Gravitation · Lesson 4 of 7
Gravitational Potential and Potential Energy
“Derive gravitational energy, define potential and apply superposition without confusing zero field with zero potential.”
• Derive the gravitational potential-energy difference by integration. • Relate exact inverse-distance energy to the near-surface mgh approximation. • Distinguish potential from potential energy and force. • Count pair energies and analyse zero-force points.
Work done by gravity versus work done against gravity
Gravity is conservative: its work between fixed endpoints is independent of the path. If gravity pulls a body inward, it can increase the body’s kinetic energy while reducing gravitational potential energy. To lift the body slowly outward, an external agent does positive work against gravity. These two work signs are opposite. The source uses W for potential energy in part of the discussion; here U denotes potential energy and φ denotes potential to avoid confusing them with work.
Deriving the inverse-distance expression
Take radial coordinate r outward from a central mass M. The gravitational radial force is −GMm/r². A slow lifting force balances it with +GMm/r². Integrating that external force from r1 to r2 gives GMm(1/r1−1/r2). This equals the increase in potential energy. Therefore a potential-energy function with the required differences is −GMm/r plus an arbitrary additive constant.
Set C=0 when choosing U=0 at infinity. Then any finite attractive pair has negative potential energy. The negative sign expresses that positive work is required to separate the pair to infinity. It is not a negative mass or a direction. Adding a constant changes no energy difference and no force. One must keep the same reference throughout a calculation.
How mgh emerges near Earth
Near the surface, gravity changes little over the motion, so the exact work integral becomes almost constant force times height change. At large height, treating the same g as constant overestimates lifting work. For two arbitrary exterior radii, use the inverse-radius difference rather than a constant-g shortcut. Neither formula should silently change the energy zero midway through a solution.
Potential is energy per unit test mass
The source’s potential depends on the attracting mass and location, not on which small test mass is placed there. Multiplying φ by a test mass gives that mass’s potential energy. A vector field can vanish when opposite force contributions cancel while the scalar potential contributions add. Thus zero force does not imply zero potential. The reference φ(∞)=0 makes all ordinary positive-source point-mass potentials negative.
Adding potentials and counting pair energies
Problem
Source Example 7.3 puts four equal masses m at the vertices of a square of side l. Find the system energy and the potential at its centre.
- 1.There are six distinct pairs: four sides of length l and two diagonals of length √2 l.
- 2.The pair energy is −4Gm²/l − 2Gm²/(√2 l) = −(4+√2)Gm²/l ≈ −5.414Gm²/l.
- 3.Each corner is l/√2 from the centre. The centre potential is φC = −4Gm/(l/√2) = −4√2 Gm/l.
- 4.The centre potential is energy per unit additional test mass; the six-pair system energy is a different quantity. Symmetry makes the field at the centre zero, yet its potential is negative.
A midpoint can have zero force but negative potential
Problem
In source Exercise 7.21, take the two sphere centres to be 1.0 m apart. Each sphere has radius 0.10 m and mass 100 kg. Find the field and potential at the midpoint and discuss equilibrium.
- 1.The midpoint is outside both spheres and 0.50 m from each centre. Their equal fields have opposite directions, so the net field and force on a test object are zero.
- 2.With zero potential at infinity, φ = −G(100/0.50+100/0.50) = −400G ≈ −2.67×10⁻⁸ J/kg.
- 3.The negative sign is required by the stated reference. The answer appendix’s positive sign is inconsistent with its own potential convention.
- 4.Move a small test mass a little along the joining line. The nearer sphere pulls more strongly, driving it farther from the midpoint. The equilibrium is therefore unstable; stability in every direction would be needed for full stable equilibrium.
- 5.The phrase “1.0 m apart” is interpreted as centre separation, matching the appendix’s intended potential magnitude. If it meant a 1.0 m surface gap, the centre separation and numerical potential would differ. Spheres remain fixed in this idealisation.
Check your understanding
Quiz
The work done by gravity equals
Conservative-force work reduces potential energy when it is positive.
With U(∞)=0, the potential energy of an attractive mass pair is
U=−GMm/r.
The quantity −GM/r has units of
It is gravitational potential, not total pair energy.
Four corner masses of a square have how many distinct interaction pairs?
There are 4 side pairs and 2 diagonal pairs.
At the midpoint of two equal fixed attracting masses, zero field implies
Vector fields cancel, while scalar negative potentials add.
Key Takeaways
• Use opposite signs for gravity work and potential-energy change. • Potential is energy per unit test mass. • mgh is a small-height approximation to an energy difference. • Count each interacting pair once and keep the potential reference fixed.