Re-entry

[MeteorWayne]

It still doesn't help until you are well beyond the moon (since the moon is in orbit about the earth.) If you get to the moon's distance and stop accelerating, you'll come straight back down and hit the atmosphere at 25,000 mph. That's an ouchie <img src="/images/icons/smile.gif" /><br /><br />I'm not sure what the actual edge of the earth's gravitation is, but it's more than halfway to Venus, much more than halfway to mars, etc. In the moon's direction, of course it's 90% (?) of the way to the moon. Of course, subject to all the other gravitational influences in the solar system.<br /><br /> Whereas you can orbit (stay up there) once you get just above the atmosphere if you're going fast enough around. <div class="Discussion_UserSignature"> <p><font color="#000080"><em><font color="#000000">But the Krell forgot one thing John. Monsters. Monsters from the Id.</font></em> </font></p><p><font color="#000080">I really, really, really, really miss the "first unread post" function</font><font color="#000080"> </font></p> </div>

 

[mrmorris]

<font color="yellow">"If you get to the moon's distance and stop accelerating, you'll come straight back down ..."</font><br /><br />Your terminology is a bit off unless I'm missing a tengent. The problem was that there never <b>was</b> any acceleration -- only constant velocity. If we assume there was acceleration, then whether you come back down depends on exactly how much velocity you gained before ending the acceleration. If you substitute 'moving' for 'accelerating'... then maybe. There may be other vectors involved that prevent the fall all the way back... depending. If the mysterious accelerationless motion were to stop at one of the Earth-moon LaGrange points for example...<br /><br /><font color="yellow">"I'm not sure what the actual edge of the earth's gravitation is, but it's more than halfway to Venus..."</font><br /><br />Given the bizarre method of motion (assuming we're still talking about the 1MPH constant velocity here), you'll get into some other weirdnesses long before getting that far from Earth that would trump the itsy-bitsy remnants of Earth's gravitational field. For one -- the solar orbital velocity inherited from the launchpad (i.e. Earth) will be making for a curving path. The physics are bizarre, sine this isn't standard orbital mechanics. My <b>guess</b> would be that if the craft is launched *toward* the sun, the reduced circumference of the orbit would mean that the path of the craft would curve in the prograde direction of Earth's orbit, whereas movement away from the sun would curve retrograde. By the time the craft were 'halfway to Venus' (presuming by that you mean halfway between Earth's orbital path and Venus' -- rather than halfway to the planet itself) the craft and Earth would no longer be on a straight-line path from the sun. Earth might well be on the opposite side of the sun, and its gravity would be negligible when compared to nearer objects.<br /><br /><font color="yellow">"...once you get just above</font>

 

[nyarlathotep]

<font color="yellow">In fact, carrying the dead weight of propellant just for SSTO vehicles such as the Delta Clipper is an engineering problem still beyond the best and brightest engineers the world has to offer.</font><br /><br />That's why the clipper was to carry most of it's ascent propellent in drop tanks.

 

[nyarlathotep]

<font color="yellow">If it were possible to 'launch' something to orbital altitudes in the fashion you describe (100km to ~32000km), once it was *at* the intended altitude, if the mysterious force providing constant velocity were to cease -- the craft would simply fall back to earth from gravitational acceleration. It would not have the orbital velocity to counteract the pull of the Earth.</font><br /><br />If we were to assume that you continued 1mph acceleration out of earths gravity well, you're going to get to the escape altitude for that velocity. Good luck with fitting that much fuel in your rocket though, even with ludicrous amounts of staging. <br /><br />A purple star for anyone whos calculator has the required precision:<br />Assuming only gravity losses, a constant isp of 430s, and a trajectory straight up, what mass fraction will be required for a single vehicle to reach escape altitude from sea level?

 

[qso1]

I do not recall seeing any version of the Delta Clipper with drop tanks. That would make it not much different than the shuttle which has one big drop tank. The SSTO orbital version of the clipper was the DC-Y which was a full scale version of the clipper Graham (DC-X) that actually flew. Do you have a link to the drop tank Delta Clipper? <div class="Discussion_UserSignature"> <p><strong>My borrowed quote for the time being:</strong></p><p><em>There are three kinds of people in life. Those who make it happen, those who watch it happen...and those who do not know what happened.</em></p> </div>

 

[nyarlathotep]

<font color="yellow">I do not recall seeing any version of the Delta Clipper with drop tanks.</font><br /><br />They weren't in the pretty viewgraphs, but the actual engineers made it pretty clear that there was SFA chance of building the thing with sufficient mass fraction to carry any significant payload without putting the LH2 in drop tanks.<br /><br />They had the right idea, and with a couple more years of development I think they could have made the system work. Tankage is cheap, it's the engines and operations costs that kill you.

 

[qso1]

That gets right back to my comment about the enginbeering challenges being beyond our best and brightest. Delta Clipper operational and Venture Star operational vehicles are simply still beyond us. <div class="Discussion_UserSignature"> <p><strong>My borrowed quote for the time being:</strong></p><p><em>There are three kinds of people in life. Those who make it happen, those who watch it happen...and those who do not know what happened.</em></p> </div>

 

[mrmorris]

<font color="yellow">"If we were to assume that you continued 1mph acceleration out of earths gravity well, you're going to get to the escape altitude for that velocity. "</font><br /><br />One more time... 1MPH <b>is not an acceleration</b>! 1MPH is a velocity. You cannot <i>achieve</i> escape velocity when the figure supplied *is* a constant (and ridiculously low) velocity.

 

[qso1]

There is one possible exception. If you could go 1 mph straight up from earth...how far could you go before your effectively out of earths gravity well. That is, go 1 mph for 20 million miles. Earths gravity should be pretty much overtaken by solar gravity at that point. But I have never seen any data on this.<br /><br />To illustrate further. It has generally been publicized in the press that the Pioneer and Voyager outer planet craft have escaped the Suns influence. Yet the velocity of the fastest one on a Jupiter flypast was 107,000 mph. Escape velocity from the Sun is something like 1.3 million mph. So by that reckoning, did the probes really escape the suns influence? Or did the probes just go into a deep hyperbolic comet like orbit? Or are they overtaken by galactic gravitational influence once they get so far from the Sun they are no longer effectively affected by solar gravity?<br /><br />And do you know of any references that verify this? <div class="Discussion_UserSignature"> <p><strong>My borrowed quote for the time being:</strong></p><p><em>There are three kinds of people in life. Those who make it happen, those who watch it happen...and those who do not know what happened.</em></p> </div>

 

[Boris_Badenov]

<font color="yellow"> Escape velocity from the Sun is something like 1.3 million mph. </font><br /><br /> That's the escape velocity from the surface of the Sun.<br /><br />However, escape velocity from the solar system from the orbit of<br />the earth (not surface of the sun) is much less because r (in<br />this case earth's orbit) is about 200 times greater than the<br />sun's radius. This reduces the escape velocity from the solar<br />system to about 4 times that of earth's escape velocity, or<br />about 94,000 miles an hour.<br /><br />http://www.virtuallystrange.net/ufo/updates/2006/oct/m29-006.shtml<br /> <div class="Discussion_UserSignature"> <font color="#993300"><span class="body"><font size="2" color="#3366ff"><div align="center">. </div><div align="center">Never roll in the mud with a pig. You'll both get dirty & the pig likes it.</div></font></span></font> </div>

 

[qso1]

Thanks...good point.<br /><br />Now it still means that if you could go 1 mph straight up from earth...you would reach a point where earth escape velocity is much less. The example being what is earth escape velocity at 20 million miles from earth?<br /><br />Having said that, there is no current practical way to sustain 1 mph against earths gravitational pull without massive amounts of propellant. <div class="Discussion_UserSignature"> <p><strong>My borrowed quote for the time being:</strong></p><p><em>There are three kinds of people in life. Those who make it happen, those who watch it happen...and those who do not know what happened.</em></p> </div>

 

[willpittenger]

Aggh, where is a 24th century shuttlecraft when you need one. We could just go out there and try it. <div class="Discussion_UserSignature"> <hr style="margin-top:0.5em;margin-bottom:0.5em" />Will Pittenger<hr style="margin-top:0.5em;margin-bottom:0.5em" />Add this user box to your Wikipedia User Page to show your support for the SDC forums: <div style="margin-left:1em">{{User:Will Pittenger/User Boxes/Space.com Account}}</div> </div>

 

[willpittenger]

Then it wouldn't be SSTO, would it? <div class="Discussion_UserSignature"> <hr style="margin-top:0.5em;margin-bottom:0.5em" />Will Pittenger<hr style="margin-top:0.5em;margin-bottom:0.5em" />Add this user box to your Wikipedia User Page to show your support for the SDC forums: <div style="margin-left:1em">{{User:Will Pittenger/User Boxes/Space.com Account}}</div> </div>

 

[spacester]

From my archives, Nov 2001:<br />***<br />Escape Velocity is the speed you need to have when departing directly away from a massive body (starting at zero velocity from a particular altitude above the center of mass of that body) so that, in the absence of other forces (such as drag or other nearby massive bodies) as you travel, you will steadily give up your kinetic energy (velocity) in exchange for steadily gaining potential energy due to getting further from that body, so that you end up traveling at zero velocity relative to the body, but yet you will not fall back to that body. Such a flight cannot actually take place in our solar system, it is strictly a theoretical concept.<br /><br />Because the point you arrive at where you have completely escaped the body's influence is infinitely far away, the number given by "escape velocity" is strictly a theoretical value, which is one reason why it can be confusing. If you start off with even a little excess velocity, you will truly escape the gravitational field of the body. If you start off with just a bit less, you will eventually fall back to the body - unless you change the velocity a bit with another burn of your rocket engine, which is how you achieve orbit.<br /><br />Escape velocity is given by:<br />Vesc = (2*G*M/R)^1/2<br />where<br />G = Universal Gravitational Constant = 6.673 x 10^-11 N-m^2 / kg^2<br />M = Mass of the planet<br />For the moon:<br />M = 7.35 x 10^22 kg<br />R = 1738 km<br />Vesc = [2* (6.673 x 10^-11 N-m^2 / kg^2)* 7.35 x 10^22 kg / (1738000 m)]^1/2<br />(note that N/kg = m/s^2)<br />Vesc = 2376 m/s<br />escape velocity = 2376 m/s<br /><br />omega_quantum, on your last post of 08 November, the answer is yes, the change of gravity varies as the inverse square (1/D^2), and tuvas is mostly correct with the answer, but to clarify: If D is the separation distance (usually R is used instead of D), then tuva's equation is correct, where Ri is the average radius of the planet and h is the altitude above that radius <div class="Discussion_UserSignature"> </div>

 

[willpittenger]

I agree someone is confusing acceleration with velocity. However, you forget that the escape velocity falls as you climb the gravity well. If you have a constant velocity, you have found a way to ignore gravity. <div class="Discussion_UserSignature"> <hr style="margin-top:0.5em;margin-bottom:0.5em" />Will Pittenger<hr style="margin-top:0.5em;margin-bottom:0.5em" />Add this user box to your Wikipedia User Page to show your support for the SDC forums: <div style="margin-left:1em">{{User:Will Pittenger/User Boxes/Space.com Account}}</div> </div>

 

[scottb50]

SSTO would be great, but with what we have to work with it is impractical. Throw away tanks would work, but that really defeats the purpose, while the cost is "insignificant" it really isn't. What we need is a vehicle that can takeoff, return, refuel and takeoff over and over again.<br /><br />The only way to meet this, with current technology is TSTO, precisely what Rutan has proven. Then it becomes a matter of vertical launch or runway launch, and if you want to get to orbit vertical launch is the only alternative. <div class="Discussion_UserSignature"> </div>

 

[qso1]

Thanks for the info. That Embry Riddle data is especially useful where 140 Rearth is concerned. <div class="Discussion_UserSignature"> <p><strong>My borrowed quote for the time being:</strong></p><p><em>There are three kinds of people in life. Those who make it happen, those who watch it happen...and those who do not know what happened.</em></p> </div>

 

[qso1]

willpittenger:<br />you have found a way to ignore gravity.<br /><br />Me:<br />In essence...antigravity. <div class="Discussion_UserSignature"> <p><strong>My borrowed quote for the time being:</strong></p><p><em>There are three kinds of people in life. Those who make it happen, those who watch it happen...and those who do not know what happened.</em></p> </div>

 

[willpittenger]

<blockquote><font class="small">In reply to:</font><hr /><p>antigravity<p><hr /></p></p></blockquote><br />I have that now (as do you). I don't recall falling through my chair. <div class="Discussion_UserSignature"> <hr style="margin-top:0.5em;margin-bottom:0.5em" />Will Pittenger<hr style="margin-top:0.5em;margin-bottom:0.5em" />Add this user box to your Wikipedia User Page to show your support for the SDC forums: <div style="margin-left:1em">{{User:Will Pittenger/User Boxes/Space.com Account}}</div> </div>

 

[willpittenger]

I wasn't argueing for or against SSTO. Rather, I was noting that Delta Clipper variants were billed as SSTO. <div class="Discussion_UserSignature"> <hr style="margin-top:0.5em;margin-bottom:0.5em" />Will Pittenger<hr style="margin-top:0.5em;margin-bottom:0.5em" />Add this user box to your Wikipedia User Page to show your support for the SDC forums: <div style="margin-left:1em">{{User:Will Pittenger/User Boxes/Space.com Account}}</div> </div>

 

[qso1]

Antigravity for me works only after a bottle of Jack, lol. <div class="Discussion_UserSignature"> <p><strong>My borrowed quote for the time being:</strong></p><p><em>There are three kinds of people in life. Those who make it happen, those who watch it happen...and those who do not know what happened.</em></p> </div>

 

[scottb50]

Billed as and being capable of are two different things. When you look at the mass fraction of a launch vehicle then look at the amount of propellant needed for a vertical landing it just doesn't add up. Maybe with a fuel station in orbit it would work.<br /><br />That's why TSTO works, the payload can be higher and a powered flyback orbital vehicle would be much easier to design. <div class="Discussion_UserSignature"> </div>

 

[qso1]

I'm all for a fully reusable TSTO. <div class="Discussion_UserSignature"> <p><strong>My borrowed quote for the time being:</strong></p><p><em>There are three kinds of people in life. Those who make it happen, those who watch it happen...and those who do not know what happened.</em></p> </div>

 

[willpittenger]

My only problem with some TSTO systems is docking the parts. Most of the time you either need a big crane system like the STS mate/demate device or to dock in midair. The former requires equipment on hand and the later is dangerous. <div class="Discussion_UserSignature"> <hr style="margin-top:0.5em;margin-bottom:0.5em" />Will Pittenger<hr style="margin-top:0.5em;margin-bottom:0.5em" />Add this user box to your Wikipedia User Page to show your support for the SDC forums: <div style="margin-left:1em">{{User:Will Pittenger/User Boxes/Space.com Account}}</div> </div>

 

[qso1]

Mating on the ground isn't really that much of a problem if the system is designed from the outset to accomodate it. Until we can do SSTO, TSTO is probably the best we can possibly do.<br /><br />If the Rutan system were orbital, it would be a TSTO and a well designed one from a mating standpoint at that and eventually, a Rutan orbital system is supposed to become operational. <div class="Discussion_UserSignature"> <p><strong>My borrowed quote for the time being:</strong></p><p><em>There are three kinds of people in life. Those who make it happen, those who watch it happen...and those who do not know what happened.</em></p> </div>