Difficulty: Introductory review | Prerequisites: P211 Newton's law of gravitation, vector components
This topic drills the superposition principle for forces using gravity as the model. You add up the individual force vectors from each source mass to find the net force on a test mass. In P212 you will do exactly the same thing with Coulomb's law and point charges, so this is direct preparation for electric-field problems. If you are comfortable resolving vectors into x and y components, you have the tools you need.
To find the net gravitational force on a test mass from several source masses, compute each individual force vector and add them component by component. Symmetry can kill entire components before you do any algebra. At large distances the whole cluster looks like a single object with the total mass, which gives a quick approximation and a useful limiting-behaviour check.
Superposition (of forces)
The principle that the net force on a particle is the vector sum of the individual forces from every other particle, each calculated as though the others were absent. In simple terms, you work out each force one at a time and then add them all up.
Newton's law of universal gravitation
The gravitational force between two masses m1 and m2 separated by distance r12 is F = G m1 m2 / r12^2, directed along the line joining them. Think of it as: bigger masses or shorter distances mean a stronger pull.
Unit vector (r-hat)
A vector of length 1 pointing from one object to another, used to specify direction without carrying magnitude. In simple terms, it tells you which way the force points.
Limiting behaviour
The practice of checking your symbolic answer by sending variables to extremes (zero, infinity) and confirming the physics makes sense. Covered in Discussion 1A as well; it appears again here with four distinct checks.
Long-distance approximation
When the test mass is very far from a cluster of sources compared with the cluster's own size, the cluster acts as if all its mass were concentrated at one point. Think of it as zooming out until the square of masses is just a dot.
Four equal masses M sit at the corners of a square with side length a. A fifth particle of mass m moves under their combined gravitational pull. You need to find the net force on m at several positions.
By symmetry, each corner mass exerts a force of the same magnitude on m.
The four force vectors point from the centre toward each corner, forming two pairs of equal and opposite vectors.
They cancel completely: Fx = 0 and Fy = 0.
Lesson: always sketch the vectors first. Symmetry can eliminate work before any calculation.
Place m at (a/2, 0). The two right-hand masses (M1 and M2) are each a distance a/2 from m, sitting directly above and below it.
Their x-components cancel (they are symmetric about the x-axis). Their y-components also cancel for the same reason. So the two nearest masses contribute nothing to the net force.
The two left-hand masses (M3 and M4) are each at distance r = sqrt((a/2)^2 + a^2) = a sqrt(5)/2.
Each exerts a force of magnitude G M m / (5a^2/4) = 4 G M m / (5a^2).
Resolving into components: cos(theta) = a / (a sqrt(5)/2) = 2/sqrt(5).
The y-components of M3 and M4 cancel (symmetry again), leaving only an x-component pointing toward the centre.
Final result: Fy = 0, Fx = -16 G M m / (5 sqrt(5) a^2).
Four checks on the expression for Fx:
Fx increases without bound as m or M go to infinity. (Correct: more mass means more force.)
Fx goes to zero as m or M go to zero. (Correct: no mass, no gravitational force.)
Fx increases without bound as a goes to zero. (Correct: the masses crowd together and the force diverges.)
Fx goes to zero as a goes to infinity. (Correct: infinitely separated masses exert no force.)
Key lesson: never substitute numbers until the end. With M = 3 kg, m = 1 kg, a = 5 cm, the numerical answer is 1.15 x 10^-7 N. You cannot check any limiting behaviour from a number.
When m is at distance X >> a along the x-axis, the four corner masses look like a single point mass of 4M.
The net force is approximately Fx = -4 G M m / X^2 (the minus sign indicates attraction).
The constant C in the approximation F = C G M m X^-2 is therefore C = -4.
Quantity | Formula | Notes |
|---|---|---|
Gravitational force (two masses) | F = G m1 m2 / r^2 | Directed along the line joining them |
Force at centre of square | F_net = 0 | Symmetry cancellation |
Force at midpoint of right side | Fx = -16 G M m / (5 sqrt(5) a^2), Fy = 0 | Only the two far masses contribute |
Long-distance approximation | Fx approx -4 G M m / X^2 | Valid when X >> a |
Students sometimes add force magnitudes without resolving into components. You must break each force into x and y parts before summing.
A common error is forgetting that the two nearest masses at the midpoint of the right side contribute zero net force (their contributions cancel by symmetry). Students waste time computing what symmetry eliminates for free.
Students often plug in numbers at the start. This makes limiting-behaviour checks impossible. Keep everything symbolic until the final step.
Superposition of inverse-square forces is exactly how you compute the electric field from a collection of point charges in P212. Replace G m1 m2 with k q1 q2 (Coulomb's law) and the mathematics is identical. The symmetry arguments transfer directly.
Superposition is the single most-used technique in the first half of P212. Master it here with gravity and it carries over to electric fields and potentials.
Always draw a vector diagram before computing. Label every mass M1 through M4 and every force F1 through F4.
Limiting-behaviour analysis will appear on exams. You should be able to list at least three checks for any symbolic answer.
The long-distance approximation (treating a cluster as a point mass) reappears in P212 as the far-field approximation for charge distributions.
True or false: the net gravitational force on a test mass at the exact centre of four equal corner masses is zero. (True)
Fill in the blank: the distance from the midpoint of the right side to a left-hand corner mass is r = a ___/2. (sqrt(5))
True or false: when the test mass is very far from the square, the net force falls off as 1/X^2. (True)
True or false: plugging in numbers early helps you check limiting behaviour. (False)
Fill in the blank: the constant C in the long-distance approximation Fx = C G M m / X^2 is ___. (-4)
Q: Four identical masses M are placed at the corners of a square of side a. A test mass m sits at the centre. What is the net force on m? Explain using a symmetry argument.
A: Each corner mass exerts a force of the same magnitude on m. The four forces form two pairs pointing in opposite directions. Each pair cancels, giving a net force of zero in both x and y.
Q: The test mass is now moved to the midpoint of the right-hand side of the square. Which masses contribute to the net x-component of the force, and why?
A: Only the two left-hand masses (M3 and M4) contribute. The two right-hand masses are equidistant above and below m, so their x-components cancel by symmetry.
Q: State three limiting-behaviour checks for the expression Fx = -16 G M m / (5 sqrt(5) a^2).
A: (1) Fx goes to infinity as m or M go to infinity (more mass, more force). (2) Fx goes to zero as m or M go to zero (no mass, no force). (3) Fx goes to infinity as a goes to zero (masses on top of each other produce infinite force).
Q: Why is the long-distance approximation constant C equal to -4 and not +4?
A: The gravitational force is attractive. At large distance the cluster of four masses pulls the test mass toward it, so the force points in the -x direction. The minus sign encodes the attractive nature of gravity.
This connects directly to Coulomb's law problems in P212: swap G for k and masses for charges, and every technique here (superposition, symmetry cancellation, limiting behaviour) applies unchanged. It also connects to the potential-energy discussion in Question 1C, where the same four-mass setup is analysed using scalar superposition instead of vector superposition.
Superposition principle, Newton's law of gravitation, vector addition, component method, symmetry cancellation, limiting behaviour, long-distance approximation, point-mass approximation, Coulomb's law analogy, P212 Week 1, P211 review, PHYS 212 UIUC, gravitational field, inverse-square law