PhysicsICSE Class 9 · Gravitation

Universal Law of Gravitation: F = Gm₁m₂/r², Ratio Problems and Calculations (ICSE Class 9)

Updated 17 October 2026 · Fact-checked

Every body attracts every other body with a force F = G m₁m₂/r², proportional to the product of the masses and inversely proportional to the square of the distance between their centres. G = 6.67 × 10⁻¹¹ N m² kg⁻² everywhere. The forces on the two bodies are equal and opposite.

Notice the two force arrows are equal and opposite; doubling r divides F by 4, halving r multiplies it by 4, doubling both masses gives 4 times.

Illustration drawn for QuestionBuddy.

Understand Universal Law of Gravitation: F = Gm₁m₂/r² and Ratio Problems

Statement. Every body in the universe attracts every other body with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.

$F = G \frac{m _{1} m _{2}}{r ^{2}}$

$G$ is the universal gravitational constant, $6.67 \times 1 0^{- 11}$ N m² kg⁻², the same everywhere in the universe. It is not $g$ .
$r$ is the distance between the centres of the bodies.
The force is always attractive, acts along the line joining the centres, and is the same size on both bodies (third law), though the lighter one accelerates more.

**Why $G$ has that unit.** From $F = \frac{G m _{1} m _{2}}{r ^{2}}$ : $G = \frac{F r ^{2}}{m _{1} m _{2}}$ , so the unit is $\frac{N m ^{2}}{kg ^{2}}$ .

Why we do not feel the pull of nearby objects. $G$ is tiny. Two 1 kg masses 1 m apart attract with only $6.67 \times 1 0^{- 11}$ N. Gravity is noticed only when at least one body is as massive as a planet.

**Ratio problems (no $G$ needed).** Write the new force as the old force times the change factors:

$\frac{F _{2}}{F _{1}} = \frac{m _{1}^{'} m _{2}^{'}}{m _{1} m _{2}} \times (\frac{r _{1}}{r _{2}})^{2} .$

Quick results: double the distance, force $\frac{1}{4}$ ; half the distance, $4 \times$ ; both masses doubled, $4 \times$ ; one mass tripled and distance tripled, $\frac{3}{9} = \frac{1}{3}$ .

Powers of ten. For numerical work keep the mantissas and the powers separate:
1. multiply the masses (add the powers);
2. square $r$ (double the power);
3. divide, then multiply by $G$ .
Quote the answer in standard form with the unit N.

Point of zero force. For masses $m_{1}$ and $m_{2}$ at distance $d$ , a small body between them feels no resultant force where the two pulls are equal: $\frac{m _{1}}{x ^{2}} = \frac{m _{2}}{( d - x ) ^{2}}$ . Taking square roots gives $\frac{x}{d - x} = \frac{m _{1}}{m _{2}}$ . The point lies nearer the lighter mass.

What the law explains. The fall of bodies, the Moon going round the Earth, the planets going round the Sun, ocean tides and the Earth holding on to its atmosphere. Gravity provides the force towards the centre needed to keep a body on a circular path.

Gravity vs gravitation. Gravitation is the attraction between any two bodies; gravity is the Earth's pull on a body near it.

Universal Law of Gravitation: F = Gm₁m₂/r² and Ratio Problems: formulas and key facts

Universal law: $F = G \frac{m _{1} m _{2}}{r ^{2}}$; r between centres; attractive.
Gravitational constant: $G = 6.67 \times 1 0^{- 11} N m^{2} kg^{- 2}$; Same everywhere.
Ratio form: $\frac{F _{2}}{F _{1}} = \frac{m _{1}^{'} m _{2}^{'}}{m _{1} m _{2}} (\frac{r _{1}}{r _{2}})^{2}$; No G needed.
Null point: $\frac{x}{d - x} = \frac{m _{1}}{m _{2}}$; x measured from m₁.

How to solve Universal Law of Gravitation: F = Gm₁m₂/r² and Ratio Problems questions

Decide first whether G is needed.

1If the question only compares forces, use the ratio form and cancel G.
2Otherwise convert masses to kg and distances to m, and write them in standard form.
3Substitute in $F = G m_{1} m_{2} / r^{2}$ , handling the powers of ten separately.
4Give the answer in N with a sensible number of significant figures.
5Remember the force on each body is the same size.

Universal Law of Gravitation: F = Gm₁m₂/r² and Ratio Problems in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Two friends on a bench

Suppose two students of mass 50 kg and 60 kg sit 1 m apart (centre to centre).

What is the gravitational force between them?

Try it first, then show the solutionSolution

$F = \frac{6.67 \times 1 0 ^{- 11} \times 50 \times 60}{1 ^{2}} = 2.0 \times 1 0^{- 7}$ N.

Answer: About $2 \times 1 0^{- 7}$ N, far too small to feel.

Memory hook: G is tiny, so gravity needs a planet.

Tricks to remember Universal Law of Gravitation: F = Gm₁m₂/r² and Ratio Problems

Short memory aids. Say each one aloud once, then use it on a question the same day.

Product up, square down
Masses multiply on top; distance squares at the bottom.
Use it when: Setting up the formula.
G is a constant, g changes
Capital G is universal; small g depends on the planet.
Use it when: Avoiding confusion.
Square root for the null point
Distances from the two bodies are in the ratio √m₁ : √m₂.
Use it when: Zero-force problems.

Universal Law of Gravitation: F = Gm₁m₂/r² and Ratio Problems: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Both masses and distance doubled · 1 mark · MCQBoard level

Both masses are doubled and the distance between them is also doubled. The gravitational force becomes:
(a) four times
(b) twice
(c) unchanged
(d) one-fourth

Try it first, then show the solutionSolution

Masses give a factor $2 \times 2 = 4$ ; distance gives $\frac{1}{2 ^{2}} = \frac{1}{4}$ . Net factor 1.

Answer: (c) unchanged

Example 2: Two small masses · 2 marks · Short answerBoard level

Find the gravitational force between masses of 5 kg and 8 kg whose centres are 0.5 m apart ( $G = 6.67 \times 1 0^{- 11}$ N m² kg⁻²).

Try it first, then show the solutionSolution

$F = \frac{G m _{1} m _{2}}{r ^{2}} = \frac{6.67 \times 1 0 ^{- 11} \times 5 \times 8}{0. 5 ^{2}}$ (1)
$= \frac{2.668 \times 1 0 ^{- 9}}{0.25} \approx 1.07 \times 1 0^{- 8}$ N. (1)

Answer: About $1.07 \times 1 0^{- 8}$ N

Example 3: Two planets · 3 marks · Short answerHarder board level

Two planets of masses $3 \times 1 0^{22}$ kg and $5 \times 1 0^{23}$ kg have their centres $2 \times 1 0^{8}$ m apart. Find the force between them.

Try it first, then show the solutionSolution

$m_{1} m_{2} = 3 \times 1 0^{22} \times 5 \times 1 0^{23} = 1.5 \times 1 0^{46}$ . (1)
$r^{2} = 4 \times 1 0^{16}$ . (½)
$F = \frac{6.67 \times 1 0 ^{- 11} \times 1.5 \times 1 0 ^{46}}{4 \times 1 0 ^{16}} \approx 2.5 \times 1 0^{19}$ N. (1½)

Answer: About $2.5 \times 1 0^{19}$ N

Example 4: Several changes at once · 3 marks · Short answerHarder board level

The force between two bodies A and B is 18 N. The mass of A is tripled, the mass of B is halved and the distance is tripled. Find the new force.

Try it first, then show the solutionSolution

$\frac{F _{2}}{F _{1}} = \frac{3 \times \frac{1}{2}}{1} \times (\frac{1}{3})^{2} = \frac{1.5}{9} = \frac{1}{6}$ . (2)
$F_{2} = \frac{18}{6} = 3$ N. (1)

Answer: 3 N

Example 5: Equal force, unequal motion · 4 marks · Case-basedHOTS / competency

Bodies of masses 2 kg and 8 kg are 0.4 m apart in space, far from other bodies.
(a) Find the gravitational force between them.
(b) Find the acceleration of each.
(c) State the ratio of the accelerations and the reason.
(d) The distance is increased to 1.6 m. Find the new force.

Try it first, then show the solutionSolution

(a) $F = \frac{6.67 \times 1 0 ^{- 11} \times 16}{0.16} = 6.67 \times 1 0^{- 9}$ N. (1)
(b) $a_{1} = \frac{F}{2} = 3.3 \times 1 0^{- 9}$ m s⁻²; $a_{2} = \frac{F}{8} = 8.3 \times 1 0^{- 10}$ m s⁻². (1)
(c) $a_{1} : a_{2} = 4 : 1$ , because the forces are equal and $a \propto \frac{1}{m}$ . (1)
(d) Distance ×4, so $F = \frac{6.67 \times 1 0 ^{- 9}}{16} \approx 4.2 \times 1 0^{- 10}$ N. (1)

Answer: (a) 6.67 × 10⁻⁹ N (b) 3.3 × 10⁻⁹ and 8.3 × 10⁻¹⁰ m s⁻² (c) 4:1 (d) 4.2 × 10⁻¹⁰ N

Where students lose marks in Universal Law of Gravitation: F = Gm₁m₂/r² and Ratio Problems

The slips examiners see most often, and what to write instead.

Mistake: Distance not squared
The law is inverse square.
Wrong working
F = Gm₁m₂/r
Right working
F = Gm₁m₂/r²
Do this: Write the formula first.
Usually costs: The answer.
Mistake: Radius used instead of distance between centres
r is centre to centre.
Wrong working
Two spheres of radius 0.1 m touching: r = 0.1 m.
Right working
r = 0.2 m.
Do this: Add the radii when bodies touch.
Usually costs: The answer.
Mistake: Powers of ten mishandled
Powers add on multiplication and double on squaring.
Wrong working
(4 × 10⁸)² = 1.6 × 10¹⁶
Right working
1.6 × 10¹⁷
Do this: Square the number and double the power separately.
Usually costs: 1 to 2 marks.

Exam tips

Write the law with symbols explained before substituting.
Use the ratio method when no masses are given.
Give G with its unit when you quote it.
State that the force is the same on both bodies when asked.

Practice questions on Universal Law of Gravitation: F = Gm₁m₂/r² and Ratio Problems

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Which statement about the gravitational constant G is correct?
(a) It depends on the masses of the two bodies
(b) It has different values on the Moon and the Earth
(c) It has the same value everywhere in the universe
(d) Its SI unit is m s⁻²
Hint
Universal.
Answer
(c)
Q2 · 1 mark · Very short answerBoard level
State the SI unit of G.
Hint
From $G = F r^{2} / (m_{1} m_{2})$ .
Answer
N m² kg⁻².
Q3 · 2 marks · Short answerHarder board level
The force between two bodies is 4 N. One mass is tripled and the distance between them is reduced to one-third. Find the new force.
Hint
Ratio method.
Answer
$\frac{F _{2}}{F _{1}} = 3 \times 3^{2} = 27$ ; $F_{2} = 4 \times 27 = 108$ N.
Q4 · 3 marks · Short answerHarder board level
A planet of mass $8 \times 1 0^{24}$ kg and its moon of mass $6 \times 1 0^{22}$ kg are $4 \times 1 0^{8}$ m apart. Find the force on the planet and on the moon.
Hint
Same force on both.
Answer
$F = \frac{6.67 \times 1 0 ^{- 11} \times 8 \times 1 0 ^{24} \times 6 \times 1 0 ^{22}}{( 4 \times 1 0 ^{8} ) ^{2}} = \frac{3.2 \times 1 0 ^{37}}{1.6 \times 1 0 ^{17}} \approx 2.0 \times 1 0^{20}$ N on each (third law).
Q5 · 2 marks · Assertion-reasonHarder board level
Assertion (A): The gravitational force between two 1 kg masses 1 m apart is extremely small.
Reason (R): The value of G is very small.
(a) Both true, R explains A.
(b) Both true, R does not explain A.
(c) A true, R false.
(d) A false, R true.
Hint
F = G here.
Answer
(a). $F = G \times \frac{1 \times 1}{1 ^{2}} = 6.67 \times 1 0^{- 11}$ N.
Q6 · 3 marks · Short answerHOTS / competency
Bodies of mass $m$ and $16 m$ are 5 m apart. At what point on the line joining them would a small body feel no resultant gravitational force? Give the distance from the lighter body.
Hint
Equal pulls.
Answer
$\frac{x}{5 - x} = \frac{m}{16 m} = \frac{1}{4}$ , so $4 x = 5 - x$ and $x = 1$ m from the lighter body.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Value of G?Show answer
6.67 × 10⁻¹¹ N m² kg⁻².
2.Is gravitational force attractive or repulsive?Show answer
Always attractive.
3.Distance doubled, force becomes?Show answer
One-fourth.
4.Is G the same on the Moon?Show answer
Yes; it is universal.
5.What does r measure?Show answer
Distance between the centres of the bodies.

Universal Law of Gravitation: F = Gm₁m₂/r² and Ratio Problems: frequently asked questions

What is the universal law of gravitation?

Every body attracts every other body with a force F = Gm₁m₂/r², proportional to the product of the masses and inversely proportional to the square of the distance between their centres.

Why can we not feel the gravitational pull of everyday objects?

Because G is only 6.67 × 10⁻¹¹ N m² kg⁻², so for ordinary masses the force is minute.

What is the difference between G and g?

G is the universal gravitational constant (the same everywhere); g is the acceleration due to gravity, which depends on the planet.