Introduction
Gravitational potential energy represents energy stored in the gravitational interaction between masses. For two masses separated by distance r, the potential energy is negative, indicating a bound system. Understanding this energy is essential for analyzing orbital mechanics, escape velocity, and the binding energy of astronomical systems.
Potential Energy of Two Masses
For two point masses m1 and m2 separated by distance r: U = -G(m1m2)/r. The negative sign indicates that gravity is attractive - work must be done to separate the masses. Zero of potential energy is conventionally set at infinite separation (U(∞) = 0). As r decreases, U becomes more negative (system more tightly bound). The force is F = -dU/dr = -G(m1m2)/r2 (attractive, directed toward decreasing r).
\nIntroduction
Gravitational potential energy represents energy stored in the gravitational interaction between masses. For two masses separated by distance r, the potential energy is negative, indicating a bound system. Understanding this energy is essential for analyzing orbital mechanics, escape velocity, and the binding energy of astronomical systems.
Potential Energy of Two Masses
For two point masses m1 and m2 separated by distance r: U = -G(m1m2)/r. The negative sign indicates that gravity is attractive - work must be done to separate the masses. Zero of potential energy is conventionally set at infinite separation (U(∞) = 0). As r decreases, U becomes more negative (system more tightly bound). The force is F = -dU/dr = -G(m1m2)/r2 (attractive, directed toward decreasing r).
\nBinding Energy
Binding energy is the energy required to separate two masses to infinity: E_b = GMm/R for masses initially at distance R. This equals |U(R)| since U(∞) = 0. Larger binding energy means more tightly bound system. For Earth-Sun system: E_b ≈ 2.6×10^33 J. For hydrogen atom (using Coulomb potential analogously): 13.6 eV binding energy.
Escape Velocity
Escape velocity is the minimum speed needed for an object to escape from a mass M starting at distance R: v_esc = √(2GM/R). Derived from energy conservation: 1/2mv2 - GMm/R = 0 at infinity (zero energy boundary). For Earth: v_esc ≈ 11.2 km/s. Independent of escaping mass m. Independent of direction (for non-rotating body). Objects with v > v_esc follow hyperbolic orbits; v = v_esc gives parabolic trajectory.
Total Mechanical Energy in Orbits
For a body of mass m in orbit around mass M: Total energy E = K + U = 1/2mv2 - GMm/r. Using v2 = GM/r for circular orbit: E = -GMm/(2r) = -1/2|U|. Negative total energy indicates bound orbit. Virial theorem for circular orbits: K = -E = 1/2|U|, meaning kinetic energy equals half the magnitude of potential energy.
\nPotential Energy Diagrams
The effective potential for orbital motion includes centrifugal barrier: U_eff = -GMm/r + L2/(2mr2). Plot shows: (1) At large r, gravitational term dominates (negative); (2) At small r, centrifugal term dominates (positive, repulsive); (3) Minimum at stable circular orbit; (4) E < 0: bound elliptical orbits; (5) E = 0: parabolic escape trajectory; (6) E > 0: hyperbolic unbound trajectory.
\nSolved Example: Escape from Earth
A 1000 kg rocket needs to escape Earth (M_E = 5.98×10^24 kg, R_E = 6.37×10^6 m). (a) Find escape velocity. (b) Find initial KE needed. (c) If launched at 15 km/s, what is its speed very far from Earth? Solution: (a) v_esc = √(2GM_E/R_E) = √(2×6.67×10¹×5.98×10^24/6.37×10^6) = √(1.25×10^8) = 11,190 m/s ≈ 11.2 km/s. (b) Initial KE = 1/2mv_esc2 = 1/2×1000×(11,190)2 = 6.25×10^10 J = 62.5 GJ. This equals |U| at surface = GM_Em/R_E. (c) Launched at v0 = 15 km/s = 15,000 m/s. Initial total energy E = 1/2mv02 - GM_Em/R_E = 1/2×1000×(15,000)2 - 6.25×10^10 = 1.125×10^11 - 6.25×10^10 = 5.0×10^10 J. Very far away, U → 0, so 1/2mv2 = 5.0×10^10 → v = √(2×5.0×10^10/1000) = 10,000 m/s = 10 km/s. Excess velocity over escape velocity: 152 - 11.22 = 225 - 125 = 100, giving v_∞ = √100 = 10 km/s ✓.
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