Falling to exit velocity

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PJay_A

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<p>QUESTION:</p><p>Let's say we build a tower one mile tall. The Dubai Tower under construction will be 1/2 mile tall when completed, so a mile tall tower is within the realm of probability. Now, let's say we build this tower near the highest point of the Grand Canyon (which I believe is eight miles high). At the bottom of the this location at the Grand Canyon, drill a sloapped tunnel as deep as we can drill down (I believe this would be eight or nine miles down). At the top of the tower, going down a mile to the top of the Grand Canyon, then continuing down eight miles along the sloapping walls of the Grand Canyon to its deepest elevation, and continuing down eight miles through the drilled tunnel, and continuing on an eventual sloaping angle until the path becomes totally flat (not rounded according to the curvature of the earth), continuing underground until the tunnel meets up to a tunnel exit at the horizon point where surface-level is met by earth's curvature, followed by an upwards ramp that ends 90 degrees pointing skyward... That entire path would be enclosed and air-tight, surrounded by electromagnets. A vehicle would fall through the top starting point. Gravity would push it down and the electromagnets would assist in assisting veleocity increase faster than gravity. When it reaches its lowest point underground, the electromagnets would assist in keeping its momemtum as it leaves the tunnel and turns upward. Rockets on the vehicle would keep the vehicle's momentum at exit velocity until LEO is achieved. Since exit velocity had been achieved at ground level without propellent, the amount of fuel needed to maintain that momentum to orbit should be very minimum, thereby drastically reducing cost per pound to get any kind of payload to orbit!</p><p>&nbsp;Doable? </p>
 
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Cygnus_2112

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<p>Not doable.&nbsp;&nbsp; The energy gained by falling would be the same as the energy at the top.&nbsp; All that is happening is trading potential energy for kinetic and then trading it back.&nbsp; Also there still is the atmosphere. &nbsp; </p><p>&nbsp;</p><p>&nbsp;</p>
 
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samkent

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Not doable.&nbsp;&nbsp; The energy gained by falling would be the same as the energy at the top.&nbsp; All that is happening is trading potential energy for kinetic and then trading it back.&nbsp; Also there still is the atmosphere. &nbsp; &nbsp;&nbsp; <br />Posted by Cygnus_2112</DIV></p><p>&nbsp;</p><p>Plus the perigee of the orbit would intersect with the Earth. Slpat right back where you started.<br /></p>
 
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keermalec

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<p>It is an interesting concept in that you are using electromagnetic forces to accelerate a payload to launch velocities. As Samkent pointed out however, the downslope part is not necessary as all the energy gained would simply be lost again on the upslope part. You would achieve the same result if you were simply accelerating your load horizontally on the surface of the earth, at much lower cost. The problem you will face is keeping the launch tube in a vaccum, as the payload must come out somewhere and at that point, if it is travelling at several km/s and "hits" a wall of atmospheric pressure, it will be sort of like hitting a surface of water at several hundred m/s: you can imagine the result.</p><p>If you choose to accelerate your payload without a vacuum, you will face increased trouble with speed, jsut imagine passing the sound barrier on the surface of the Earth, and that is only 300 m/s while you need close to 8'000 m/s for orbital velocity...</p><p>The closest we have&nbsp;ever come to accelerating a load to space on the ground is the HARP gun project.&nbsp;</p><p>&nbsp;</p><p>&nbsp;</p> <div class="Discussion_UserSignature"> <p><em>“An error does not become a mistake until you refuse to correct it.” John F. Kennedy</em></p> </div>
 
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