D. Gravitational, Electric, and Magnetic Fields · Gravitational, Electric and Magnetic FieldsNewton's Law of Universal Gravitation
D2.1 D2.2About 70 minutes
Learning goals- I can state Newton's law of universal gravitation and explain each symbol in F = Gm₁m₂/r².
- I can use the inverse-square relationship to predict how the force changes when masses or distances change.
- I can calculate the gravitational force between two objects, including planets, moons and people.
- I can find the net gravitational force on an object from two or more masses by adding vectors.
Words to know
gravitational force
The attractive force that every object with mass exerts on every other object with mass.
Newton's law of universal gravitation
F = Gm₁m₂/r²: the force between two masses is proportional to the product of the masses and inversely proportional to the square of the distance between their centres.
universal gravitational constant (G)
G = 6.67 × 10⁻¹¹ N·m²/kg², the same everywhere in the universe. It was first measured with a torsion balance by Henry Cavendish in 1798.
inverse-square law
A relationship in which a quantity is proportional to 1/r². Doubling r makes the quantity one quarter as large.
superposition
The net force (or field) from several sources is the vector sum of the forces (or fields) from each source acting alone.
action at a distance
A force between objects that are not touching. Gravity, electric and magnetic forces all act at a distance; fields are the modern way to explain how.
The apple and the Moon
Drop a pencil and it falls. The Moon, 384 000 km away, never falls onto Earth. Isaac Newton's big idea was that these are the same force. The Moon is falling toward Earth all the time, but it is also moving sideways so fast that it keeps missing. In 1687 Newton published one law that covers falling pencils, the Moon, the planets and (as we now know) galaxies.
In SPH3U you used Fg = mg with g = 9.8 N/kg near Earth's surface. In this unit you will find out where that 9.8 comes from, why it changes as you go up, and how the same idea leads to electric and magnetic fields.
Remember from Grade 11: a force is a push or pull measured in newtons (N), forces add as vectors, and Newton's third law says forces come in equal and opposite pairs.
The law
Every particle of matter in the universe attracts every other particle 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.
In symbols:
F = G m₁ m₂ / r²
- F is the size of the gravitational force on each object, in newtons (N).
- m₁ and m₂ are the two masses, in kilograms (kg).
- r is the distance between their centres, in metres (m). For a planet or a uniform sphere, measure from the centre, not from the surface.
- G = 6.67 × 10⁻¹¹ N·m²/kg² is the universal gravitational constant.
The force is always attractive, and it acts along the line joining the two centres. By Newton's third law, the force on the Moon from Earth is exactly as large as the force on Earth from the Moon. The Moon accelerates much more only because its mass is much smaller (a = F/m).
How tiny is G?
G is very small. Two 70 kg students standing 1.0 m apart attract each other with only about 3 × 10⁻⁷ N, roughly the weight of a few grains of fine sand. You never notice it because Earth, with its enormous mass, pulls on each of you about two billion times harder.
That is why G was so hard to measure. In 1798 Henry Cavendish hung a light rod with small lead spheres on a thin fibre and brought large lead spheres close to them. The tiny attraction twisted the fibre by a measurable angle. This torsion balance gave the first value of G, and with it, the mass of Earth. Modern versions of the experiment are still done today, and G is one of the least precisely known fundamental constants.
The inverse-square law
Because r is squared in the denominator:
- double the distance → the force is 1/2² = 1/4 as large;
- triple the distance → 1/3² = 1/9 as large;
- halve the distance → 2² = 4 times as large.
Because F is proportional to each mass, doubling one mass doubles the force, and doubling both multiplies it by 4.
A quick way to handle a change is a ratio: F_new / F_old = (m₁ new/m₁ old)(m₂ new/m₂ old) ÷ (r new/r old)². You do not need G at all.
Centre to centre
Always use the distance between centres. For an object at height h above Earth's surface, r = R_E + h, where R_E = 6.37 × 10⁶ m. Forgetting to add Earth's radius is the most common error in this unit.
Worked exampleExample 1: Earth and Moon
Find the gravitational force between Earth (m = 5.97 × 10²⁴ kg) and the Moon (m = 7.35 × 10²² kg). The distance between their centres is 3.84 × 10⁸ m.
- Write the law: F = G m₁ m₂ / r².
- Substitute: F = (6.67 × 10⁻¹¹)(5.97 × 10²⁴)(7.35 × 10²²) / (3.84 × 10⁸)².
- Numerator: 6.67 × 5.97 × 7.35 = 292.7, and 10⁻¹¹ × 10²⁴ × 10²² = 10³⁵, so the numerator is 2.927 × 10³⁷.
- Denominator: (3.84 × 10⁸)² = 1.4746 × 10¹⁷.
- Divide: F = 2.927 × 10³⁷ / 1.4746 × 10¹⁷ = 1.98 × 10²⁰ N.
F ≈ 1.98 × 10²⁰ N, attractive, along the line joining the centres. The same size of force acts on each body.
Worked exampleExample 2: Using ratios
Two asteroids attract with a force of 2.4 × 10⁶ N. One asteroid's mass is tripled and the distance between their centres is doubled. What is the new force?
- F is proportional to m₁m₂/r². Write the ratio: F_new/F_old = (3)(1) ÷ (2)².
- F_new/F_old = 3/4 = 0.75.
- F_new = 0.75 × 2.4 × 10⁶ N = 1.8 × 10⁶ N.
1.8 × 10⁶ N. No need for G: the ratio method is faster and avoids rounding.
Worked exampleExample 3: Net force in two dimensions
Three small spheres lie on a frictionless table. A (50 kg) is at the origin, B (20 kg) is 2.0 m east of A, and C (30 kg) is 1.5 m north of A. Find the net gravitational force on A.
- Force from B (east): F_AB = G(50)(20)/(2.0)² = (6.67 × 10⁻¹¹)(1000)/4.0 = 1.67 × 10⁻⁸ N [E].
- Force from C (north): F_AC = G(50)(30)/(1.5)² = (6.67 × 10⁻¹¹)(1500)/2.25 = 4.45 × 10⁻⁸ N [N].
- The two forces are perpendicular, so use Pythagoras: F_net = √((1.67 × 10⁻⁸)² + (4.45 × 10⁻⁸)²) = 4.75 × 10⁻⁸ N.
- Direction: θ = tan⁻¹(4.45/1.67) = 69° north of east.
F_net ≈ 4.75 × 10⁻⁸ N [E 69° N]. Gravitational forces obey superposition: find each force as if the others were absent, then add them as vectors.
Superposition: each sphere pulls A toward itself. The two pulls are perpendicular, so the net force is the diagonal of the rectangle they form, found with Pythagoras.Worked exampleExample 4: Where do the pulls balance?
A spacecraft travels from Earth toward the Moon. At what distance from Earth's centre do the two gravitational forces on it cancel?
- Let x be the distance from Earth's centre. Then the distance to the Moon's centre is d − x, with d = 3.84 × 10⁸ m.
- Set the forces equal: G M_E m / x² = G M_M m / (d − x)². G and m cancel.
- Rearrange and take square roots: (d − x)/x = √(M_M/M_E) = √(7.35 × 10²² / 5.97 × 10²⁴) = 0.1110.
- So d/x − 1 = 0.1110, giving x = d/1.1110 = 3.84 × 10⁸ / 1.1110 = 3.46 × 10⁸ m.
About 3.46 × 10⁸ m from Earth's centre, roughly 90% of the way to the Moon. The spacecraft's mass does not matter.
Just for funRight now you are pulling Earth up toward you with exactly the same force that Earth pulls you down: your weight. Earth just doesn't seem to mind.
Hands-on activityModel the inverse-square law with a flashlight
Gravity, light from a small source and (as you will see) electric force all spread out in three dimensions, which is why they follow an inverse-square law.
You need: a small flashlight or phone light, a sheet of graph paper (or paper with a 1 cm grid drawn on it), a piece of cardboard with a 2 cm × 2 cm square hole cut in it (a grown-up should help with the cutting), tape, a ruler and a dark room.
- Tape the graph paper to a wall. Hold the light so it shines through the square hole onto the paper.
- Put the light 10 cm behind the hole, and put the hole 10 cm from the paper (light-to-paper distance 20 cm). Count how many 1 cm squares the bright patch covers.
- Move the paper so the light-to-paper distance is 30 cm, then 40 cm (keep the light 10 cm behind the hole). Count the squares each time.
- Make a table of distance from the light and area lit. Then calculate area ÷ distance².
- The same light energy spreads over the bright area, so the brightness per square is proportional to 1/area.
What to look for: the patch side grows in proportion to the distance from the light, so the area grows with distance². At 20 cm the patch is about 4 cm × 4 cm (16 squares), at 30 cm about 6 cm × 6 cm (36 squares), at 40 cm about 8 cm × 8 cm (64 squares). Area ÷ distance² stays about the same (0.04 cm²/cm²), so brightness falls off as 1/r², just like gravity.
Your turn
1Write Newton's law of universal gravitation and give the SI unit of each symbol.
F = Gm₁m₂/r². F in newtons (N), m₁ and m₂ in kilograms (kg), r in metres (m), G in N·m²/kg².
2Find the gravitational force between two 70 kg students whose centres are 1.0 m apart.
F = (6.67 × 10⁻¹¹)(70)(70)/(1.0)² = 3.3 × 10⁻⁷ N.
3Find the gravitational force between the Sun (1.99 × 10³⁰ kg) and Earth (5.97 × 10²⁴ kg) if their centres are 1.50 × 10¹¹ m apart.
F = (6.67 × 10⁻¹¹)(1.99 × 10³⁰)(5.97 × 10²⁴)/(1.50 × 10¹¹)² = 3.52 × 10²² N.
4Two objects attract with force F. What is the new force if the distance between them becomes one third as large?
F is proportional to 1/r². r → r/3 gives 1/(1/3)² = 9, so the force becomes 9F.
5Two objects attract with force F. Both masses are doubled and the distance is doubled. What is the new force?
(2)(2)/(2)² = 4/4 = 1, so the force is still F.
6Use F = Gm₁m₂/r² to find the force of gravity on a 60 kg person standing on Earth's surface (M_E = 5.97 × 10²⁴ kg, R_E = 6.37 × 10⁶ m). Compare with mg.
F = (6.67 × 10⁻¹¹)(5.97 × 10²⁴)(60)/(6.37 × 10⁶)² = 589 N. With g = 9.8 N/kg, mg = 588 N: the same, to rounding. Fg = mg is just the law of universal gravitation near Earth's surface.
7How far apart must the centres of two 1000 kg spheres be for them to attract with 1.0 × 10⁻⁶ N?
r = √(Gm₁m₂/F) = √((6.67 × 10⁻¹¹)(1000)(1000)/(1.0 × 10⁻⁶)) = √66.7 = 8.2 m.
More practice: warm-up
1Is the gravitational force ever repulsive?
No. Gravity between masses is always attractive.
2If the distance between two masses doubles, the force becomes ___ as large.
One quarter (1/2² = 1/4).
3If one mass is tripled (distance unchanged), the force becomes ___ as large.
4Earth pulls on you with 600 N. How hard do you pull on Earth?
600 N, upward toward you (Newton's third law).
5What is the value of G, with units?
More practice: on your own
1A 1500 kg satellite is 2000 km above Earth's surface. Find the gravitational force on it.
r = 6.37 × 10⁶ + 2.00 × 10⁶ = 8.37 × 10⁶ m. F = (6.67 × 10⁻¹¹)(5.97 × 10²⁴)(1500)/(8.37 × 10⁶)² = 8.53 × 10³ N (about 8.5 kN). Remember to add Earth's radius.
2Two spheres attract with 4.0 × 10⁻⁹ N when their centres are 0.50 m apart. At what separation is the force 1.0 × 10⁻⁹ N?
The force is one quarter as large, so the distance must be doubled: 1.0 m. (F ∝ 1/r², so r ∝ 1/√F, and √4 = 2.)
3Find the mass of an object that is attracted to a 5.0 kg mass with a force of 3.0 × 10⁻¹⁰ N when their centres are 0.20 m apart.
m = Fr²/(G × 5.0) = (3.0 × 10⁻¹⁰)(0.040)/((6.67 × 10⁻¹¹)(5.0)) = 0.036 kg (36 g).
4A 1000 kg spacecraft is on the line from Earth to the Moon, 2.00 × 10⁸ m from Earth's centre. Find the net gravitational force on it from Earth and the Moon.
From Earth: G(5.97 × 10²⁴)(1000)/(2.00 × 10⁸)² = 9.95 N toward Earth. From the Moon: distance = 3.84 × 10⁸ − 2.00 × 10⁸ = 1.84 × 10⁸ m, F = G(7.35 × 10²²)(1000)/(1.84 × 10⁸)² = 0.145 N toward the Moon. Net = 9.81 N toward Earth.
5Explain why a feather and a hammer fall with the same acceleration on the Moon even though the Moon pulls harder on the hammer.
The force is proportional to the falling object's mass (F = GMm/r²), and a = F/m, so the mass m cancels: a = GM/r² for both objects. With no air on the Moon, nothing else acts, so they fall together.
Challenge
1The Sun is about 390 times as far from the Moon as Earth is, but about 330 000 times as massive as Earth. Which pulls harder on the Moon: the Sun or Earth? By roughly what factor?
Ratio = (330 000)/(390)² = 330 000/152 100 ≈ 2.2. The Sun pulls on the Moon about twice as hard as Earth does. (Using the data: Sun on Moon ≈ 4.3 × 10²⁰ N, Earth on Moon ≈ 2.0 × 10²⁰ N.) The Moon still orbits Earth because Earth and Moon orbit the Sun together, accelerating almost equally toward it.
2Three 2.0 kg masses sit at the corners of an equilateral triangle with sides 0.30 m. Find the size of the net gravitational force on any one mass.
Each pair force: F = (6.67 × 10⁻¹¹)(2.0)(2.0)/(0.30)² = 2.96 × 10⁻⁹ N. The two forces on one mass are 60° apart. By symmetry the net is along the bisector: F_net = 2F cos 30° = 2(2.96 × 10⁻⁹)(0.866) = 5.1 × 10⁻⁹ N, pointing toward the centre of the triangle.
3A planet has twice Earth's mass and twice Earth's radius. How does the gravitational force on you at its surface compare with your weight on Earth?
F ∝ M/R²: (2)/(2)² = 1/2. You would weigh half as much there.
Quiz
1. Two masses attract with a force F. If the distance between their centres is halved, the force becomes
2. Which quantity in F = Gm₁m₂/r² is the same everywhere in the universe?
3. A 2 kg book rests on Earth. Compared with the force Earth exerts on the book, the force the book exerts on Earth is
4. A satellite orbits 6.37 × 10⁶ m above Earth's surface (one Earth radius up). The gravitational force on it is what fraction of its weight on the surface?
5. Henry Cavendish's torsion balance experiment (1798) was the first to measure
6. Two forces act on a mass: 3.0 × 10⁻⁸ N east and 4.0 × 10⁻⁸ N north. The net force has size
Big idea (D2.2): one law, F = Gm₁m₂/r², describes the pull between any two masses. The key skills are substituting carefully in scientific notation, using centre-to-centre distance, and reasoning with ratios.
How to help: Ask your teen to explain out loud what happens to the force when the distance doubles or triples, before touching a calculator. Then have them check one worked example on the calculator, entering powers of ten with the EXP or ×10ˣ key (not by typing × 10 ^).
Common difficulties:
- Forgetting to add Earth's radius to the altitude (r = R_E + h).
- Squaring only the number and not the power of ten, or the other way round: (3.84 × 10⁸)² = 1.47 × 10¹⁷.
- Thinking the bigger object exerts the bigger force. The forces are always equal; the effects (accelerations) differ.
- Adding forces from two sources as numbers instead of vectors.
The flashlight activity uses a dark room and a cutting step: help with the cardboard hole.