Gravitational field intensity g = GM - r²: Worked examples

Before anything: recall the two tools we use over and over.

The scenario matrix

Every question about $g$ falls into one of these boxes. The examples below are tagged with the box they fill.

Cell	Case class	What makes it tricky	Example
C1	Single mass, $r = R$ (on surface)	pick the right distance	Ex 1
C2	Single mass, $r > R$ (altitude) via ratio	don't recompute — scale by $1/ r^{2}$	Ex 2
C3	Two fields, same line, opposite	vectors subtract; sign matters	Ex 3
C4	Two fields, null point ( $g = 0$ )	degenerate: net field vanishes	Ex 4
C5	Two fields at right angles	Pythagoras, not addition	Ex 5
C6	Limiting behaviour: $r \to \infty$ and $r \to 0$	field → 0 vs. field "blows up"	Ex 6
C7	Inverse problem: given $g$ , find $M$ or $r$	rearrange the formula	Ex 7
C8	Word problem (real world, altitude)	translate words → $r = R + h$	Ex 8
C9	Exam twist: new planet, ratios of $M$ and $R$	combine two scalings at once	Ex 9

We now fill every cell.

Ex 1 — Surface field (Cell C1)

Worked example Field at the surface of Mars

Mars: $M = 6.42 \times 1 0^{23} kg$ , radius $R = 3.39 \times 1 0^{6} m$ . Find $g$ at its surface.

Forecast: Mars is much lighter than Earth but also smaller. Guess: bigger or smaller than Earth's $9.8$ ? (Write your guess down.)

Step 1 — Choose the distance. On the surface, your distance from the centre of Mars is its radius, so $r = R = 3.39 \times 1 0^{6} m$ . Why this step? $r$ in $GM / r^{2}$ is always measured from the centre of mass, and "surface" means exactly one radius out.

Step 2 — Plug into Tool A. $g = \frac{( 6.67 \times 1 0 ^{- 11} ) ( 6.42 \times 1 0 ^{23} )}{( 3.39 \times 1 0 ^{6} ) ^{2}}$ Why this step? One mass, one distance — this is the direct-substitution case, no vectors needed.

Step 3 — Arithmetic. Numerator $= 4.28 \times 1 0^{13}$ ; denominator $= (3.39 \times 1 0^{6})^{2} = 1.149 \times 1 0^{13}$ . $g = \frac{4.28 \times 1 0 ^{13}}{1.149 \times 1 0 ^{13}} \approx 3.7 N kg^{- 1}$

Verify: Units: $\frac{N m ^{2} kg ^{- 2} \cdot kg}{m ^{2}} = N kg^{- 1}$ ✔. Magnitude ≈ $3.7$ , about $0.38 \times$ Earth's — matches the known "Martian gravity is about a third of Earth's." Your forecast: smaller was correct.

Ex 2 — Altitude by ratio (Cell C2)

Ex 3 — Two fields, opposite directions (Cell C3)

Now we need Tool B — vectors. See the picture: two masses pull the point $P$ in opposite directions along one line.

Figure — Gravitational field intensity g = GM - r²

Worked example Net field between two masses on a line

Mass $A$ sits to the left of point $P$ and its field there points left (toward $A$ ) with magnitude $g_{A} = 6.0 \times 1 0^{- 6} N kg^{- 1}$ . Mass $B$ sits to the right; its field at $P$ points right with $g_{B} = 2.0 \times 1 0^{- 6} N kg^{- 1}$ . Find the net field.

Forecast: Which way does the net arrow point, and is it bigger or smaller than $6.0 \times 1 0^{- 6}$ ?

Step 1 — Pick a positive direction. Let "right" $= +$ , so "left" $= -$ . Why this step? A vector on a line is just a signed number once we choose which way is positive. This turns arrow-addition into ordinary $\pm$ arithmetic.

Step 2 — Assign signs. $g_{A}$ points left $\Rightarrow - 6.0 \times 1 0^{- 6}$ . $g_{B}$ points right $\Rightarrow + 2.0 \times 1 0^{- 6}$ . Why this step? Look at the figure: the two red arrows fight each other, so they must carry opposite signs.

Step 3 — Add. $g_{net} = (- 6.0 + 2.0) \times 1 0^{- 6} = - 4.0 \times 1 0^{- 6} N kg^{- 1}$ Why this step? Adding signed numbers is adding the arrows. The negative result tells us the direction.

Step 4 — Interpret the sign. Negative = points left, toward $A$ .

Verify: The stronger field ( $A$ ) wins, so the net must point toward $A$ — it does ✔. Magnitude $4.0 < 6.0$ because $B$ partially cancelled it ✔.

Ex 4 — The null point (Cell C4, degenerate zero)

Worked example Where does the net field vanish?

Two masses: Earth $M_{E} = 5.97 \times 1 0^{24} kg$ and Moon $M_{m} = 7.35 \times 1 0^{22} kg$ , separated by $d = 3.84 \times 1 0^{8} m$ . Along the line joining them, find the distance $x$ from Earth's centre where the net field is zero.

Forecast: Will the null point sit closer to Earth or to the Moon?

Step 1 — Set the two magnitudes equal. At the null point the two fields point opposite ways and have equal size, so they cancel: $\frac{G M _{E}}{x ^{2}} = \frac{G M _{m}}{( d - x ) ^{2}}$ Why this step? "Net field zero" for two opposite vectors means their magnitudes match exactly — that's the definition of the degenerate case.

Step 2 — Cancel $G$ and take the square root. $\frac{M _{E}}{x ^{2}} = \frac{M _{m}}{( d - x ) ^{2}} \Rightarrow \frac{d - x}{x} = \frac{M _{m}}{M _{E}}$ Why this step? $G$ is common to both and cancels; the square root turns the awkward squares into a clean linear ratio.

Step 3 — Compute the mass ratio. $\frac{7.35 \times 1 0 ^{22}}{5.97 \times 1 0 ^{24}} = 0.01231 = 0.1109$

Step 4 — Solve for $x$ . From $d - x = 0.1109 x$ : $d = 1.1109 x \Rightarrow x = \frac{3.84 \times 1 0 ^{8}}{1.1109} \approx 3.457 \times 1 0^{8} m$

Verify: $x \approx 3.46 \times 1 0^{8}$ m out of $3.84 \times 1 0^{8}$ — that's $90%$ of the way to the Moon, i.e. close to the Moon because the Moon is much lighter and can only balance Earth when you stand almost on top of it ✔. Forecast: closer to Moon was correct.

Ex 5 — Fields at right angles (Cell C5)

The two masses no longer sit on one line through $P$ . See the figure: the arrows meet at $9 0^{\circ}$ , so we use Pythagoras.

Worked example Perpendicular fields

At point $P$ , mass $A$ produces a field $g_{A} = 3.0 \times 1 0^{- 6} N kg^{- 1}$ pointing east, and mass $B$ produces $g_{B} = 4.0 \times 1 0^{- 6} N kg^{- 1}$ pointing north. Find the magnitude and direction of the net field.

Forecast: The answer is not $7.0 \times 1 0^{- 6}$ . Guess whether it's bigger or smaller than $5.0 \times 1 0^{- 6}$ .

Step 1 — Recognise they can't be added directly. East and north are perpendicular. Why this step? Only fields on the same line add or subtract as numbers. At $9 0^{\circ}$ the arrows form the two legs of a right triangle (see figure), and the net is the hypotenuse.

Step 2 — Apply Pythagoras. $g_{net} = g_{A}^{2} + g_{B}^{2} = (3.0)^{2} + (4.0)^{2} \times 1 0^{- 6}$ Why this step? The hypotenuse length of a right triangle is $leg_{1}^{2} + leg_{2}^{2}$ — this is exactly how perpendicular vectors combine.

Step 3 — Compute. $g_{net} = 9 + 16 \times 1 0^{- 6} = 25 \times 1 0^{- 6} = 5.0 \times 1 0^{- 6} N kg^{- 1}$

Step 4 — Direction. The angle north-of-east: $θ = arctan (\frac{g _{B}}{g _{A}}) = arctan (\frac{4.0}{3.0}) = 53. 1^{\circ}$ Why $arctan$ ? $arctan$ answers "which angle has this rise-over-run?" — here the north leg over the east leg.

Verify: $3^{2} + 4^{2} = 5$ (the classic 3-4-5 triangle) ✔, and $5.0 < 7.0$ so direct addition would've over-counted ✔. $tan 53. 1^{\circ} = 1.33 = 4/3$ ✔.

Ex 6 — Limiting behaviour (Cell C6)

Worked example What happens as

$r \to \infty$ and as $r \to 0$ ? Using $g = \frac{GM}{r ^{2}}$ , describe the field very far away and very close to a point mass. Then compute $g$ at $r = 10 R$ for Earth to show the far limit numerically ( $g_{0} = 9.8$ ).

Forecast: Does the field ever reach exactly zero at a finite distance?

Step 1 — Far limit, $r \to \infty$ . As $r$ grows, $r^{2}$ grows without bound, so $\frac{GM}{r ^{2}} \to 0$ . Why this step? A fixed numerator over an ever-larger denominator shrinks toward zero — but only reaches zero at infinity, never at any finite $r$ .

Step 2 — Numerical taste of the far limit. At $r = 10 R$ : $g = \frac{g _{0}}{1 0 ^{2}} = \frac{9.8}{100} = 0.098 N kg^{- 1}$ Why this step? Ratio trick again: distance $\times 10 \Rightarrow$ field $\div 100$ . Small but non-zero — confirming Step 1.

Step 3 — Close limit, $r \to 0$ . As $r \to 0$ , $r^{2} \to 0$ and $\frac{GM}{r ^{2}} \to \infty$ : the formula "blows up." Why this step? Dividing by a vanishing number gives an unbounded result — mathematically the field diverges for an idealised point mass.

Step 4 — The physics caveat. For a real planet you cannot actually stand at $r = 0$ ; once you go inside, only the mass beneath you pulls, and $g$ actually falls to zero at the centre (see Variation of g with Altitude and Depth). The $1/ r^{2}$ blow-up is only for an external point mass.

Verify: $0.098 \times 100 = 9.8$ ✔. Far away → tiny but positive; the field is zero only in the limit ✔.

Ex 7 — Inverse problem (Cell C7)

Worked example Find the mass from a measured field

A field probe at $r = 2.0 \times 1 0^{7} m$ from an unknown planet's centre reads $g = 5.0 N kg^{- 1}$ . Find the planet's mass $M$ .

Forecast: Bigger or smaller than Earth's $5.97 \times 1 0^{24}$ kg?

Step 1 — Rearrange Tool A for $M$ . From $g = \frac{GM}{r ^{2}}$ : $M = \frac{g r ^{2}}{G}$ Why this step? We know $g$ , $r$ , $G$ and want $M$ — multiply both sides by $r^{2}$ and divide by $G$ to isolate the unknown.

Step 2 — Substitute. $M = \frac{( 5.0 ) ( 2.0 \times 1 0 ^{7} ) ^{2}}{6.67 \times 1 0 ^{- 11}}$ Why this step? Direct plug-in; the only trap is squaring the distance first.

Step 3 — Arithmetic. $r^{2} = 4.0 \times 1 0^{14}$ ; numerator $= 5.0 \times 4.0 \times 1 0^{14} = 2.0 \times 1 0^{15}$ . $M = \frac{2.0 \times 1 0 ^{15}}{6.67 \times 1 0 ^{- 11}} \approx 3.0 \times 1 0^{25} kg$

Verify: Plug back: $g = \frac{( 6.67 \times 1 0 ^{- 11} ) ( 3.0 \times 1 0 ^{25} )}{( 2.0 \times 1 0 ^{7} ) ^{2}} = \frac{2.0 \times 1 0 ^{15}}{4.0 \times 1 0 ^{14}} = 5.0$ ✔. Larger than Earth — a heavier planet, forecast dependent on your guess.

Ex 8 — Word problem, altitude (Cell C8)

Worked example A mountaintop observatory

An observatory sits on a peak $h = 4.8 km$ above Earth's surface. Given surface $g_{0} = 9.80 N kg^{- 1}$ and $R = 6.37 \times 1 0^{6} m$ , find $g$ at the observatory.

Forecast: By roughly how much does $g$ drop — a few parts in a thousand, or a few percent?

Step 1 — Translate the words into a distance. "Above the surface" means altitude, so the distance from the centre is $r = R + h = 6.37 \times 1 0^{6} + 4.8 \times 1 0^{3} = 6.3748 \times 1 0^{6} m$ Why this step? The single most common exam mistake is using $h$ alone. $r$ is measured from the centre — see the parent note's "radius from the centre" warning.

Step 2 — Use the ratio. $g = g_{0} (\frac{R}{R + h})^{2} = 9.80 (\frac{6.37 \times 1 0 ^{6}}{6.3748 \times 1 0 ^{6}})^{2}$ Why this step? $GM$ is common to surface and peak; the ratio avoids ever needing $M$ .

Step 3 — Compute. $\frac{6.37}{6.3748} = 0.999247$ ; squared $= 0.998495$ . $g = 9.80 \times 0.998495 \approx 9.785 N kg^{- 1}$

Verify: Drop $= 9.80 - 9.785 = 0.015 N kg^{- 1}$ , about $0.15%$ — a "few parts in a thousand," which matches the fact that $4.8$ km is tiny next to Earth's $6370$ km radius ✔.

Ex 9 — Exam twist: a new planet (Cell C9)

Worked example Two scalings at once

Planet Zorp has twice Earth's mass and half Earth's radius. Earth's surface $g_{0} = 9.8 N kg^{- 1}$ . Find Zorp's surface gravity.

Forecast: More mass and smaller radius both make gravity stronger. Guess the multiple: ×2? ×4? ×8?

Step 1 — Write Zorp's field in terms of Earth's. $g_{Z} = \frac{G ( 2 M )}{( R /2 ) ^{2}}$ Why this step? Substitute the two changes into Tool A so we can factor out Earth's $\frac{GM}{R ^{2}} = g_{0}$ .

Step 2 — Separate the two scalings. $g_{Z} = \frac{2 GM}{R ^{2} /4} = 2 \times 4 \times \frac{GM}{R ^{2}} = 8 g_{0}$ Why this step? The mass doubling multiplies by $2$ ; halving the radius divides by $(1/2)^{2} = 1/4$ , i.e. multiplies by $4$ . The two effects multiply, giving $8$ .

Step 3 — Number. $g_{Z} = 8 \times 9.8 = 78.4 N kg^{- 1}$

Verify: Independent check — factors combine as $2/ (1/2)^{2} = 2/0.25 = 8$ ✔. So $78.4 N kg^{- 1}$ . Forecast: ×8, because both effects pile up and the radius one is squared.

Active recall

Recall Which cell? Match the question to its method

"Field at $r = 5 R$ from surface $g_{0}$ " ::: C2 — ratio, divide by $5^{2} = 25$
"Point where Earth and Moon fields cancel" ::: C4 — set magnitudes equal, null point
"Two fields at $9 0^{\circ}$ " ::: C5 — Pythagoras on the hypotenuse
"Given $g$ and $r$ , find $M$ " ::: C7 — rearrange to $M = g r^{2} / G$
" $g$ on a peak $h$ above ground" ::: C8 — use $r = R + h$ , not $h$

Connections

Parent topic note — the definition and derivation these examples exercise
Newton's Law of Universal Gravitation — the force law behind every $g$
Newton's Second Law (F=ma) — why $g$ doubles as free-fall acceleration
Variation of g with Altitude and Depth — Ex 6 & Ex 8 generalised for $r \neq = R$
Gauss's Law for Gravity — the deep reason the field lines give $1/ r^{2}$
Gravitational Potential and Potential Energy — the energy view of these same fields
Electric Field Intensity E=kQ/r² — every vector-addition trick here transfers verbatim