Difficulty: Introductory review | Prerequisites: P211 potential energy, conservation of energy, Discussion 1B (gravitational superposition)
This topic shifts from forces (vectors) to potential energy (a scalar). You will see that superposition is far easier with scalars because there are no components to worry about. Potential energy maps are introduced as topographical landscapes: the particle always "rolls downhill." In P212, this directly prepares you for electric potential and voltage, which are the scalar counterparts of the electric field.
Gravitational potential energy is a scalar, so superposition means simple addition rather than vector decomposition. A potential energy map works like a topographic map: the particle always moves toward lower potential energy. By comparing potential energy values at two positions and using conservation of energy, you can determine whether a particle can reach a given point and how fast it will be travelling when it gets there.
Gravitational potential energy (U)
For two masses m1 and m2 separated by distance r: U = -G m1 m2 / r. It is always negative (bound systems have less energy than separated ones) and approaches zero as r goes to infinity. In simple terms, the closer the masses are, the more negative (lower) the potential energy.
Potential energy map U(x, y)
A function that assigns a potential energy value to every point in space. Think of it as a landscape: hills are regions of high potential energy, valleys are low. A particle released from rest always rolls downhill toward lower U.
Scalar superposition
The total potential energy from multiple sources is the ordinary sum of the individual potential energies: U = U1 + U2 + U3 + U4. No components, no direction, just addition. This is much simpler than vector superposition of forces.
Conservation of energy
The total mechanical energy (kinetic plus potential) of a particle remains constant if only conservative forces act. In this context: the loss in potential energy equals the gain in kinetic energy. In simple terms, whatever energy the particle loses by "falling" in the potential landscape shows up as speed.
Six candidate 3-D surface plots were shown. Three key observations eliminate five of them:
Gravity is attractive, so potential energy must be lower (more negative) near the masses. The three plots showing peaks near the corners are wrong.
The bottom-left plot shows the potential for a single point mass, not four. One well, not four.
The bottom-right plot shows ridges that a simple collection of four point masses cannot produce.
The correct plot (bottom middle) is the superposition of four single-mass potential wells, one centred on each corner. It shows four deep wells at the corners and a shallow saddle point at the centre.
Each corner mass M is the same distance from the centre: r = a / sqrt(2).
Each contributes U_i = -G M m / (a / sqrt(2)) = -G M m sqrt(2) / a.
By scalar superposition, U(0,0) = 4 x (-G M m sqrt(2) / a) = -4 sqrt(2) G M m / a.
Numerically, -4 sqrt(2) is approximately -5.657.
At infinity, U = 0 and v = 0, so total energy E = 0.
At the origin, E = (1/2) m v^2 + U(0,0) = 0.
Solving: v = sqrt(8 sqrt(2) G M / a).
The two right-hand masses are each at distance r1 = a/2 from the point.
The two left-hand masses are each at distance r2 = a sqrt(5) / 2.
U = -2 G M m / (a/2) - 2 G M m / (a sqrt(5)/2) = -4 G M m / a (1 + 1/sqrt(5)).
Numerically, -4(1 + 1/sqrt(5)) is approximately -5.789.
U at the midpoint (-5.789 G M m / a) is lower than U at the origin (-5.657 G M m / a).
The midpoint is a deeper potential well than the centre.
A particle released from rest at the midpoint would need to climb uphill in potential energy to reach the origin. With zero kinetic energy to start, it cannot do so.
The particle will not reach the origin.
U is negative everywhere and approaches zero at large |x|.
U has deep wells near x = +/- a/2 (close to the right-hand and left-hand masses).
U at the origin is a local maximum (a shallow hump between the two wells).
The overall shape is a double-well curve: two dips near the masses with a slight rise at the centre.
The force at the midpoint points toward the origin (from Discussion 1B), confirming the particle sits on a slope that descends toward the centre, but the centre is still higher than the midpoint.
Quantity | Formula | Notes |
|---|---|---|
Gravitational PE (two masses) | U = -G m1 m2 / r | Always negative; zero at infinity |
U at centre of square | U(0,0) = -4 sqrt(2) G M m / a | All four distances equal a/sqrt(2) |
Speed from infinity to origin | v = sqrt(8 sqrt(2) G M / a) | From conservation of energy with E = 0 |
U at midpoint of right side | U = -4 G M m / a (1 + 1/sqrt(5)) | Two masses at a/2, two at a sqrt(5)/2 |
Students often think potential energy is higher near an attractive mass. It is the opposite: gravity pulls things together, so PE is lower (more negative) close to a mass.
A common error is treating potential energy as a vector. It is a scalar. You add numbers, not components.
Students sometimes forget that U = 0 at infinity is a convention, not a physical law. But it is the standard convention for gravitational and electrostatic PE, and the problems assume it.
Students confuse "the force points toward the origin" with "the particle can reach the origin." A force pointing in some direction does not guarantee the particle has enough energy to get there. The potential energy landscape tells the full story.
In P212, electric potential V plays the same role as gravitational PE per unit mass. Equipotential lines on a circuit diagram are the electrical analogue of contour lines on a topographic map. The reasoning about whether a particle can reach a point (comparing PE values) is exactly how you determine whether a charged particle can escape a potential well or pass through a potential barrier.
Scalar superposition of potential energy is a core exam technique in P212. It is simpler than force superposition and often preferred.
Know how to identify the correct potential energy map: PE must decrease near attractive sources.
Energy conservation arguments (comparing PE at two points to determine reachability or velocity) appear repeatedly.
The double-well shape of U(x) with a local maximum at the centre is the kind of qualitative sketch you may be asked to draw.
True or false: gravitational potential energy between two masses is always positive. (False, it is always negative.)
Fill in the blank: the distance from the centre of a square of side a to any corner is ___. (a / sqrt(2) or a sqrt(2)/2)
True or false: a particle released from rest will always move toward a region of lower potential energy. (True)
True or false: if the force on a particle points toward the origin, the particle can always reach the origin. (False, it also needs sufficient energy.)
Fill in the blank: the numerical coefficient in U(0,0) = ___ G M m / a is -4 sqrt(2). (-4 sqrt(2))
Q: Show that the net gravitational potential energy of mass m at the centre of the square is U(0,0) = -4 sqrt(2) G M m / a.
A: The distance from the centre to each corner is r = sqrt((a/2)^2 + (a/2)^2) = a/sqrt(2). Each mass contributes U_i = -G M m sqrt(2)/a. Summing four identical terms gives U = -4 sqrt(2) G M m / a.
Q: A particle of mass m falls from rest at infinity to the centre of the square. What is its speed at the origin?
A: Total energy E = 0 (rest at infinity). At the origin: 0 = (1/2)mv^2 + U(0,0). Solving: v = sqrt(-2U(0,0)/m) = sqrt(8 sqrt(2) G M / a).
Q: The potential energy at the midpoint of the right-hand side is lower than at the centre. A particle starts from rest at the midpoint. Can it reach the centre? Why or why not?
A: No. Moving from the midpoint to the centre would require the particle to climb to a higher potential energy. Starting from rest, it has no kinetic energy to spend on that climb. It will oscillate in the potential well near the right-hand side.
Q: Why is scalar superposition of potential energy simpler than vector superposition of forces?
A: Potential energy is a scalar (a single number), so you just add contributions from each source mass. Forces are vectors, requiring you to decompose each into x and y components, sum those separately, and recombine. Scalar addition skips all of that.
This connects to electric potential (voltage) in P212: replace -G M m / r with k q1 q2 / r and the entire scalar-superposition framework carries over. It also connects to Discussion 1B (the same four-mass geometry analysed via forces rather than energy) and to the concept of potential barriers and wells that appears in both electrostatics and later in quantum mechanics.
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