Gravitational Potential Energy Beyond mgh & Escape Speed
- Derive U = −GMm/r by integrating the gravitational force
- Explain the sign convention and why U is negative for bound systems
- Compute escape speed and relate total energy to whether an orbit is bound
Why mgh fails at large distances
U = mgh assumes g is constant, which is true only over heights small compared with Earth's radius. Gravity actually falls off as 1/r², so for a satellite, a comet or an escaping rocket the constant-g formula is badly wrong. The correct expression comes from integrating the true force: U(r) = −∫F dr with F = −GMm/r², giving U = −GMm/r, with the reference point taken at infinity where U = 0.
The sign is doing real work
The negative sign is not a bookkeeping quirk. It encodes boundness. A system with total energy E = K + U < 0 does not have enough kinetic energy to reach infinity, so it is bound and follows a closed orbit. A system with E ≥ 0 escapes. That single inequality classifies every trajectory in the two-body problem: E < 0 gives an ellipse, E = 0 a parabola, E > 0 a hyperbola. Getting the sign wrong reverses the physical conclusion, not merely the arithmetic.
Escape speed
Escape means arriving at infinity with at least zero speed, so set the total energy to zero: ½mv² − GMm/R = 0, giving v_esc = √(2GM/R). The escaping mass m cancels — a marble and a spacecraft need the same escape speed. For Earth this is about 11.2 km/s. Note carefully what escape speed does not mean: a rocket with continuous thrust can leave Earth at any speed whatever. Escape speed is the launch speed required for an unpowered projectile, which is why it applies to ballistic problems rather than to actual rockets.
Using G = 6.67 × 10⁻¹¹ N·m²/kg², M_Earth = 5.97 × 10²⁴ kg and R_Earth = 6.37 × 10⁶ m, compute the escape speed from Earth's surface.
- 1.Set total energy to zero for marginal escape: ½mv² + (−GMm/R) = 0.
- 2.The mass m cancels: v = √(2GM/R).
- 3.Numerator: 2GM = 2(6.67 × 10⁻¹¹)(5.97 × 10²⁴) ≈ 7.97 × 10¹⁴.
- 4.Divide and take the root: 7.97 × 10¹⁴ / 6.37 × 10⁶ ≈ 1.25 × 10⁸, so v = √(1.25 × 10⁸) ≈ 1.12 × 10⁴ m/s.
A satellite in a bound circular orbit has total mechanical energy that is —
Escape speed is not "the speed needed to leave Earth." A rocket under continuous thrust can climb away arbitrarily slowly. Escape speed answers a narrower question: how fast must an object be launched so that, coasting with no further thrust, it never returns.
The escape speed from a planet does not depend on —
Answer the 2 checkpoints as you read.
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