Fez

So the constant 1878968 is only for Kerbin? If I want to do a Hohmann with another gravitational body, I have to use a different constant in my equation?

The transfer orbit is from a lower altitude of 13,381,911,000 meters above the sun to 20,000,000,000 above the sun. When I plug those numbers in, I get 17 m/s as an answer.

I have an equation that shows the target speed you need to reach to have a successful hohmann transfer: V = 1878968 * sqrt(2/RInitial-2/(RLowest + RHighest)) It's always worked in the past, but I recently tried to use it to find the target velocity for a transfer around the sun. When I plug in the numbers, I get a ridiculous answer of 16 m/s, and I don't know why. The weird thing is, I tried looking this formula up on Google to see if I missed something, and I cannot find it anywhere. Has anyone else used this formula?

So the radial velocity outward doesn't decrease to the point that the craft starts descending as soon as the crafts tangential velocity drops below the amount needed for circular orbit? Is that because it keeps wanting to move outward due to newtons first law?

According to my knowledge, a spacecraft climbs because it has enough tangential velocity, that for every x feet itbdrops due to gravity, it goes so much farther tangentially, that it misses the surface, and stays in orbit. Therefore, a faster tangential speed than that needed for a circular orbit would lead to a less curved trajectory, and thus the craft would climb. Vice versa if the craft is traveling slower than the velocity needed for a circular orbit. With that in mind, I'm confused as to how the craft that I was using in KSP was climbing when it had a velocity lower than that needed for circular orbitm. Thanks

That definitely makes sense, but what about when in other parts of hr orbit, besides apoapsis? Like when you're climbing to apoapsis, but you're speed is below the amount necessary for a circular orbit. Sincerely Rita slower, shouldn't it be descending?

Would that also explain why a spacecraft starts climbing faster and faster while losing speed, while going towards the apoapsis? Because the vector is facing out? Still kinda confused...

Butbi calculated the speed needed for that elliptical orbit I was in, and got the speed i was traveling at, so it was accurate. But how does that equation account for different vectors of velocity?

But isn't that outward radial velocity caused by the faster tangential velocity? In other words, you climb because you go fast

Hey, sorry to revive this, but I've got another question I was in an elliptical orbit, and I calculated the speed i would've needed to be in a circular orbit at the altitude I was at. I found that the speed i was at in the elliptical orbit, while climbing to apoapsis, was slower than the speed needed for a circular orbit. How can that be, if a faster speed means I'll climb, but I was going slower than if I had been in a circular orbit? It's been bugging me, thanks.

Thanks so much for the help!!

That makes sense, thanks. The way I always imagined orbits working was that gravity pulls the spacecraft down a bit, but the spacecraft moves tangentially at the same time. This combination of movements creates a curved trajectory. If the tangential velocity is fast enough, it will climb, since it's travelling so far horizontallybin the time it takes to fall a little

Actually, nevermind, I see how energy conservation would work. Bit is there a physical way to imagine it?

This doesn't make sense to me though. According to that argument, there is no one velocity that accounts for a particular altitude. You just lose some speed as you climb, and gain some speed as you descend. Also, is there a physical way to imagine it?

No problem Is there a reason though, that you can maintain a higher orbit at a slower speed, but need more speed for a lower orbit?