Joshua F Thomas
Well-Known Member
- Joined
- Jun 4, 2019
- Messages
- 664
- Reaction score
- 342
To get up to 100 km, the change in potential energy is:
PE = mgh = 1kg * 9.8 m/s^2 * 100,000m = 1x10^6 N-m
To accelerate to 8 km/s, the change in kinetic energy is:
KE = 1/2 mv^2 = 0.5 * 1kg * (8000 m/s)^2 = 32x10^6 N-m
So, it takes 32 times the energy to reach orbital velocity as it does to reach orbital altitude on earth, ignoring the weight of your motors/propellant and aerodynamics. Total energy would then be 33x10^6 N-m.
Treating this as an impulse problem, we can consider it as a change in momentum (which is the definition of impulse). At least one source I checked considered 9.4 km/s (9400 m/s) as the delta-v needed to get to and sustain LEO.
Assuming 1 kg mass, J = p2 - p1 = m(v2 - v1) = 1 kg (9.4 km/s - 0) = 9.4 * 10^3 N-s
So one would need 9400 Ns. That's an M-class motor, but it also assumes your entire rocket plus motor is only 1 kg! The problem with having to haul your motor mass around quickly becomes apparent.