3 ms·

One can also look at it this way, though it does not explain the energies' relation to each other: The total gravitational potential energy you start from, fro

One can also look at it this way, though it does not explain the energies' relation to each other: The total gravitational potential energy you start from, from Earth's surface, is far from zero, as you are already at over 6000 km height. The 300 km height is a small blip on top of that. A 5 % change. The velocity you start with is only the spinning of the earth (lower velocity closer to poles, vector sum too because you must launch to an inclined orbit from there) which is about 0.4 km/s. So you need a 20x change. (Coincidentally reciprocal of 5%.)

ColinWright 13y ago
But that's really misleading, as the energy involved in increases in velocity go as a square, whereas increases in height are linear. In fact, if you go to 600 km height then you double the PE change, but don't much change the KE requirement. I'm not sure your comment would really help someone who doesn't already know what's going on, but I'd appreciate replies from people who are trying to understand more about this. Maybe I'm just the wrong audience.
- lutusp 13y ago
  The easy way to solve this class of problem is to compute two quantities -- one, the object's energy at rest on earth's surface, and two, the object's energy when it gets into orbit. The first is mostly gravitational potential energy (GPE), the second is a mixture of GPE and kinetic energy (KE). They're both easy to compute. Finally, to figure out how much energy is required to go from the launch pad to orbit, simply subtract one energy number from the other, remembering that the total energy is the sum of KE and GPE. A complicating factor is the fact that a rocket's mass decreases as it ascends (as the fuel is burned), but this isn't a deal-breaker -- all one needs to do is profile the ascent, the increase in velocity alongside the decrease in mass, using a numerical integral. That's not difficult at all (and it has to be a numerical integral). More here: http://arachnoid.com/gravitation_equations/orbital_mechanics.html http://arachnoid.com/gravitation_equations/orbital_mechanics...
- Gravityloss 13y ago
  Yeah, you're right.

ColinWright 13y ago

But that's really misleading, as the energy involved in increases in velocity go as a square, whereas increases in height are linear. In fact, if you go to 600 km height then you double the PE change, but don't much change the KE requirement. I'm not sure your comment would really help someone who doesn't already know what's going on, but I'd appreciate replies from people who are trying to understand more about this. Maybe I'm just the wrong audience.

lutusp 13y ago
The easy way to solve this class of problem is to compute two quantities -- one, the object's energy at rest on earth's surface, and two, the object's energy when it gets into orbit. The first is mostly gravitational potential energy (GPE), the second is a mixture of GPE and kinetic energy (KE). They're both easy to compute. Finally, to figure out how much energy is required to go from the launch pad to orbit, simply subtract one energy number from the other, remembering that the total energy is the sum of KE and GPE. A complicating factor is the fact that a rocket's mass decreases as it ascends (as the fuel is burned), but this isn't a deal-breaker -- all one needs to do is profile the ascent, the increase in velocity alongside the decrease in mass, using a numerical integral. That's not difficult at all (and it has to be a numerical integral). More here: http://arachnoid.com/gravitation_equations/orbital_mechanics.html http://arachnoid.com/gravitation_equations/orbital_mechanics...
Gravityloss 13y ago
Yeah, you're right.