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.

1. Increase launch speed gradually.
2. Read K, U and E; find the zero-energy threshold.
3. Monitor numerical energy error over a bound orbit.

What to measure

Specific energy ε=E/m in J/kg and numerical energy error.

The idea underneath

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.

Build the argument

1. Start from F = GMm/r² and divide by satellite mass: a = μ/r².
2. For circular motion set v²/r = μ/r². Then v = √(μ/r) and T = 2π√(r³/μ).
3. Combine kinetic and potential energy per mass: ε = v²/2 − μ/r. Its sign separates bound and unbound trajectories.

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.

Applications

Orbital mechanics connects satellite navigation, communications and interplanetary transfers. A launch is a choice of position and velocity, not a command to follow a circle.

Common misconception

Zero weight is not zero gravity. Orbital radius is measured from the centre. Numerical energy drift must not be mistaken for real energy production.

A further investigation

Why does a smaller timestep reduce numerical error without changing launch conditions?

Check your understanding

Escape speed is what multiple of circular speed?

• √2
• 2
• 1/2

Model limits

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