Gravitational Escape — Breaking Free of a Central Mass
1 · Predict
A satellite is launched sideways past a central mass at just above the escape speed for its distance. Does it fall back into orbit, circle forever, or leave and never return?
- It escapes — its distance keeps growing and it never comes back.
- It settles into a closed orbit around the central mass.
- It slows, turns around, and falls back toward the central mass.
2 · Set Up
- Open the Gravitational Escape preset and press Reset. Note the central-mass parameter GM and the satellite's starting distance r from the centre.
- Confirm the speed probe is attached to the satellite — it reports the satellite's speed relative to the central mass in m/s.
- For each trial below, set GM and the launch distance r to the listed values, press Play, and record the satellite's speed at the instant of launch.
3 · Collect Data
| Central-mass parameter GM (m³/s²) | Launch distance r (m) | Escape speed v_esc (m/s) | Measured launch speed v (m/s) |
|---|---|---|---|
| 40 | 2.8 | ||
| 40 | 4 | ||
| 60 | 3 |
Plot the measured launch speed v (y-axis) against the escape speed v_esc you computed (x-axis) for your three trials. Describe the line and estimate its slope.
4 · Analyze
- For each row, divide the measured launch speed v by the computed escape speed v_esc. What do you notice about this ratio across all three trials, and what does a ratio greater than 1 tell you about whether the satellite escapes?
- The specific orbital energy is ε = v²/2 − GM/r. Compute ε for your first trial using the measured launch speed. Show that it is positive, and explain why a positive ε means the satellite is unbound.
5 · Extend
- Predict what path the satellite would follow if it were launched at exactly the escape speed (ratio = 1.00) instead of 1.08×. Would it ever truly stop, and at what distance? Explain using ε.
- A real spacecraft launched from a planet's surface must also climb out through atmospheric drag before reaching space. Would that make the speed actually needed to escape larger or smaller than the ideal v_esc = √(2GM/r) computed here? Why?
The Physics Behind This Experiment
Escape Speed
The minimum speed at which a body's kinetic energy equals the depth of the gravitational well at distance r, so its total energy is exactly zero and it can coast to infinity. Any faster and it escapes with energy to spare.
Circular-Orbit Speed
The speed needed to hold a circular orbit at the same distance r. The escape speed is exactly √2 times this — a useful benchmark for judging how far above 'staying in orbit' each launch really is.