Gravitational field intensity g = GM - r²: Visual walkthrough

Step 1 — Two masses and the arrow between them

WHAT. We place a big blob of stuff — call its amount of "stuff" $M$ (the source mass, measured in kilograms) — at a point. Some distance away we place a tiny speck, amount $m$ (the test mass). Between their centres we draw a straight line and measure its length $r$ (in metres).

WHY. Before any formula, gravity is just this: how strongly does the blob tug the speck? We need names for the three things that could possibly matter — how much stuff pulls ( $M$ ), how much stuff is pulled ( $m$ ), and how far apart they are ( $r$ ). Everything else is built from these three.

PICTURE. Look at the blue blob $M$ on the left and the small orange dot $m$ on the right. The gray line labelled $r$ is the distance centre to centre — not surface to surface. That detail returns later.

Step 2 — Newton hands us the force

WHAT. Newton's Law of Universal Gravitation says the pulling force $F$ (measured in newtons, N) between the two masses is $F = fixed number of nature G \cdot more stuff, closer = stronger pull \frac{M \cdot m}{r ^{2}}$

Reading it term by term:

$M \cdot m$ on top — double either mass, double the force. More stuff on either end means a bigger tug.
$r^{2}$ on the bottom — push them apart, the force weakens fast (we prove the "squared" visually in Step 6).
$G = 6.67 \times 1 0^{- 11} N m^{2} kg^{- 2}$ — the gravitational constant. It is just the conversion factor that turns "kilograms and metres" into "newtons." It never changes.

WHY. This is our single starting assumption — the one law we take on faith (it is measured, not derived). See Newton's Law of Universal Gravitation. Everything on this page flows out of this one equation.

PICTURE. The orange arrow shows the force on the speck: it points back toward the blob (gravity always attracts). Its length is the force's strength.

Step 3 — The problem: the answer keeps changing

WHAT. Suppose I keep the blob $M$ fixed and the distance $r$ fixed, but I swap the speck for a heavier one — double $m$ . From Step 2, the force $F$ doubles. Swap in a triple speck, force triples.

WHY. This is annoying. The force depends on who is visiting. If I want to describe the location — "how gravitationally dangerous is this spot?" — I do not want an answer that changes every time a different object walks in. I want a number that belongs to the space itself.

PICTURE. Three test masses ( $m$ , $2 m$ , $3 m$ ) sit at the same distance $r$ . Their force arrows have different lengths (1×, 2×, 3×). The spot did not change — only the visitor did. That is the thing we must fix.

Step 4 — The trick: force per kilogram

WHAT. Divide the force by the test mass. We invent a new quantity, the gravitational field intensity $g$ : $g = \frac{F}{m} (newtons per kilogram, N kg^{- 1})$

WHY. In Step 3 the force was proportional to $m$ : double $m$ , double $F$ . So the ratio $F / m$ stays the same for every visitor. Dividing by $m$ is exactly the operation that cancels the visitor and leaves a property of the location. This is the whole idea of a "field": stop describing forces on objects, start describing the space.

PICTURE. The same three test masses from Step 3, but now each force arrow is scaled down by its own mass. All three shrink to the same length — the field $g$ is one arrow, shared by the spot regardless of who stands there.

Step 5 — Substitute and cancel

WHAT. Put Newton's force (Step 2) into the definition of field (Step 4): $g = \frac{F}{m} = \frac{1}{m} \cdot \frac{G M m}{r ^{2}} = \frac{G M m}{r ^{2} m} = \frac{GM}{r ^{2}}$

Term by term in the final result:

$G$ — nature's constant, still along for the ride.
$M$ — the source mass. Only the blob's mass survives on top.
$r^{2}$ — the centre-to-centre distance, squared, on the bottom.
$m$ — gone. It appeared once on top (from Newton's $F$ ) and once on the bottom (from dividing by $m$ ), so it cancels perfectly.

WHY. The cancellation of $m$ is the payoff of Step 4. It proves the field does not care about the visitor — a feather and a bowling ball at the same point feel the identical $g$ .

PICTURE. The $m$ in the numerator and the $m$ in the denominator are drawn on a balance beam, striking each other out, leaving $GM / r^{2}$ standing alone.

Step 6 — WHY the square? The spreading-sphere picture

WHAT. We now see where the $r^{2}$ comes from, not just accept it. Imagine a fixed number of "field lines" bursting out of $M$ in all directions. Every line that leaves $M$ must pierce every sphere drawn around it — the count of lines is fixed. The field strength is the crowding of these lines: how many pierce each square metre.

A sphere of radius $r$ has surface area $A = 4 π r^{2} .$ Same lines, bigger area, so the density (= field) is $g \propto \frac{fixed lines}{4 π r ^{2}} \propto \frac{1}{r ^{2}} .$

WHY $r^{2}$ and not $r$ or $r^{3}$ ? Because we live in three dimensions and a sphere's area grows as the square of its radius. Double $r$ → area ×4 → lines four times as spread out → field quartered. The exponent 2 is literally the "2" in "area of a surface." This is the geometric heart of Gauss's Law for Gravity.

PICTURE. Two nested spheres, radii $r$ and $2 r$ . The same lines cross both, but on the outer sphere they are four times as sparse — arrows drawn one-quarter as dense.

Step 7 — Edge cases: what happens at the extremes?

WHAT. A formula is only trustworthy if we check its boundaries.

$r \to \infty$ (very far away): $g = GM / r^{2} \to 0$ . The pull fades to nothing but never exactly reaches zero — gravity has infinite reach, just vanishingly weak.
$r \to 0$ (approaching a point mass): $g = GM / r^{2} \to \infty$ . The formula blows up. This is fine for an idealised point mass, but a real planet has size — inside it, you must use Variation of g with Altitude and Depth instead, because $r < R$ changes the story.
On a real planet's surface ( $r = R$ ): here $r$ equals the planet's radius $R$ , so $g = GM / R^{2}$ . This is the everyday $9.8 N kg^{- 1}$ .
At altitude $h$ : the distance from the centre is $r = R + h$ , not $h$ . This is the single most common slip.

WHY. Every scenario a reader might meet — deep space, the surface, up a mountain, the mathematical singularity — is covered so no case can ambush you.

PICTURE. A curve of $g$ versus $r$ : it shoots up toward infinity as $r \to 0$ , passes through the surface value at $r = R$ , and decays toward zero as $r$ grows. The surface, altitude, and far-field points are all marked.

Step 8 — When two sources pull at once (vectors)

WHAT. Fields add as vectors — by direction, not by plain numbers. If Earth's field at a point is $g_{E}$ pointing left and the Moon's is $g_{M}$ pointing right, the net field is $g_{net} = g_{E} - g_{M} (opposite directions subtract) .$ Where $g_{E} = g_{M}$ , they cancel completely: the null point, where $g_{net} = 0$ .

WHY. Because $g = F / m$ inherits the arrow-nature of force. Two arrows head-to-tail give the resultant. Same-direction → add; opposite → subtract; at right angles → Pythagoras, $g = g_{1}^{2} + g_{2}^{2}$ .

PICTURE. Earth (blue) and Moon (gray) with their opposing field arrows. Between them, one point where the arrows are equal and cancel — marked in red as the null point.

The one-picture summary

Everything above, compressed into one frame: Newton's force law feeds the definition $g = F / m$ ; the visitor mass $m$ cancels; out drops $g = GM / r^{2}$ ; the spreading sphere explains the square; and the $g$ -vs- $r$ curve shows every regime at once.

Recall Feynman retelling — the whole walkthrough in plain words

Picture a heavy ball and a tiny bead near it (Step 1). Newton tells us how hard they tug: more stuff or closer distance means a stronger pull, with a fixed nature-number $G$ setting the scale (Step 2). Trouble: swap in a heavier bead and the pull grows — so "pull" describes the bead, not the spot (Step 3). Fix: divide the pull by the bead's own weight-in-kilograms. Now every bead — feather or brick — gives the same answer at that spot; we've captured the space, not the visitor (Step 4). Do the division and the bead's mass literally cancels top-and-bottom, leaving $GM / r^{2}$ (Step 5). Why the square on the bottom? Because the heavy ball's "pulling rays" spread over a sphere, and a sphere's area grows as radius-squared — go twice as far, the rays are four times as thin (Step 6). Then we sanity-check the ends: infinitely far is nearly zero, sitting on the point is infinite, and altitude is measured from the centre (Step 7). Finally, two balls pulling at once add like arrows, and where their arrows match they cancel — the null point (Step 8).

Recall Predict before you check

Triple $r$ . What happens to $g$ ?
Why did $m$ vanish from the final formula?
On a sphere twice as far, why is the field a quarter, not a half?

Answers: 1. $g \to g /9$ (inverse-square, $3^{2}$ ). 2. It appeared once on top from $F$ and once on the bottom from dividing by $m$ ; they cancel. 3. Sphere area $= 4 π r^{2}$ scales with $r^{2}$ , so double $r$ → area ×4 → field ÷4.

Connections

Newton's Law of Universal Gravitation — the single first principle we started from (Step 2)
Newton's Second Law (F=ma) — why $g$ is also the free-fall acceleration
Gauss's Law for Gravity — the formal engine behind the spreading-sphere picture (Step 6)
Variation of g with Altitude and Depth — what to do when $r \neq = R$ (Step 7)
Gravitational Potential and Potential Energy — the field's "height map," $g = - d V / d r$
Electric Field Intensity E=kQ/r² — identical algebra, charge instead of mass (Step 5)