Orbital Energy
Introduction
When you launch a payload into space, your launch mechanism does work (i.e., requires energy) both to lift it and to accelerate it to orbital velocity. This calculator tells you the theoretical minimum required to do this (in practice, especially for a chemical rocket, the energy required is much much larger).
Calculator
Latitude of launch point | |
Radius of Earth's surface at launch point | (Estimate from launch latitude) |
Altitude of target circular orbit above launch point |
|
Gravitational acceleration at launch point | (Estimate from launch point) |
Mass to be launched | |
Electricity rate |
¢ per kWhr (10.31 = USA all sectors, 2008) |
Required orbital velocity: | ||
Change in velocity (Δv): | ||
Change in gravitational potential energy: | ||
Change in kinetic energy: | ||
Total change in energy: | ||
Cost of energy in electricity ($): |
Discussion
For low Earth orbit (LEO), the kinetic energy is the most significant component. Indeed, at the International Space Station's 330–435 km altitude (inclination 51.6°), potential energy is just 10.1–12.9% of the energy required. As we get higher, the satellite orbits slower and the potential energy gets higher. The breakeven point for a prograde equatorial orbit seems to be about 2758.71528 km.
It's interesting to note that the theoretical minimum energy is extremely reasonable if thought of in terms of electricity. It has been claimed that Arthur C. Clarke once did a calculation, achieving a figure of $10/kg. I haven't found that reference, but a similar quote is attested in his 1979 The Fountains of Paradise:
"With the efficiency of the latest fusion plants, it will take only twenty dollars' worth of electricity to lift a ton up to orbit."
Hence, a thought experiment: suppose I have a 1kg mass, and I want to put it into orbit around Earth. How much energy would that require (theoretical minimum)? We have to take into account the change in gravitational potential energy (orbits are higher) and kinetic energy (orbits are faster). If you calculate the energy cost in terms of electricity, it turns out that the cost is surprisingly small (although the amount of energy required is in the several-to-tens of kilowatt-hours range). This says to me that electricity is massively undervalued.
TODO
Should make valid for elliptical orbits and for centers other than Earth.