From: alain245@videotron.ca   
      
   On Aug/15/2020 at 17:04, JF Mezei wrote :   
   >   
   >   
   > When on the ground, I understand that launching from equator due east   
   > gets you started with 1,666km/h delta V due to rotation of earth.   
   > (circumference of earth at equator / 24 hour).   
   >   
   >   
   > At 28", you get 1471km delta-V at start, a difference of 195kmh versus   
   > equator.   
   >   
   > at 45°, you get 1178kmh at start, a difference of 488kmh vs equator launch.   
   >   
   >   
   > Considering that in the end, you want to have something akin to   
   > 25,000kmh speed, is the 200 to 500kmh difference in what you get from   
   > Earth at launch really significant?   
      
   It isn't crucial. The Russians launch from a high latitude and it isn't   
   a big problem. But the difference is significant. Remember that the   
   energy required for a speed increase isn't proportional to the speed   
   increase but to the square of the speed increase. If you have a rocket   
   that can bring you to 24,500 km/h, you don't just need a little extra   
   fuel that can give you the additional 500 km/h to reach the desired   
   25,000 km/h, because that extra fuel that would give you the 500 km/h   
   has weight and you need to bring it to 24,500 km/h if you want to use it   
   for that extra push.   
      
   > Here is the other thing I don't understand.   
   >   
   > Much has been said about how the ISS being at 51° forced a lot more   
   > shuttle launches than if it had been at 28°. I still do not understand this.   
   >   
   > At equator, ascending node for 51°, that 25,000kmh heading north east   
   > would have both a north and and east speed compoennt that is way higher   
   > than what was imparted during launch at 28°.   
   >   
   > So since launching to 51° results in both horizontal and vertical   
   > vectors being greater than what was imparted by the ppanet at launch,   
   > why would there be losses?   
   >   
   > I can understand launching to equatorial orbit from 28° where you have   
   > to spend fuel to kill the vertical component that makes you travel from   
   > +28 to -28.   
   >   
   > I also undertand that launching to polar orbit means that you need to   
   > spend fuel to kill all the horizontal speed given to you at launch, and   
   > have just the little vertical boost when launching from 28".   
   >   
   >   
   > But when you launch to an inclination greater than launch one but where   
   > both horizontal and vertical vectors are greater than what Earth gives   
   > you at launch, why would it be considered inefficient and require more   
   > fuel?   
      
   There is a lost. It isn't all that big but it is significant. If you   
   want to reach a speed vector (x,y) the delta-v needed is given by square   
   root(x^2 + y^2) [Pythagorean theorem]. If you do it by first   
   accelerating to the speed vector (x,0) then from there, accelerating to   
   the speed vector (x,y) you need a total delta-v of x+y which is greater   
   than square root(x^2 + y^2) unless x or y is zero. Something similar   
   happens if you start with Earth's rotational speed due east, and then go   
   for an orbit for which your speed shouldn't be due east, a significant   
   portion of your initial speed isn't used in an optimal way.   
      
      
   Alain Fournier   
      
   --- SoupGate-Win32 v1.05   
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