Gravitational field intensity g = GM - r²

WHAT it is

WHY "per unit mass"? Because the force on a test mass is proportional to $m$ . Dividing by $m$ cancels the visitor's mass, leaving a quantity that depends only on the source and where you stand. That is the whole trick of a field.

HOW we derive $g = \dfrac{GM}{r^2}$ (from scratch)

Formula Derivation from Newton's law of gravitation

Step 1 — Start with the force law. Newton's universal law for two point masses $M$ and $m$ separated by $r$ : $F = \frac{GMm}{r^2}$ Why this step? This is our first principle — the only assumption. $G=6.67\times10^{-11}\ \text{N m}^2\text{kg}^{-2}$ .

Step 2 — Apply the definition of field. Field intensity is force per unit mass: $g = \frac{F}{m} = \frac{1}{m}\cdot\frac{GMm}{r^2}$ Why this step? We are removing the test mass's influence to describe the location alone.

Step 3 — Cancel $m$ . $\boxed{g = \frac{GM}{r^2}}$ Why this step? The $m$ in numerator and denominator cancel — proving the field is independent of the test mass. ✔

As a vector (with $\hat{r}$ pointing away from $M$ ): $\vec{g} = -\frac{GM}{r^2}\,\hat{r}$ The minus sign means it points inward, toward $M$ (gravity attracts).

Figure — Gravitational field intensity g = GM - r²

Worked examples

Worked example 1. Field at Earth's surface

$M_\oplus = 5.97\times10^{24}\ \text{kg}$ , $R_\oplus = 6.37\times10^{6}\ \text{m}$ . $g = \frac{GM}{R^2} = \frac{(6.67\times10^{-11})(5.97\times10^{24})}{(6.37\times10^{6})^2}$ Why use $R$ ? On the surface, your distance from Earth's centre is its radius. $g = \frac{3.98\times10^{14}}{4.06\times10^{13}} \approx 9.8\ \text{N kg}^{-1}$ This recovers the familiar $9.8\ \text{m s}^{-2}$ — the field intensity is the free-fall acceleration. WHY equal? Because $g=F/m$ and $F=ma \Rightarrow a = F/m = g$ .

Common mistakes (Steel-manned)

Active recall

Recall Forecast-then-Verify: predict before reading

If you triple your distance from $M$ , what happens to $g$ ? → (forecast)
What are the units of $g$ , and why two equivalent forms?
Why doesn't the test mass appear in the final formula?

Answers: 1. $g \to g/9$ (inverse-square, $3^2$ ). 2. $\text{N kg}^{-1}=\text{m s}^{-2}$ because $g$ equals free-fall acceleration. 3. It cancels in $F/m$ ; field is a property of source + location.

Recall Feynman: explain to a 12-year-old

Imagine a huge magnet ball (but for weight, not metal). It makes invisible "pulling rays" all around it. Close up, the rays are crowded and pull hard; far away they've spread out over a giant bubble, so they pull weakly. The pulling-strength is the field. The cool part: it pulls a marble and a bowling ball with the same strength per kilogram — only the planet and the distance decide how strong, not who's visiting.

Flashcards

Gravitational field intensity is defined as

force per unit mass,

\vec g = \vec F/m

Formula for

g

from a point/spherical mass

M

at distance

r

g = GM/r^2

, directed toward

M

Units of gravitational field intensity

\text{N kg}^{-1}

\text{m s}^{-2}

Why does the test mass cancel out of

g

Force

F\propto m

; dividing by

m

leaves a source-only quantity

Distance

r

is measured from where?

The centre of the source mass, so

r=R+h

at altitude

h

At distance

2R

the field is what fraction of surface value?

1/4

(inverse-square,

2^2

)

Why is gravity an inverse-square law geometrically?

Field lines spread over sphere area

4\pi r^2 \propto r^2

Relation between field intensity and free-fall acceleration

They are equal:

a = F/m = g

At the Earth–Moon null point the net field is

zero (opposing fields equal in magnitude)

Vector form of the field with

\hat r

pointing away from

M

\vec g = -\dfrac{GM}{r^2}\hat r

Connections

Newton's Law of Universal Gravitation — the force law we derive $g$ from
Newton's Second Law (F=ma) — why $g$ equals free-fall acceleration
Gravitational Potential and Potential Energy — $g = -dV/dr$
Gauss's Law for Gravity — formal reason for the $1/r^2$ field-line picture
Variation of g with Altitude and Depth — extends this for $r\ne R$
Electric Field Intensity E=kQ/r² — identical structure (force per unit charge)

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, gravitational field intensity ka matlab simple hai: kisi bhi point pe agar tum ek chhota sa test mass rakho, to us pe kitna force per kilogram lagega — wahi $\vec g$ hai. Formula $g = GM/r^2$ . Yahan $M$ source ka mass hai (jaise Earth) aur $r$ centre se distance. Important baat: test mass $m$ formula mein dikhta hi nahi, kyunki force $F=GMm/r^2$ ko $m$ se divide karne pe $m$ cancel ho jaata hai. Isliye feather ho ya patthar, ek hi jagah pe dono ko same $g$ feel hota hai.

Inverse-square wali baat ko aise samjho: field lines source se nikalke ek sphere pe phailti hain, aur sphere ka area $4\pi r^2$ hota hai. Distance double karo to area chaar guna, isliye field chaar guna kam — yani $1/r^2$ . Yahi reason hai ki $r=2R$ pe $g$ surface ka sirf $1/4$ reh jaata hai.

Ek aur key insight: $g$ bilkul free-fall acceleration ke barabar hota hai, kyunki $a = F/m = g$ . Isliye Earth surface pe $g \approx 9.8\ \text{m s}^{-2}$ . Exam mein hamesha $r$ ko centre se lena — altitude $h$ ho to $r = R+h$ , sirf $h$ mat lena, ye sabse common galti hai.

Jab do masses ka field ho (jaise Earth aur Moon), to $\vec g$ vector hai — direction dekh ke add/subtract karo. Jis point pe dono fields barabar aur opposite ho, wahan net field zero — usko null point kehte hain. Ratios use karna seekho, har baar full calculation karne ki zaroorat nahi.

Gravitational field intensity g = GM - r²

WHAT it is

HOW we derive $g = \dfrac{GM}{r^2}$ (from scratch)

Worked examples

Common mistakes (Steel-manned)

Active recall

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Newton's Laws & Dynamics

Connections