Newton's law of gravitation — universal, action at distance: Worked examples

Before anything, let us re-earn every symbol so a fresh reader is never stranded.

The scenario matrix

Every gravitation problem is one (or a mix) of these case classes. Each row is covered by at least one worked example below.

#	Case class	What is weird about it	Covered by
A	Two ordinary masses, plain plug-in	nothing — the baseline	Ex 1
B	Surface gravity $g$ from $M, R$	which $r$ ? mass cancels	Ex 2
C	Distance scaling (double/halve $r$ )	inverse-square, no numbers	Ex 3
D	Two forces as vectors (3 bodies)	must add arrows, not numbers	Ex 4 (figure)
E	Degenerate: $r \to 0$	formula blows up — real meaning?	Ex 5
F	Limiting: $r \to \infty$	force fades to zero, never exactly 0	Ex 5
G	Real-world word problem	pick numbers out of prose	Ex 6
H	Exam twist: change both mass & distance at once	combine A + C carefully	Ex 7
I	Ratio / "which is bigger" trap	Sun's pull vs Moon's pull on you	Ex 8

The examples

Worked example Ex 2 — Case B: surface gravity, mass cancels

Estimate $g$ on Earth from $M_{\oplus} = 5.97 \times 1 0^{24} kg$ , $R_{\oplus} = 6.37 \times 1 0^{6} m$ .

Forecast: which $r$ do you use — the height of the object, or Earth's radius?

A test mass $m$ at the surface feels $F = \frac{G M _{\oplus} m}{R _{\oplus}^{2}}$ . Why this step? By the shell theorem, Earth pulls as if all its mass sat at its centre, so $r = R_{\oplus}$ (centre-to-surface), not zero.
But weight is also $F = m g$ . Set them equal: $m g = \frac{G M _{\oplus} m}{R _{\oplus}^{2}}$ . Why this step? Two names for the same downward force.
The $m$ cancels: $g = \frac{G M _{\oplus}}{R _{\oplus}^{2}}$ . Why this step? This is why every object falls at the same rate — the falling mass appears on both sides and vanishes.
Plug in: $g = \frac{( 6.674 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} )}{( 6.37 \times 1 0 ^{6} ) ^{2}} \approx 9.82 m/s^{2}$ .

Verify: Known value is $\approx 9.8 m/s^{2}$ ✓. Units: $\frac{N m ^{2} k g ^{- 2} \cdot kg}{m ^{2}} = N/kg = m/ s^{2}$ ✓. See Weight vs Mass.

Worked example Ex 4 — Case D: two pulls as arrows (never as plain numbers)

Mass $M_{C} = 4 kg$ sits at a corner. Mass $M_{A} = 3 kg$ is $2 m$ to its west; mass $M_{B} = 3 kg$ is $2 m$ to its south. Find the total gravitational force on $M_{C}$ .

Forecast: two equal pulls at $9 0^{\circ}$ — will the total be $F + F$ , or something smaller?

Force from $A$ on $C$ : it points west (toward $A$ ), magnitude $F_{A} = \frac{( 6.674 \times 1 0 ^{- 11} ) ( 4 ) ( 3 )}{2 ^{2}} = 2.00 \times 1 0^{- 10} N$ . Why this step? Each pair uses the same scalar formula; direction is along the line toward the pulling mass.
Force from $B$ on $C$ : points south, and by symmetry $F_{B} = F_{A} = 2.00 \times 1 0^{- 10} N$ . Why this step? Same masses, same distance ⇒ same magnitude, but a different direction.
These are perpendicular arrows, so add by Pythagoras (look at the red resultant in the figure): $F_{tot} = F_{A}^{2} + F_{B}^{2} = F_{A} 2 = 2.83 \times 1 0^{- 10} N$ . Why this step? You cannot just add $F_{A} + F_{B}$ because they point different ways — forces are vectors. Perpendicular vectors combine like the sides of a right triangle.
Direction: exactly halfway between west and south, i.e. $4 5^{\circ}$ south-of-west (points toward the midpoint of $A$ and $B$ ). Why this step? Equal legs ⇒ the resultant bisects the right angle.

Verify: $F_{A} 2 = 2.00 \times 1 0^{- 10} \times 1.414 = 2.83 \times 1 0^{- 10}$ ✓. The total ( $2.83$ ) is less than the naive sum ( $4.00$ ) — correct, because the two pulls partly work across each other. See Newton's Third Law for why each pull has a matching reaction on $A$ and $B$ .

Worked example Ex 5 — Cases E & F: the two dangerous ends,

$r \to 0$ and $r \to \infty$ What does $F = \frac{G m _{1} m _{2}}{r ^{2}}$ predict as the masses (a) touch, and (b) fly infinitely far apart?

Forecast: does the formula "break," or just tell you something extreme?

As $r \to 0$ : the denominator $r^{2} \to 0$ , so $F \to \infty$ . Why this step? Dividing a fixed top by a shrinking bottom makes the result explode (see the left spike in the figure).
But this is a degenerate case: the formula is only valid for point masses or for the region outside solid spheres. For real spheres you can never get their centres to $r = 0$ without them overlapping — once you burrow inside a sphere, the shell theorem says the mass outside your radius stops pulling, so $F$ actually drops back to $0$ at the exact centre. Why this step? Always ask "is the formula's assumption still true here?" It is not at $r = 0$ for real bodies.
As $r \to \infty$ : $r^{2} \to \infty$ , so $F \to 0$ . Why this step? A fixed top over an exploding bottom shrinks toward zero (right tail of the figure).
But $F$ is never exactly zero for any finite $r$ : gravity has infinite reach, just vanishingly weak. Why this step? $1/ r^{2}$ approaches $0$ but never lands on it — this is what "universal, action at a distance" really means.

Verify: Numerically, halving $r$ four times ( $r \to r /16$ ) multiplies $F$ by $1 6^{2} = 256$ — confirming the blow-up trend. Doubling $r$ ten times shrinks $F$ by $2^{20} \approx 1 0^{6}$ — confirming the fade ✓. See Shell Theorem and Gravitational Field & Potential.

Worked example Ex 6 — Case G: real-world word problem

A $70 kg$ astronaut orbits at altitude $400 km$ (the ISS height). What gravitational force does Earth exert on her? Use $M_{\oplus} = 5.97 \times 1 0^{24} kg$ , $R_{\oplus} = 6.37 \times 1 0^{6} m$ .

Forecast: she "floats," so is the force (a) zero, or (b) almost as strong as on the ground?

Find the true centre-to-centre distance: $r = R_{\oplus} + h = 6.37 \times 1 0^{6} + 0.40 \times 1 0^{6} = 6.77 \times 1 0^{6} m$ . Why this step? $r$ is measured from Earth's centre, so add altitude to the radius — the classic word-problem trap.
Substitute: $F = \frac{( 6.674 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} ) ( 70 )}{( 6.77 \times 1 0 ^{6} ) ^{2}}$ . Why this step? Standard plug-in with the corrected $r$ .
Compute: numerator $= 2.789 \times 1 0^{16}$ ; denominator $= 4.583 \times 1 0^{13}$ ; $F \approx 608 N$ .

Verify: On the ground she'd weigh $70 \times 9.8 = 686 N$ . At ISS height $608 N$ is about 89% of that — she is not weightless because gravity vanished; she floats because she is in free fall around Earth. Ratio check: $(R / r)^{2} = (6.37/6.77)^{2} = 0.885 \Rightarrow 686 \times 0.885 = 607 N$ ✓. See Circular Motion & Centripetal Force.

Worked example Ex 8 — Case I: the "which pull is bigger" trap

On your $70 kg$ body, which is stronger: the Sun's pull or the Moon's pull? Sun: $M_{S} = 1.99 \times 1 0^{30} kg$ , $r_{S} = 1.50 \times 1 0^{11} m$ . Moon: $M_{M} = 7.35 \times 1 0^{22} kg$ , $r_{M} = 3.84 \times 1 0^{8} m$ .

Forecast: the Moon is so much closer — surely it wins?

Sun's pull: $F_{S} = \frac{( 6.674 \times 1 0 ^{- 11} ) ( 1.99 \times 1 0 ^{30} ) ( 70 )}{( 1.50 \times 1 0 ^{11} ) ^{2}} \approx 0.413 N$ . Why this step? Direct plug-in with the Sun's numbers.
Moon's pull: $F_{M} = \frac{( 6.674 \times 1 0 ^{- 11} ) ( 7.35 \times 1 0 ^{22} ) ( 70 )}{( 3.84 \times 1 0 ^{8} ) ^{2}} \approx 0.00233 N$ . Why this step? Same formula, Moon's numbers.
Ratio: $\frac{F _{S}}{F _{M}} \approx \frac{0.413}{0.00233} \approx 177$ . Why this step? The Sun's enormous mass ( $1 0^{7}$ times the Moon's) beats its greater distance, even after the distance is squared.

Verify: The Sun pulls you about 180× harder than the Moon ✓ — so the Moon does not win. (Tides are ruled by the Moon not because its total pull is bigger, but because its pull varies more across Earth's diameter — a different, gradient effect from Gravitational Field & Potential.)

Active Recall

Recall Cover the answers

When you double $r$ , force becomes... ::: one quarter (inverse-square: $1/ 2^{2}$ ). Two equal perpendicular pulls $F$ combine to... ::: $F 2$ , at $4 5^{\circ}$ between them (Pythagoras, not $2 F$ ). What is $r$ for an astronaut $400$ km up? ::: $R_{\oplus} + 400 km$ , from Earth's centre. As $r \to \infty$ , does $F$ ever reach exactly $0$ ? ::: No — it fades toward $0$ but stays nonzero (infinite reach). Planet with $2 M$ and $2 R$ : surface $g$ is... ::: half of Earth's, because the radius is squared. Sun vs Moon pull on you — winner? ::: the Sun, by about $180 \times$ (mass beats distance).

Connections

Read this in Hinglish →
Newton's Third Law — every pull in Ex 4 has an equal, opposite partner.
Shell Theorem — justifies centre-to-centre $r$ and the $r \to 0$ resolution.
Gravitational Field & Potential — the tide/gradient idea in Ex 8.
Circular Motion & Centripetal Force — why the Ex 6 astronaut floats.
Weight vs Mass — the $m g = GM m / R^{2}$ cancellation in Ex 2.