Source: P212 Discussion Worksheet, UIUC
Tags: gravitational potential energy, potential energy map, superposition of potentials, conservation of energy, energy landscape, scalar vs vector, work-energy theorem, P211 review, P212
Difficulty: Intermediate Prerequisites: Newton's law of gravitation, kinetic energy, work-energy theorem, Discussion 1B (forces on m in the same square configuration).
Forces are vectors and can be painful to add. Potential energy is a scalar, and scalars just add as numbers, no direction to worry about. This problem introduces the idea of a potential energy map U(x,y), which tells you everything about the forces in one tidy function. In P212, the electric potential V(x,y) plays exactly the same role for electric forces that U(x,y) plays here for gravity. Learning to read and reason about energy landscapes now will pay off immediately when you encounter voltage maps, equipotential surfaces, and the relationship E = −∇V.
The gravitational potential energy of a test mass near four fixed corner masses is the sum of four individual −GMm/r terms (one per corner). At the centre of the square it equals −4√2 GMm/a. You can use conservation of energy to find speeds and determine whether the particle can travel between two points. The potential energy landscape has deep wells at the corners and a saddle point at the centre.
Gravitational potential energy (between two point masses)
U₁₂ = −Gm₁m₂ / r₁₂, where r₁₂ is the separation. Negative because gravity is attractive: you must do positive work to pull the masses apart to infinity, where U = 0. Think of it as: the deeper the well (more negative U), the more tightly bound the two masses are.
Potential energy map U(x,y)
A function that gives the potential energy of the test mass at every point in space. Acts like a topographical map: the particle "rolls downhill" toward lower U. The force at any point is the negative gradient of U. In simple terms, this means if you can draw the energy landscape, you can read off which direction the particle will accelerate without computing any forces.
Superposition (of potential energies)
Because potential energy is a scalar, the total PE of mass m due to multiple sources is the algebraic sum of the individual pair contributions. No vector components needed.
Conservation of energy
K₁ + U₁ = K₂ + U₂. If no non-conservative forces act, the total mechanical energy is constant. A particle released from rest (K = 0) at one point can reach another point only if U at the destination is less than or equal to U at the start.
For two point masses separated by distance r:
U = −Gm₁m₂ / r
Convention: U → 0 as r → ∞.
The origin is the centre of the square. The distance from the centre to each corner is a/√2 (half the diagonal, which is a√2).
Each corner mass M contributes:
U_i = −GMm / (a/√2) = −√2 GMm / a
There are four such contributions, so by superposition:
U(0,0) = −4√2 GMm / a
This is a single number (a scalar), which is vastly simpler than the vector-force calculation in Discussion 1B.
At infinity: K = 0, U = 0. Total energy E = 0.
At the origin: K = ½mv², U = −4√2 GMm/a. Total energy E = ½mv² − 4√2 GMm/a.
By conservation of energy:
0 = ½mv² − 4√2 GMm/a
v = √(8√2 GM / a)
Or equivalently v = (8√2 GM/a)^(1/2). Note that the test mass m cancels, just as it does in free-fall problems.
The midpoint of the right side, with the square centred at the origin, is at (a/2, 0). Corners are at (±a/2, ±a/2).
Distances from (a/2, 0) to each corner:
To (a/2, a/2): distance = a/2
To (a/2, −a/2): distance = a/2
To (−a/2, a/2): distance = √(a² + a²/4) = a√5/2
To (−a/2, −a/2): distance = a√5/2
Potential energy contributions:
Two near corners: each contributes −GMm/(a/2) = −2GMm/a
Two far corners: each contributes −GMm/(a√5/2) = −2GMm/(a√5)
U(a/2, 0) = 2(−2GMm/a) + 2(−2GMm/(a√5))
U(a/2, 0) = −4GMm/a − 4GMm/(a√5)
U(a/2, 0) = −(4GMm/a)(1 + 1/√5)
Numerically, 1 + 1/√5 ≈ 1.447, so U at the midpoint of the right side is less negative (higher, closer to zero) than U at the origin (where the factor is 4√2 ≈ 5.657 times GMm/a).
If the particle is released from rest at (a/2, 0):
Initial energy: E = U(a/2, 0) + 0 (zero kinetic energy).
At the origin: E = U(0,0) + K.
Since U(0,0) is more negative than U(a/2, 0), the particle would need to lose potential energy (go downhill), gaining kinetic energy. That is allowed.
Yes, the particle reaches the origin. It arrives with kinetic energy K = U(a/2, 0) − U(0,0) > 0.
The energy landscape slopes downward from the midpoint of the side toward the centre, so the particle accelerates inward. (This is consistent with the Discussion 1B result that the force at that point is in the −x direction.)
Key features of U(x) along y = 0:
At x = 0 (centre): U = −4√2 GMm/a, a local maximum of U along the x-axis (a saddle point in 2D, since it is a minimum along the diagonal direction toward the corners).
At x = ±a/2 (midpoints of the sides): U is more negative than at infinity but less negative than at the centre? No, let's reconsider. We showed U(a/2,0) is less negative than U(0,0). So along the x-axis, U decreases (becomes more negative) as you move from the side toward the centre. The centre is the deepest point along the x-axis.
As x → ∞: U → 0 from below.
Near the corners (x = ±a/2, y = ±a/2): U → −∞ (the 1/r terms diverge). But along the x-axis, the corners are not on the axis, so U stays finite.
The sketch: starting from large positive x, U rises from a very negative value (not quite −∞, since you are not at a corner) as you approach x = a/2, continues to become more negative as you move toward the centre, reaches its most negative value at x = 0, and then mirrors on the left side. The curve is a smooth well with its deepest point at the origin (for motion restricted to the x-axis).
The correct 3D surface plot must show:
U is everywhere negative (gravity is attractive, and U = −GMm/r < 0 for all finite r).
Deep wells (U → −∞) at the four corners where the fixed masses sit.
A saddle region near the centre.
The correct figure is the one with four deep negative wells at the corners and negative values everywhere. Plots that show positive U or positive peaks at the corners are wrong.
Quantity | Expression |
|---|---|
Pair potential energy | U = −Gm₁m₂/r |
U at centre of square | U(0,0) = −4√2 GMm/a |
Speed at origin (from rest at ∞) | v = √(8√2 GM/a) |
U at midpoint of right side | U = −(4GMm/a)(1 + 1/√5) |
In P212, the electric potential V plays the same role as U/m (potential energy per unit charge). Equipotential contour plots for charge distributions work identically to the gravitational energy maps here. Engineers use potential maps to design electron optics, ion traps, and particle accelerators, all of which require understanding where charged particles will and will not go based on energy landscapes.
Trying to add potential energies as vectors. Potential energy is a scalar. No direction, no components. Just add the numbers.
Forgetting the negative sign in U = −GMm/r. Gravitational PE is always negative for bound systems. A correct potential map for attractive masses has deep negative wells, not positive peaks.
Confusing "lower potential energy" with "smaller number." A value of −10 is lower than −5. The particle moves toward more negative U.
Assuming that because the force at the midpoint of the right side points toward the centre, U must decrease monotonically all the way to the centre. In this case it does along the x-axis, but in general, force direction and energy slope can have subtleties in 2D (saddle points).
⚠️ The scalar nature of potential energy (and electric potential) is a massive computational shortcut compared with vector force calculations. Exams frequently ask you to choose between computing forces and computing energies, and energy is often faster.
⚠️ Conservation of energy problems ("is the particle released from rest at point A fast enough to reach point B?") appear throughout P212 in the context of electric potential.
⚠️ Knowing how to read a potential energy map, and in particular identifying wells, saddle points, and the direction of force from the slope, is directly tested.
⚠️ The centre-of-square result U(0,0) = −4√2 GMm/a is a clean application of superposition that could appear as a short exam question.
True or false: gravitational potential energy between two masses is always negative.
Fill in the blank: the total potential energy from multiple sources is found by ____ (vector addition / scalar addition).
True or false: a particle released from rest can only move to regions of lower potential energy.
Fill in the blank: the speed of a particle released from rest at infinity and arriving at a point where U = U₀ is v = √(____).
True or false: the test mass m appears in the final expression for the speed at the origin.
Answers: 1. True (for finite separation, with the convention U = 0 at infinity). 2. Scalar addition. 3. True. 4. −2U₀/m (since U₀ is negative, this is positive under the square root). 5. False (m cancels).
Q: Show that U(0,0) = −4√2 GMm/a for the four-corner-mass configuration.
A: Each corner is at distance a/√2 from the centre. Each contributes U_i = −GMm/(a/√2) = −√2 GMm/a. By superposition, U_total = 4 × (−√2 GMm/a) = −4√2 GMm/a.
Q: A particle is released from rest at the midpoint of the right side. Does it reach the centre? Explain using energy arguments.
A: U at the midpoint is −(4GMm/a)(1 + 1/√5), which is less negative (higher) than U(0,0) = −4√2 GMm/a. Since the centre has lower potential energy, the particle can reach it, converting the PE difference into kinetic energy.
Q: Why does the test mass m cancel out of the expression for speed when released from infinity?
A: Both the kinetic energy (½mv²) and the potential energy (−GMm × constant) are proportional to m. When you set their sum equal to zero and solve for v, the m divides out. This is the gravitational analogue of "all objects fall at the same rate."
This topic connects directly to electric potential in P212: replace −GMm/r with kq₁q₂/r (or −Gm₁m₂/r → kq₁q₂/r, noting the sign difference for repulsive charges). The potential-energy map U(x,y) becomes the voltage map V(x,y), and the relationship F = −∇U becomes E = −∇V. The conservation-of-energy reasoning here also appears in circuit analysis (energy gained/lost by charges moving through potential differences).
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