04 · The energy to escape
What separates a bound orbit from an escape trajectory?
Predict the energy sign at 1.4 and 1.5 times circular speed.
- Increase launch speed gradually.
- Read K, U and E; find the zero-energy threshold.
- Monitor numerical energy error over a bound orbit.
Measurement
Specific energy ε=E/m in J/kg and numerical energy error.
With potential zero at infinity, negative energy is bound, zero is marginal escape, and positive energy is unbound. Numerical drift is not physical energy production. Near zero total energy, compare error against the initial binding-energy scale to avoid a misleading relative error.
ε in J/kg; K/m=v²/2; U/m=−μ/r; potential zero at infinity.
Worked example 1
At r=7×10⁶ m, v_esc≈10672 m/s, 41.4% above circular speed.
Worked example 2
Circular specific energy is −28.47 MJ/kg. A 1000 kg body has E=−28.47 GJ.
Why does a smaller timestep reduce numerical error without changing launch conditions?
Assumptions
Newtonian two-body model with fixed central mass. No drag, third bodies, oblateness or relativity. Visual sizes are illustrative; initial conditions are fictional, not live ephemerides.
University Physics Volume 1, 13.4 Satellite Orbits and Energy