D.1 · IB Physics
Escape velocity and the potential well
Launch a rocket straight up and watch it climb the gravitational potential well. Below the escape speed the total energy is negative and it falls back; at the escape speed it only just gets away.
Predict first
Start from the default settings (Earth, launch speed 9.0 km s⁻¹, which is bound). Change Planet to Moon, keeping 9.0 km s⁻¹. What happens?
The Moon's escape speed is only about 2.4 km s⁻¹, so ½v² − GM/R is positive for 9.0 km s⁻¹: E ≥ 0 and the rocket is unbound.
Check it in the model: Choose Moon in the Planet list and read the Total energy per kg readout, then press Launch.
potential energy total E < 0 (bound) total E ≥ 0 (escapes) kinetic energy
E ÷ m = ½v² − GM ÷ r
bound (E < 0): r_max = R ÷ (1 − v² ÷ v_esc²)
Try this
- Choose Earth and launch at 9.0 km s⁻¹, then at 11.0 km s⁻¹. Compare the maximum distance, and check it against rmax = R ÷ (1 − v²/vesc²).
- Find the smallest launch speed (in steps of 0.5 km s⁻¹) that escapes from Earth, then from the Moon. Why are the two values so different even though the rocket is the same?
- Launch at 6.0 km s⁻¹ on Earth and watch the gap between the potential curve and the energy line. What does that gap represent, and where is it zero?
Exam tip: Setting the total energy to zero, ½mv² = GMm/R, gives v_esc = √(2GM/R), which does not depend on the mass launched; if E < 0 the object is bound and stops where the energy line meets the potential curve.