04 / CHAPTER 10
Orbits and escape
Use energy and angular momentum to classify trajectories.
ANCHOR EQUATION
Specific orbital energy
In symbols: ε = v²/2 − GM/r
The sign of specific orbital energy classifies bound and unbound trajectories.
Orbits trade kinetic and potential energy
An orbit is continuous free fall: gravity bends the velocity so the object keeps missing the surface. Energy and angular momentum together distinguish circular, elliptical, parabolic, and hyperbolic motion.
Classify by specific energy
Divide total mechanical energy by mass: ε = v²/2 − GM/r. For a circular orbit, gravity supplies centripetal acceleration v²/r = GM/r², so v_c = √(GM/r). Substitution gives ε_c = −GM/(2r), negative because the orbit is bound.
At r = 7.0×10⁶ m around Earth, v_c = √(3.986×10¹⁴/7.0×10⁶) = 7.55 km/s. A low circular orbit takes T = 2πr/v ≈ 97 minutes.
Escape is a speed threshold, not a lack of gravity
Setting ε = 0 gives v_escape = √(2GM/r) = √2 v_c. A spacecraft above escape speed still feels gravity; it follows an unbound trajectory rather than stopping being attracted.
Work it: At altitude 400 km, use Earth radius 6.37×10⁶ m to compute circular and escape speeds. Explain why the satellite does not need continuous upward thrust to stay aloft.