a non-linear, non-subjective viewpoint it's more like a big ball of wibbly wobbly . . . timey wimey . . . stuff."</em>  -- The Tenth Doctor, "Blink"</font></p> </div>

 

[askold]

OK, so it's mainly a matter of convention - 3g's means you're accelerating at a rate of 96 f/s/s comapred with how fast you were going before.<br /><br />How much you feel you "weigh", depends on where you're at - if you normally weigh 150lbs and you leaving the Earth, you'll feel like you weigh 600lbs; if you're leaving the moon you'll feel like you're 475lbs, etc.

 

[newtonian]

askold - To understand how important a slow but constant acceleration can be to space travel, you may find interesting researching ion drive.<br /><br />This type of engine accelerates slowly, but steadily, to incredible ultimate speed.<br /><br />I forget the numbers, though!

 

[igorsboss]

IIRC, the force provided by an ion drive is on the same order as the force required to hold up one sheet of ordinary paper.

 

[spacester]

Orbital velocity is a different animal than the velocity you see elsewhere. This is because the energy associated with that velocity is "bound" to the orbit.<br /><br />An object in a non-circular orbit is continuously trading back and forth between Kinetic Energy and Potential Energy. If you fire your rocket engines, you change the orbital energy, and the effect on the actual velocity is complicated.<br /><br />The key is to realize that "velocity" in an orbit is a measure of the energy bound up in the orbit. Use the definition of Kinetic Energy to get a handle on this:<br />KE = E = 1/2 m * v^2<br />or<br />v = sqrt(2*E/m)<br />where E is the total orbital energy and v is the equivalent circular velocity (the velocity in a circular orbit with the same orbital period)<br /><br />Here's the real trick: when you raise an orbit, you add energy, but the velocity decreases! This can only make sense when you consider Potential Energy. (Look up the 'Vis-Viva Equation')<br /><br />When you raise an orbit, you add energy. You actually add twice as much Potential Energy to the bound energy as you reduce the Kinetic Energy by reducing the velocity. So 1 unit of added orbital energy adds 2 units of PE and subtracts 1 unit of KE.<br /><br />OK, I'm gonna quit now before I get too mathematical. Ooops, too late! <img src="/images/icons/laugh.gif" /><br /><br />HTH; did you get an answer to your question yet? <div class="Discussion_UserSignature"> </div>

 

[askold]

I don't mind the math - that's my undergrad degree. I actually understand orbital mechanics pretty well - thanks for the additional info.<br /><br />I think my original question has been answered - it was actually simpler than a lot of the answers! The basic question was - when you're talking about launching from Earth, what does a "3g launch" mean? Does it mean that the vehicle is accelerating such that the astronauts feel they're in a 3g gravitational field (feel 3 times as heavy), or that the vehicle is accelerating at 96 f/s/s realtive to the Earth? <br /><br />I think that the consensus is the latter.

 

[igorsboss]

<font color="yellow">what does a "3g launch" mean? Does it mean that the vehicle is accelerating such that the astronauts feel they're in a 3g gravitational field (feel 3 times as heavy), or that the vehicle is accelerating at 96 f/s/s realtive to the Earth? I think that the consensus is the latter.</font><br /><br />...uh, oh... Don't you mean the former...? Or was that a typo...? Or was I wrong...? <br /><br />3g means:<br />* Astronauts feel they're in a 3g gravitational field.<br /><br />* Astronauts accelerate at 96 fs^-2 relative to an inertial frame of reference. (Not Earth!)<br /><br />* Astronauts <i>appear</i> to accelerate at 64 f/s^2 <i>relative to an observer sitting on a chair on Earth next to the launch site, for a brief time after liftoff.</i>

 

[askold]

No - not I typo - I meant "the latter" (though I could be wrong).<br /><br />Looking at the shuttle's launch profile:<br /><br />http://spaceflightnow.com/station/stage11a/fdf/113ascenttimeline.html<br /><br />I calculate that between T+1:01 and T+2:03, they're accelerating at 46 f/s/s.<br /><br />Then, between T+7:22 (3G Limiting) and T+8:24 (MECO), they're accelerating at 92 f/s/s. So, is that what they mean be a 3G launch?

 

[igorsboss]

<font color="yellow">not I typo</font><img src="/images/icons/smile.gif" /><br /><br /><font color="yellow">shuttle's launch profile</font><br /><br />My answer was overly simplified, to address your overly simplified question.<br /><br />My answer applies to a flat Earth, a vertical accent, and Gravity that does not diminish with distance, all of which are false assumptions! If you try to applly them to a real shuttle launch profile, they won't work.<br /><br />The higher the shuttle goes, the less Earth's gravitational pull will be. The more the shuttle accelerates downrange, the more the shuttle's 3g acceleration vector will include both vertical and horizontal components. Both of these make the answer more complicated.<br /><br />So, that said, how much do we agree on?<br /><br />Do we at least agree that 3g means the astronauts feel 3g, and that means that if the shuttle commander were to drop a pencil in the flight deck, the pencil would appear to the commander to accelerate towards the back wall of the cabin at 96 f/s/s?

 

[newtonian]

igorsboss - Yes - therefore ion drive shows little things mean a lot!<br /><br />As in little but constant acceleration!

 

[askold]

OK, here's what I think we have so far.<br /><br />- We don't know the shuttle's orientation during T+7:22 to T+8:24, but it's probably neither vertical or tangential to the surface of the Earth. Somewhere in between.<br /><br />- During that 62 seconds, acording to the data, the shuttle gained 3,886 mph. That's 92 f/s/s, or about 3g. Since the rockets are pushing directly from behind, they are creating a force vector (3g's) from the front of the shuttle to the back.<br /><br />- In low Earth orbit, the shuttle feels a 1g force down toward the center of the Earth. The force of gravity is not substantuially less than 1g at 200 miles up (certainly not at 80 or 100 miles at MECO - T+8:24).<br /><br />- Let's assume the shuttle is moving in more-or-less a straight line during this minute, so we don't have to account for substantial centrifugal forces.<br /><br />So, if the astronaut drops a pencil, what is the tragectory of that pencil and at what acceleration? I'd say it depends on the oreintation of the shuttle relative to the Earth. For simplicity, let's say it's at 45 degrees relative to the surface of the Earth (45 degrees relative to a vertical line through the center of the Earth or a tangent to the surface). Then you'd have a force vector that's the sum of 1g pointing down and 3g pointing opposite to the direction of travel of the shuttle. The pencil would not hit the back wall directly behind the astronaut - it'll tend to sink toward the floor [assuming the shuttle is belly down - otherwise, the pencil would move toward whatever is "down"].<br /><br />Similary, if we were to "weigh" the astronaut (let's say 150 lbs), their weight would be the sum of 450lbs pointing to the back of the shuttle and 150 lbs poiting toward Earth. How much weight they "feel" and in which direction they feel it would depend on the orientation of the shuttle.<br /><br />I guess my question is not so simple anymore ....