Source: P212 Discussion Worksheet, UIUC
Tags: gravitational force, superposition principle, vector addition, Newton's law of gravitation, inverse-square law, symmetry arguments, limiting behaviour, point charges analogy, P211 review, P212
Difficulty: Introductory to Intermediate Prerequisites: Newton's law of gravitation, vector components, Pythagorean theorem, basic trigonometry.
This problem trains the same skill set you will use constantly in P212 when computing the electric field or force from a collection of point charges. The gravitational force law has the same 1/r² structure as Coulomb's law, so every technique here transfers directly: drawing vector diagrams, using superposition to add individual contributions, exploiting symmetry to cancel components, and checking answers with limiting behaviour. Master these habits now and the transition to electrostatics will be straightforward.
Four equal masses sit at the corners of a square. You find the net gravitational force on a fifth mass at various positions by adding up four individual force vectors. Symmetry kills many terms. At the centre, everything cancels. Off-centre, you need careful geometry. At large distances, the four masses look like a single object of mass 4M.
Superposition (of forces)
The net force on a particle is the vector sum of every individual force acting on it. Each source contributes independently, and the contributions do not interfere with one another. Think of it as: you can calculate each force separately and then just add the vectors, no matter how many sources there are.
Newton's law of gravitation
The gravitational force exerted by mass m₁ on mass m₂ is F₁→₂ = −G(m₁m₂/r₁₂²) r̂₁₂, where r₁₂ is the distance between them and r̂₁₂ is the unit vector pointing from 1 to 2. The force is attractive: it pulls m₂ toward m₁. In simple terms, this means every pair of masses attracts each other with a force proportional to each mass and inversely proportional to the square of the distance.
Inverse-square law
A force (or field) whose magnitude falls off as 1/r². Gravity and the Coulomb force both follow this pattern. Doubling the distance cuts the force by a factor of four.
Limiting behaviour
Pushing a parameter to an extreme (e.g. very large distance) and checking whether the formula returns a physically sensible result. In this problem, the key limit is X ≫ a, where the four corner masses should look like a single point mass of 4M.
Four masses M sit at the corners of a square with side length a.
A fifth mass m can be placed at different positions.
Label the corner masses M₁, M₂, M₃, M₄ (e.g. starting bottom-left, going clockwise) and the corresponding forces F₁, F₂, F₃, F₄.
Geometry:
The distance from the centre to each corner is r = a√2 / 2 = a/√2.
By symmetry, each corner mass is at the same distance from the centre.
Symmetry argument:
M₁ and M₃ are diagonally opposite. Their forces on m are equal in magnitude but point in exactly opposite directions. They cancel.
M₂ and M₄ are also diagonally opposite. Same cancellation.
F_net = 0 at the centre. Both F_x = 0 and F_y = 0.
This is the power of symmetry: no calculation needed beyond recognising the geometry.
Place the origin at the centre of the square. The midpoint of the right side is at (a/2, 0).
Distances from each corner to (a/2, 0):
Right-side corners (a/2, a/2) and (a/2, −a/2): distance = a/2 each.
Left-side corners (−a/2, a/2) and (−a/2, −a/2): distance = √(a² + (a/2)²) = √(a² + a²/4) = a√5 / 2.
Forces from right-side corners:
Top-right corner pulls m straight downward (in −y). Magnitude: GMm / (a/2)² = 4GMm/a².
Bottom-right corner pulls m straight upward (in +y). Same magnitude.
These two cancel in the y-direction. Their x-components are zero (force is purely along y for each). Net contribution from the right pair: zero.
Wait, let me reconsider the geometry more carefully. The midpoint of the right side is at position (a/2, 0) if we centre the square at the origin with corners at (±a/2, ±a/2).
Top-right corner at (a/2, a/2): the vector from m to this corner points purely in the +y direction. Force on m points in +y. Magnitude: GMm/(a/2)² = 4GMm/a².
Bottom-right corner at (a/2, −a/2): force on m points in −y. Same magnitude.
These cancel exactly. Net y-contribution from the right pair: zero. Net x-contribution: zero.
Top-left corner at (−a/2, a/2): distance = a√5/2. Force direction: toward (−a/2, a/2), which has components (−a, a/2) relative to m's position, normalised.
Bottom-left corner at (−a/2, −a/2): by symmetry with the top-left, the y-components of these two forces cancel, and the x-components (both pointing in −x) add.
Working out the left-pair contribution:
For the top-left corner at (−a/2, a/2), the displacement from m at (a/2, 0) is (−a, a/2). The distance is a√5/2. The unit vector is (−a, a/2) / (a√5/2) = (−2/√5, 1/√5).
Force magnitude from each left corner: GMm / (5a²/4) = 4GMm / (5a²).
x-component from top-left: 4GMm/(5a²) · (−2/√5) = −8GMm / (5√5 a²)
By symmetry, the bottom-left contributes the same x-component and an equal but opposite y-component.
F_x = −16GMm / (5√5 a²)
F_y = 0
The force points in the −x direction (toward the centre of the square).
Three checks you should apply:
Let a → ∞ (spread the corner masses far apart): the force should go to zero because the sources are receding. The expression has a² in the denominator, so F → 0. ✓
Let M → 0: no source mass, no force. The expression is proportional to M. ✓
Let G → 0 (switch off gravity): force should vanish. The expression is proportional to G. ✓
Symmetry in y: the configuration is symmetric about the x-axis at this point, so F_y should be zero. It is. ✓
The physical argument:
When m is very far away (X ≫ a), the four corner masses are clustered together relative to the observation distance. They look like a single point mass of total mass 4M located near the origin.
The force on m from a single point mass 4M at distance X:
F_x ≈ −G(4M)m / X² = −4GMm / X²
F_y ≈ 0
So the dimensionless constant C = −4 (with the sign indicating attraction toward the origin; if stated as magnitude, C = 4).
This is the long-distance or "monopole" approximation. It gets better as X/a increases because the angular spread of the four masses, as seen from m, shrinks.
Quantity | Expression |
|---|---|
Gravitational force between two masses | F = GMm/r² (attractive) |
Distance, centre to corner | r = a/√2 |
Net force at centre | F_x = 0, F_y = 0 |
Net force at midpoint of right side | F_x = −16GMm / (5√5 a²), F_y = 0 |
Long-distance approximation | F_x ≈ −4GMm/X², F_y ≈ 0 (C = 4) |
Superposition of inverse-square forces is exactly how you compute the electric field from a group of point charges. The symmetry and limiting-behaviour techniques here are the bread and butter of electrostatics. The long-distance "looks like a single source" argument becomes the monopole approximation in multipole expansions, which is central to understanding how charge distributions produce fields at a distance.
Forgetting that force is a vector. You cannot simply add four magnitudes and call it done; you must add the x and y components separately.
Assuming the force at the centre is nonzero "because there are four masses pulling." Symmetry ensures the net force is exactly zero.
Incorrectly computing distances. A common error is using a (the side length) instead of a√2/2 for the centre-to-corner distance, or getting the distance to the left-side corners wrong in part (b).
Plugging in numbers too early. The worksheet makes this point explicitly: if you substitute M = 3 kg, m = 1 kg, a = 5 cm from the start, you get a number (1.15 × 10⁻⁷ N) that you cannot check for limiting behaviour. Always keep symbols until the end.
⚠️ Superposition is one of the most-tested principles in P212. Every Coulomb's law and electric-field problem involving multiple charges uses it.
⚠️ Symmetry arguments that eliminate components (e.g. "by symmetry, F_y = 0") save enormous amounts of time and are expected on exams.
⚠️ The long-distance approximation (a cluster of sources looks like a single combined source) is a physical insight that examiners test to see if you understand the physics beyond the algebra.
⚠️ "Never plug in numbers until the end" is a running theme in P212.
True or false: at the centre of a symmetric arrangement of equal masses, the net gravitational force on a test mass is zero.
Fill in the blank: the gravitational force between two masses is proportional to 1/r to the power of ____.
True or false: when computing the net force via superposition, you can add the magnitudes of individual forces directly.
Fill in the blank: at very large distances, four masses M clustered together behave like a single mass of ____.
True or false: if you substitute numerical values at the start of the calculation, you lose the ability to check limiting behaviour.
Answers: 1. True. 2. Two. 3. False (you must add vector components). 4. 4M. 5. True.
Q: Why is the net gravitational force on m zero when it sits at the centre of the square?
A: Each pair of diagonally opposite masses exerts forces of equal magnitude but opposite direction. Both pairs cancel, giving F_net = 0.
Q: At the midpoint of the right side of the square, why is F_y = 0?
A: The configuration is symmetric about the line y = 0 at that point. For every force component in the +y direction, there is an equal component in the −y direction from the mirror-image mass.
Q: What is the physical argument behind the long-distance approximation F_x ≈ −4GMmX⁻²?
A: When X ≫ a, the four masses are so close together (relative to the distance X) that they are indistinguishable from a single point mass of 4M. The force is then just Newton's law applied to the total mass.
Q: What is the value of the dimensionless constant C in the long-distance force F_x = CGMmX⁻²?
A: C = −4 (or 4 in magnitude). It equals the number of source masses because they all contribute equally at large distance.
This connects directly to Coulomb's law and electric field calculations for point charge distributions in P212. The symmetry cancellation technique is identical. The long-distance approximation foreshadows the idea of multipole expansions (monopole, dipole, quadrupole), which appear later when studying charge distributions. The potential-energy version of this same configuration is covered in Discussion 1C.
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