Newton's law of gravitation — universal, action at distance: Question bank

These are pure-reasoning items. No calculators — if you feel the urge to plug numbers, that item belongs on the computation pages, not here.

First — the cast of characters (define before you trap)

You cannot spot a trap in a symbol you have not met. So here is every symbol used on this page, in plain words, before it appears in any question.

Figure — Newton's law of gravitation — universal, action at distance

Notice in the figure the two red force arrows are the same length but point opposite ways — that is Newton's Third Law built into gravity, and it is the visual you should hold while reading the "product vs sum" question.

True or false — justify

True or false: If you double both masses, the gravitational force doubles.

False — force

\propto m_{1} m_{2}

(the product), so doubling each multiplies the product by

2 \times 2 = 4

. The force quadruples, not doubles.

True or false: The Earth pulls the Moon much harder than the Moon pulls the Earth.

False — the two red arrows in the figure above are equal in length:

F_{12} = F_{21}

by Newton's Third Law. The Moon just accelerates less because it is heavier to shift; the force it receives is the same.

True or false: Gravity between two objects can be switched off by placing a shield between them.

False — there is no known "anti-gravity" material. Unlike electric charge, mass comes in only one sign (always attracting), so nothing cancels the field the way a conductor cancels an electric field.

True or false: A feather and a hammer dropped on the Moon hit the ground together.

True — the object's own mass cancels (see the worked cancellation in "Why questions"), leaving

a = GM / r^{2}

, which contains no reference to the falling object. With no air on the Moon, both fall at the same

g_{Moon}

True or false: If distance

r

triples, the force drops to one-third.

False — it is inverse-square (

r^{2}

in the denominator), so force drops to

1/ 3^{2} = 1/9

. Tripling distance weakens the pull ninefold; see the falloff curve in the figure at the end.

True or false: The constant

G

is larger on Jupiter because Jupiter's gravity is stronger.

False —

G

is universal, identical everywhere in the cosmos. Jupiter feels stronger because its

g = GM / r^{2}

is large (huge

M

);

G

itself never changes.

True or false: At the exact centre of the Earth you would feel enormous gravitational pull.

False — at the centre, mass surrounds you equally in all directions and every pull cancels, so net gravity is zero (see the shell-cancellation figure below).

True or false: Since gravity is "action at a distance," it reaches distant stars instantly.

False in reality — Newton's model assumed instant action, but General Relativity shows gravitational influence travels at the speed of light

c

, not infinitely fast.

True or false: Weight and mass are two words for the same quantity.

False — mass (

m

, in kg) is the amount of matter and never changes; weight is the force

W = m g

on it (in newtons), which shrinks on the Moon because its

g

is smaller. See Weight vs Mass.

Spot the error

Spot the error: "In

F = GM m / r^{2}

for a person standing on Earth,

r

is the person's height above the ground, so

r \approx 1.7

m."

Wrong — from our definition,

r

is the centre-to-centre distance, so

r = R_{\oplus} \approx 6.37 \times 1 0^{6}

m, not the surface height. A whole sphere pulls as if its mass sat at its centre (that is the Shell Theorem, drawn below).

Spot the error: "Because

F

depends on both masses, a heavy ball falls faster than a light one."

The force is bigger for the heavy ball, but so is its inertia. Dividing force by mass,

a = F / m = GM / r^{2}

— the object's mass cancels, so both fall with the same acceleration.

Spot the error: "

g

and

G

are both gravity constants with value about

9.8

Only

g \approx 9.8 m/s^{2}

near Earth's surface, and it is local (changes with planet and altitude).

G = 6.674 \times 1 0^{- 11} N m^{2} k g^{- 2}

is a universal constant with completely different units.

Spot the error: "Astronauts float on the ISS because there is no gravity up there."

There is plenty of gravity at that altitude (roughly 90% of surface value). They float because they are in continuous free fall: as the free-body figure below shows, gravity acts but there is no floor pushing back, so no normal force — the station and astronaut fall around Earth together.

Spot the error: "The formula uses

F_{12} = + \frac{G m _{1} m _{2}}{r ^{2}} \overset{r}{^}_{12}

where

\overset{r}{^}_{12}

points from 2 to 1."

As defined at the top,

\overset{r}{^}_{12}

points from 1 toward 2 (the blue arrow). The

+

sign encodes attraction only with that convention; if you flip the arrow's direction you must flip the sign, or the force will wrongly point away.

Spot the error: "Doubling the distance and doubling one mass leaves the force unchanged."

Doubling a mass multiplies

F

2

; doubling distance divides

F

2^{2} = 4

. Net effect is

2/4 = 1/2

, so the force is halved, not unchanged.

Why questions

Why does the force depend on the product

m_{1} m_{2}

rather than the sum

m_{1} + m_{2}

Both

m_{1} m_{2}

and

m_{1} + m_{2}

are symmetric, so symmetry alone does not decide it. The deciding argument is physical: each object is built of tiny bits, and every bit of

m_{1}

pulls every bit of

m_{2}

— that is a multiplication (

m_{1}

bits

\times

m_{2}

bits), giving the product. A sum would wrongly predict a nonzero force even when one mass is zero.

Why is gravity inverse-square and not inverse-distance?

A point source's influence spreads over an expanding sphere whose surface area is

4 π r^{2}

. The same total influence diluted over area

\propto r^{2}

makes the intensity fall as

1/ r^{2}

— the same geometry as light and sound.

Why do we barely feel the gravitational pull between two everyday objects?

G

is minuscule (

\sim 1 0^{- 11}

), so between ordinary masses the force is tinier than a bacterium's weight. We only notice gravity when one mass is astronomically large, like the whole Earth.

Why did the "falling Moon" calculation convince Newton that gravity is universal?

The Moon's orbital (centripetal) acceleration matched exactly the surface

g

scaled down by

1/6 0^{2}

(since the Moon is

\approx 60

Earth-radii away). Same law, same number — proof the apple-force and Moon-force are one. See Kepler's Laws.

Why does introducing a gravitational field

g

ease the "action at a distance" worry?

Instead of two masses reaching across empty space to grab each other, each mass fills the surrounding space with a field, and any other mass simply responds to the field at its own location — the interaction becomes local rather than spooky-at-a-distance. See Gravitational Field & Potential.

Why doesn't the object's own mass appear in its free-fall acceleration near Earth?

Write the two facts side by side: gravity gives

F = G M_{\oplus} m / R_{\oplus}^{2}

, and Newton's second law gives

F = ma

. Set them equal:

ma = G M_{\oplus} m / R_{\oplus}^{2}

. The falling mass

m

sits on both sides, so divide it out:

a = G M_{\oplus} / R_{\oplus}^{2} = g

. Nothing about

m

survives — that is exactly why Galileo saw all objects fall together.

Edge cases

Edge case: What is the gravitational force when the two point masses are at the same location,

r = 0

The formula gives

F = G m_{1} m_{2} /0

, which diverges to infinity — a sign the point-mass idealisation breaks down there. Real extended bodies have their mass spread out, so this singularity never physically occurs (except in the theory of black holes).

Edge case: Two masses are

1 0^{10}

light-years apart. Is the force exactly zero?

No — inverse-square means the force shrinks toward zero but is never exactly zero at any finite distance. Every bit of matter pulls on every other bit; that is what "universal" means.

Edge case: One of the two masses is zero (empty space, no test particle). What force acts?

Zero — with

m_{2} = 0

the product

m_{1} m_{2} = 0

. The field

g = GM / r^{2}

still exists in space, but with nothing there to sample it, there is no force. See Gravitational Field & Potential.

Edge case: You are inside a uniform hollow spherical shell. What gravity do you feel from the shell?

Exactly zero everywhere inside, at any position, not just the centre. The shell-cancellation figure below shows why: the nearby wall pulls harder but covers a small patch of mass, while the far wall pulls weaker but covers proportionally more mass — the two contributions cancel exactly (Shell Theorem).

Edge case: As you go down a mineshaft toward Earth's centre, does

g

keep increasing?

No — only the mass below your current radius pulls you (the shell above contributes nothing). With less mass beneath you,

g

actually decreases with depth, reaching zero at the centre. Contrast this with going up, where

g

falls as

1/ r^{2}

— both trends are drawn in the final figure.

Edge case: Is Newton's law exact for the orbit of Mercury?

Almost, but not perfectly — Mercury's orbit precesses slightly more than Newton predicts. The tiny discrepancy is explained by General Relativity, showing Newton's law is an excellent approximation, not the final word.

Two pictures to lock in the shell results

The shell theorem is invoked by three separate traps above (centre of the Earth, inside a hollow shell, the mineshaft). Here is the cancellation drawn once, plus how $g$ actually varies as you move inward and outward — keep these images in mind while re-reading those items.

Connections

Newton's Third Law — the equal-and-opposite red arrows in figure 1.
Shell Theorem — powers the centre-to-centre, hollow-shell, and mineshaft items.
Gravitational Field & Potential — the field cure for action-at-a-distance.
Circular Motion & Centripetal Force — free-fall / orbiting astronauts.
Kepler's Laws — the falling-Moon universality check.
General Relativity — Mercury precession, finite-speed gravity.
Weight vs Mass — the mass-vs-weight trap.