Gravitational potential energy takes two fundamentally different forms depending on the scale of the system. Near Earth's surface (where \( h \ll R_E \), Earth's radius), gravity appears uniform and we use \( U = mgh \). Over planetary or cosmic distances, gravity weakens as \( 1/r^2 \) and we use \( U = -GMm/r \). This section develops both regimes and explores their physical consequences: escape velocity, binding energy, and orbital mechanics.
Uniform field: \( U_{\text{grav}} = mgh \)
When an object is lifted vertically near Earth's surface through a height \( h \), the work done against gravity is:
\[ W = mgh \]
This work is stored as gravitational potential energy. The reference point (where \( U = 0 \)) is arbitrary; we typically choose sea level or ground level.
Force check
From §8.1, we recover force via:
\[ F = -\frac{dU}{dh} = -\frac{d(mgh)}{dh} = -mg \]
Correct: a constant downward force independent of height (at least for \( h \ll R_E \)).
Example: Lifting a block
A \( 50 \) kg block is lifted from ground level to a height of \( 10 \) m in a uniform gravitational field (\( g = 9.8 \) m/s²). How much gravitational potential energy is stored?
\[ \Delta U = mgh = 50 \times 9.8 \times 10 = 4{,}900 \text{ J} \]
This energy is available if the block falls back down; gravity will do exactly \( 4{,}900 \) J of work on the block.
Non-uniform field: \( U_{\text{grav}} = -\frac{GMm}{r} \)
At distances large enough that Earth's gravity is visibly non-uniform (satellites, moons, interplanetary spacecraft), we must use Newton's law of gravitation. The gravitational force between a central mass \( M \) and a test mass \( m \) is:
\[ \mathbf{F}_{\text{grav}} = -\frac{GMm}{r^2} \hat{r} \]
where G is the gravitational constant \( 6.674 \times 10^{-11} \) N·m²/kg² and r is the distance between centers.
To find the potential energy, we integrate the force:
\[ U(r) = -\int_{\infty}^{r} F(r') \, dr' = -\int_{\infty}^{r} \left( -\frac{GMm}{r'^2} \right) dr' = -\frac{GMm}{r} \]
Note the sign convention: The reference point is placed at infinity: \( U(\infty) = 0 \). Since gravity is always attractive, \( U(r) < 0 \) for all finite \( r \). This negative energy reflects the bound nature of gravitating systems — you must add energy to escape.
Force check
\[ F = -\frac{dU}{dr} = -\frac{d}{dr}\left( -\frac{GMm}{r} \right) = -\frac{GMm}{r^2} \]
Correct: Newton's law.
Escape velocity
A key application of gravitational potential energy is finding the escape velocity — the minimum speed an object must have at Earth's surface to escape to infinity with zero final velocity.
Energy conservation argument
At Earth's surface (radius \( R_E \)):
- Initial kinetic energy: \( KE_i = \frac{1}{2}mv_{\text{esc}}^2 \) (unknown)
- Initial potential energy: \( U_i = -\frac{GMm}{R_E} \)
- Total initial energy: \( E_i = KE_i + U_i \)
At infinity (escaped):
- Final kinetic energy: \( KE_f = 0 \) (just barely escapes)
- Final potential energy: \( U_f = 0 \)
- Total final energy: \( E_f = 0 \)
By conservation of mechanical energy (no non-conservative forces):
\[ E_i = E_f \]
\[ \frac{1}{2}mv_{\text{esc}}^2 - \frac{GMm}{R_E} = 0 \]
\[ v_{\text{esc}} = \sqrt{\frac{2GM}{R_E}} \]
Example: Earth's escape velocity
For Earth: \( M_E = 5.972 \times 10^{24} \) kg, \( R_E = 6.371 \times 10^6 \) m.
\[ v_{\text{esc}} = \sqrt{\frac{2 \times 6.674 \times 10^{-11} \times 5.972 \times 10^{24}}{6.371 \times 10^6}} \]
\[ = \sqrt{\frac{7.963 \times 10^{14}}{6.371 \times 10^6}} = \sqrt{1.250 \times 10^8} \approx 11{,}200 \text{ m/s} = 11.2 \text{ km/s} \]
Any object launched from Earth's surface faster than \( 11.2 \) km/s will escape Earth's gravity and drift into space.
Binding energy
The binding energy of a system is the total mechanical energy required to disassemble it — to move all parts infinitely far apart with zero relative velocity.
For a two-body system (e.g., Earth and a satellite), if the total energy is:
\[ E = KE + U = \frac{1}{2}mv^2 - \frac{GMm}{r} \]
Bound orbit: If \( E < 0 \), the object is in a bound orbit. It cannot escape; the gravitational potential energy dominates the kinetic energy. The binding energy is:
\[ E_{\text{bind}} = -E = \frac{GMm}{r} - \frac{1}{2}mv^2 \]
Example: A satellite orbits Earth at radius \( r = 7.0 \times 10^6 \) m (about \( 600 \) km altitude). What is its binding energy?
Circular orbit speed: For a circular orbit, gravity provides the centripetal force:
\[ \frac{GMm}{r^2} = \frac{mv^2}{r} \implies v = \sqrt{\frac{GM}{r}} \]
\[ v = \sqrt{\frac{6.674 \times 10^{-11} \times 5.972 \times 10^{24}}{7.0 \times 10^6}} = \sqrt{5.71 \times 10^7} \approx 7{,}560 \text{ m/s} \]
Total energy of the satellite:
\[ E = KE + U = \frac{1}{2}m v^2 - \frac{GMm}{r} \]
For a circular orbit, \( v^2 = GM/r \), so:
\[ KE = \frac{1}{2}m \frac{GM}{r} = \frac{GMm}{2r} \]
\[ U = -\frac{GMm}{r} \]
\[ E = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r} \]
The binding energy (energy required to escape) is:
\[ E_{\text{bind}} = \frac{GMm}{2r} = \frac{6.674 \times 10^{-11} \times 5.972 \times 10^{24} \times m}{2 \times 7.0 \times 10^6} \]
\[ = \frac{3.99 \times 10^{14} m}{1.4 \times 10^7} \approx 2.85 \times 10^7 m \text{ J/kg} \]
For a \( 1000 \) kg satellite: \( E_{\text{bind}} = 2.85 \times 10^{10} \) J.
Transition between regimes
When is the uniform field approximation valid?
The uniform field formula \( U = mgh \) assumes \( h \ll R_E \). Let's check the fractional error:
\[ U_{\text{uniform}} = mgh \]
\[ U_{\text{inverse-square}} = -\frac{GMm}{R_E + h} + \frac{GMm}{R_E} = \frac{GMm h}{R_E(R_E + h)} \]
For small \( h \), this is approximately:
\[ U_{\text{inverse-square}} \approx \frac{GMm}{R_E^2} h = g h \]
where we used \( g = GM/R_E^2 \) (definition of surface gravity).
Fractional difference:
\[ \frac{U_{\text{inverse-square}} - U_{\text{uniform}}}{U_{\text{uniform}}} \approx \frac{h}{R_E} \]
- For \( h = 100 \) m: error \( \sim 0.0000158\% \) (negligible)
- For \( h = 10 \) km: error \( \sim 0.158\% \) (still small)
- For \( h = 1000 \) km: error \( \sim 15.8\% \) (significant; use inverse-square law)
Connection to Chapter 5 (Newton's Laws)
Recall that Newton's second law in the radial direction (for circular orbits) is:
\[ \sum F_r = -\frac{GMm}{r^2} = m a_r = -m\frac{v^2}{r} \]
This gives the orbital velocity \( v = \sqrt{GM/r} \), consistent with the energy argument above. The potential energy formalism provides an alternative route to the same results with less algebra.
What you should be able to do
After this section, you should be able to:
- Choose the appropriate potential energy form (uniform vs. inverse-square) based on the scale of the problem
- Calculate escape velocity for any planet or star using energy conservation
- Determine whether an orbit is bound (\( E < 0 \)) or unbound (\( E > 0 \))
- Compute binding energy and orbital speeds for satellites
- Explain why the uniform field is a good approximation for \( h \ll R_E \)
Preview of §8.3: Elastic Potential Energy
Springs store energy in their deformation, obeying Hooke's law. §8.3 develops the elastic potential energy \( U = \frac{1}{2}kx^2 \) in detail, explores the harmonic oscillator, and examines how springs and gravity can combine in realistic systems (bungee jumping, mass–spring systems).
📚 See also: Halliday Vol 1, Ch 8, §8.2 — Gravitational Potential Energy. 📖 Open reference: OpenStax University Physics Vol 1 — Chapter 8.3: Gravitational Potential Energy. 🎓 Video: MIT 8.01 Lecture 12 (Walter Lewin) — "Gravitational Potential Energy and Escape Velocity".
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