doPhysics.in
All chapters
Gravitational Force and Field · Lesson 2 of 5

Superposition of Gravitational Forces

13 min read3 ideas5 checks2 figures to playBy Asheshanand Acharya
What you will learn 0 / 3
The principle of superposition
Maths you need
Look around
A plumb line at the foot of the Himalaya

In the nineteenth century the surveyors of the Great Trigonometrical Survey mapped India with great care. They used a plumb line, a weight on a thread, to find the true vertical. Near the Himalaya the plumb line did not hang quite straight towards the Earth's centre. It leaned very slightly towards the mountains.

you will see how the pull of a mountain range adds to the pull of the whole Earth, and why a plumb line leans.

Many masses, one body

A body is rarely pulled by just one other mass. A satellite feels the Earth, the Moon and the Sun at once. A plumb line near the mountains feels the whole Earth and the mountains beside it. How do these pulls combine? Experiment gives a simple answer. Each pair of masses interacts as if the other masses were not there. The pulls then add as vectors.

principle 1.1
Principle of superposition

The net gravitational force on a body is the vector sum of the forces exerted on it by every other mass. Each of those forces is the one the pair would have alone: \(\vec F = \vec F_1 + \vec F_2 + \vec F_3 + \cdots\).

This principle of superposition has two parts, and both matter. The first is independence: a third mass nearby does not change the pull between the first two. No mass can shield another, and nothing blocks gravitation. The second is vector addition: the separate pulls add by the triangle rule, never as plain numbers.

RememberNet gravitational force on a mass m
\[ \vec F = \sum_i \dfrac{G\,m\,m_i}{r_i^2}\,\hat r_i \]
Holds whenpoint masses (or uniform spheres) acting on a point mass \(m\); \(\hat r_i\) is the unit vector from \(m\) towards the \(i\)-th mass.
Symbols\(\vec F\) net force on \(m\), \(m_i\) the \(i\)-th other mass, \(r_i\) its distance from \(m\), \(\hat r_i\) unit vector from \(m\) towards \(m_i\)
Unitsforces in N, masses in kg, distances in m
CheckpointUnanswered

Three masses pull on a body. Two of the three pulls are equal and opposite. What is the net force on the body?

Net force on a mass on a line

Masses on a line

The simplest case has all the masses on one straight line. Then every pull on a chosen body lies along that line, towards the mass that exerts it. Choose one direction as positive. A pull towards a mass on the positive side is positive, and a pull towards a mass on the other side is negative. Then add the signed pulls.

Two pulls on one massThree small spheres on a line: 6 kilograms at x = 0, 2 kilograms at x = 2 metres and 4 kilograms at x = 3 metres. On the 2 kilogram sphere the 6 kilogram sphere pulls to the left with 3G newtons and the 4 kilogram sphere pulls to the right with 8G newtons. The net force, drawn above, is 5G newtons to the right, towards the 4 kilogram sphere. O123x (m) 6 kg 4 kg 2 kg 3G 8G net 5G
Fig 2.1Spheres of \(6\,\text{kg}\), \(2\,\text{kg}\) and \(4\,\text{kg}\) at \(x = 0\), \(2\,\text{m}\) and \(3\,\text{m}\). On the \(2\,\text{kg}\) sphere the left sphere pulls with \(3G\) and the right sphere with \(8G\), in newtons. The net force, drawn above, is \(5G\) to the right.
ExampleThe pull on the middle sphere

The spheres of Fig 2.1 are small. Find the net gravitational force on the \(2\,\text{kg}\) sphere. Take \(G = 6.67 \times 10^{-11}\,\text{N m}^2/\text{kg}^2\).

Reveal one step at a time and finish the rest yourself.
CheckpointUnanswered

A \(2\,\text{kg}\) mass sits at \(x = 0\) and an \(8\,\text{kg}\) mass at \(x = 3\,\text{m}\). A \(1\,\text{kg}\) mass is placed at \(x = 1\,\text{m}\). What is the net gravitational force on it?

Forces in a plane and symmetry
Maths you need

Pulls in a plane

When the masses are not on one line, the pulls point in different directions. Resolve each pull into \(x\) and \(y\) components, add the components separately, and then combine them. Before any arithmetic, look for symmetry. Two equal pulls at equal angles on either side of a line add along that line. Their sideways parts cancel.

Adding three pulls at a cornerFour equal masses m sit at the corners of a square of side a. The mass at the lower left corner is pulled along the two sides by F1 = Gm²/a² each and along the diagonal by F1/2. The chips add the pulls one step at a time: first the two side pulls, which make √2 F1 along the diagonal, then the diagonal pull, which gives a net force of 1.91 Gm²/a² towards the centre of the square. m m m F1 F1 F1/2 m a F1 = Gm²/a² along each side
Fig 2.2Four equal masses \(m\) at the corners of a square of side \(a\). The mass at the lower left is pulled along both sides by \(F_1 = Gm^2/a^2\), and along the diagonal by \(F_1/2\). Choose the chips in order to add the pulls one step at a time.

The diagonal mass is \(\sqrt2\,a\) away, so its pull is \(Gm^2/2a^2\), half of \(F_1\). The two side pulls are equal and at right angles. By symmetry their sum lies along the diagonal, with size \(\sqrt2\,F_1\). The diagonal pull lies along the same line. So the net force points along the diagonal, towards the centre of the square, with size \((\sqrt2 + \tfrac12)\,Gm^2/a^2 = 1.91\,Gm^2/a^2\).

RememberNet force at a corner of a square of equal masses
\[ F = \left(\sqrt2 + \tfrac12\right)\dfrac{G m^2}{a^2} \approx 1.91\,\dfrac{G m^2}{a^2} \]
Holds whenfour equal point masses at the corners of a square; the force on any one of them, directed towards the centre of the square.
Symbols\(m\) each mass, \(a\) side of the square, \(F\) net force on one corner mass
Units\(F\) in N, \(m\) in kg, \(a\) in m

The same thinking settles the centre of the square at once. A mass placed at the centre is pulled equally by all four corners. The pulls come in opposite pairs, so the net force on it is zero. This holds at the centre of every regular arrangement of equal masses: an equilateral triangle, a regular hexagon or a ring.

Checkpoint · numericUnanswered

Four \(2\,\text{kg}\) spheres sit at the corners of a square of side \(0.5\,\text{m}\). Take \(G = 6.67 \times 10^{-11}\,\text{N m}^2/\text{kg}^2\). Find the net force on one sphere, in units of \(10^{-9}\,\text{N}\).

× 10⁻⁹ N± 2 %
CheckpointUnanswered

Three equal masses \(m\) sit at the corners of an equilateral triangle of side \(a\). What is the net gravitational force on one of them?

Checkpoint · tap in orderUnanswered

Four equal masses sit at the corners of a square. Rank these forces on the lower left mass, from the largest to the smallest.

Back to the plumb line

The plumb line near the Himalaya is pulled by the whole Earth, almost straight down, and by the mountains, a little sideways. It hangs along the vector sum of the two. The sideways pull is tiny compared with the Earth's, so the tilt is a very small angle. Yet the surveyors could measure it, because their instruments read angles of a few seconds of arc.

Story · John Henry Pratt, Calcutta, 1855
The mountains that pulled too little

In 1855 John Henry Pratt, Archdeacon of Calcutta, added up the pull of the Himalaya on the plumb lines of the Great Trigonometrical Survey. He cut the mountains into blocks and summed their pulls, as this lesson sums point masses. His answer was about three times the tilt the surveyors had measured. The mountains pulled too little for their size. The explanation came soon after: rock beneath the mountains is less dense than elsewhere. The high ranges float on the denser rock below, like wood on water.

Exam traps
  1. Adding the sizes of pulls that point in different directions.
    Add pulls as vectors: by components, or by symmetry along a line.
  2. Measuring each distance from the origin instead of from the body pulled.
    Every \(r\) is the distance from the body the forces act on.
  3. Thinking a nearby mass screens or weakens another mass's pull.
    Each pair interacts as if the others were absent; nothing blocks gravitation.