Newton's Law of Universal Gravitation

Equivalent forms

\vec{F} = -\frac{GMm}{r^2}\hat{r}

g = \frac{GM}{r^2}

Unified terrestrial and celestial mechanics — the same law that drops an apple steers the planets.

Symbol	Name	SI unit	Dimension	Range
$F$	Gravitational Forceoutput Attractive force between two masses	N	MLT^-2	0 – 100
$G$	Gravitational Constant Universal gravitational constant	m³·kg⁻¹·s⁻²	L³M⁻¹T⁻²	6.674e-11 – 6.674e-11
$M$	Mass 1 Mass of the first body (e.g., planet)	kg	M	1 – 1e+30
$m$	Mass 2 Mass of the second body (e.g., person)	kg	M	0.1 – 1000
$r$	Distance Distance between the centers of the two masses	m	L	1 – 100000000

Unit systems

Symbol	SI	CGS	Imperial
$F$	N	dyn	lbf
$G$	m³·kg⁻¹·s⁻²	cm³·g⁻¹·s⁻²	ft³·slug⁻¹·s⁻²
$M$	kg	g	slug
$m$	kg	g	slug
$r$	m	cm	ft

Where it holds

Valid for non-relativistic, weak-field gravity. Fails near black holes or at speeds approaching 299792458 m/s — use General Relativity instead.

Dimensional analysis

F \to [MLT^{-2}]

G\cdot M\cdot m/r^{2} \to [L^{3}M^{-1}T^{-2}]\cdot [M]\cdot [M]\cdot [L^{-2}] = [MLT^{-2}]

Both sides are

[MLT^{-2}]

= Newton.

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Discovery

Isaac Newton · 1687

Inspired by the falling apple and the Moon's orbit, Newton showed the same force governs both in Principia.

Try this

How much weaker is gravity on Mars compared to Earth? Calculate it from first principles.

Compare g on Mars (M=6.39e23 kg, r=3.39e6 m) vs Earth (M=5.97e24 kg, r=6.37e6 m) using F = GMm/r² to find surface gravity.

Research status: stable

Real-world applications

Satellite orbit calculations
Tidal force predictions
Planetary mass estimation
GPS relativistic corrections

Common misconceptions

Gravity does not require contact — it acts at a distance
Astronauts in orbit are not weightless — they are in freefall
Gravity is not 'turned off' in space — it just weakens with distance

Experimental verification

Cavendish experiment (1798) measured G using a torsion balance. Modern measurements use atom interferometry.

Derivation

Newton postulated that the force causing an apple to fall and the force keeping the Moon in orbit are the same, proportional to both masses and inversely proportional to the square of the distance.

Confirmed by showing the Moon's centripetal acceleration matches

g/r^{2}

scaling from Earth's surface.

Limiting cases

r \to \infty

⟶

F \to 0

Gravity vanishes at infinite separation.

r \to 0

⟶

F \to \infty

The formula diverges — in reality, objects have finite size.

M or

m \to 0

⟶

F \to 0

No mass means no gravitational attraction.

What if…

What if Earth were twice as massive?

Surface gravity doubles to $19.6\,\mathrm{m/s}^{2}$ . You'd weigh twice as much. Jumping height halves.

What if you're at the ISS altitude (408 km)?

$r = 6779\,\mathrm{km}$ . $g = GM/r^{2} = 8.67\,\mathrm{m/s}^{2}$ — still 88% of surface gravity. Astronauts float because they're in freefall, not because gravity vanishes.

What if G were

10\times

larger?

Stars would burn faster, planets would be denser, and life as we know it couldn't exist. G is finely tuned.

1

Your weight from first principles

Given ·

G:: 6.674e-11
M:: 5.97e+24
m:: 70
r:: 6371000

Find · F

Steps

$G = 6.674 \times 10^{-11}$ , M_Earth $= 5.97 \times 10^{24} kg$ , $r = 6.371 \times 10^{6} m$
Numerator: $6.674e-11 \times 5.97e24 \times 70 = 2.787 \times 10^{16}$
Denominator: $(6.371e_{6})^{2} = 4.059 \times 10^{13}$
$F = 2.787e16 / 4.059e13 = 686.6\,\mathrm{N}$

Result ·

F = GMm/r^{2} = 6.674e-11 \times 5.97e24 \times 70 / (6.371e_{6})^{2} = 686.6\,\mathrm{N} \approx 687\,\mathrm{N}

2

Mars surface gravity

Given ·

G:: 6.674e-11
M:: 6.39e+23
r:: 3390000

Find · g_mars

Steps

M_Mars $= 6.39 \times 10^{23} kg$ , r_Mars $= 3.39 \times 10^{6} m$
$g = GM/r^{2} = 6.674e-11 \times 6.39e23 / (3.39e_{6})^{2}$
$g = 4.264e13 / 1.149e13 = 3.71\,\mathrm{m/s}^{2}$
Mars gravity is 37.9% of Earth' $s 9.8\,\mathrm{m/s}^{2}$

Result ·

g = GM/r^{2} = 6.674e-11 \times 6.39e23 / (3.39e_{6})^{2} = 3.71\,\mathrm{m/s}^{2}

Newton's Second Law→Centripetal Acceleration→Projectile Range→