Gravitational field intensity

Two masses pull on each other — that much is Newton's law. But physicists prefer a subtler picture: instead of asking "what force does mass $A$ exert on mass $B$ ," they ask "what does mass $A$ do to the space around it?" The answer is a gravitational field, a property of every point in space near $A$ , and the strength of that field is called gravitational field intensity. Once you have the field, dropping in any test mass instantly tells you the force on it, without recomputing anything from scratch. This idea is not just bookkeeping. It is the same logic that later gives you electric and magnetic fields in Class 12, and it quietly reveals why the acceleration due to gravity $g$ is the same number for every falling object regardless of mass. This article defines gravitational field intensity, derives its formula for a point mass, connects it to $g$ , and works through the arithmetic you need for numericals.

The core insight

Gravitational field intensity $\vec E_g = \vec F/m$ is the gravitational force per unit test mass at a point, equal to $GM/r^2$ for a point mass or uniform sphere and numerically identical to $g$ .

The core idea

Field as a property of space. A source mass $M$ sets up a gravitational field in the region around it — an influence that exists whether or not any other mass happens to be there to feel it. Bring a small test mass $m$ to a point in that region and it feels a force $\vec F$ . The gravitational field intensity at that point is defined as this force per unit test mass:

\vec{E}_g = \dfrac{\vec F}{m}

where: $\vec F$ is the gravitational force on the test mass, $m$ is the magnitude of the test mass (kept small so it does not itself disturb the source), and $\vec E_g$ is the field intensity at the point in question. It is a vector, always directed toward the source mass, since gravity is purely attractive. Its SI unit is $\text{N/kg}$ .

Field of a point mass. For a point mass $M$ (or a uniform sphere treated as a point mass at its centre, valid for any external point $r \geq R$ ), Newton's law gives the force on a test mass $m$ at distance $r$ as $F = GMm/r^2$ , directed toward $M$ . Substituting into the definition,

E_g = \dfrac{F}{m} = \dfrac{GM}{r^2}

where: $G = 6.67\times10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2$ is the universal gravitational constant, $M$ is the source mass, and $r$ is the distance from the centre of $M$ to the point. Notice the test mass $m$ has cancelled out entirely — field intensity depends only on the source and the distance, never on whatever test mass you use to probe it. That is exactly the point of introducing a field in the first place: it is a property of the source and of space, not of the measuring instrument.

Field intensity is nothing but $g$ . Set $r = R$ , the Earth's radius, and $E_g = GM/R^2$ is precisely the acceleration due to gravity at the surface. "Gravitational field intensity" and " $g$ " are the same physical quantity described two ways — one as force-per-unit-mass, the other as an acceleration — and the units confirm it: $\text{N/kg} \equiv \text{m/s}^2$ dimensionally. This is also why every object, heavy or light, falls with the same acceleration $g$ near Earth: the field at a given point has one fixed value, and the force on any mass placed there is just that value times the mass, so the resulting acceleration ( $F/m = E_g$ ) is independent of $m$ .

Why bother with fields at all? The two-step separation is the real payoff. Step one: $M$ alone fixes a field $E_g(r) = GM/r^2$ everywhere around it, computed once. Step two: any test mass $m$ dropped anywhere in that field feels $F = mE_g$ immediately, with no new calculation involving $G$ or $r$ again. Compare a hundred different masses at the same point and you reuse the same $E_g$ a hundred times. This is the template Class 12 electrostatics reuses verbatim for electric field intensity — a field set up by charge, probed by a small test charge. Field intensity falls off with the inverse square of distance and always points radially inward, toward the mass creating it — the vector arrows below all converge on the source, growing shorter farther away, exactly tracking $1/r^2$ .

Figure

Field vectors around a central mass pointing inward with length shrinking as distance grows following an inverse square law

Derivation AField intensity of a point mass from Newton's law

The goal is to derive $E_g = GM/r^2$ starting only from the universal law of gravitation and the definition of field intensity as force per unit mass, showing explicitly why the test mass cancels out and the result depends only on the source mass and the distance from it.

Step 1. Write Newton's law of gravitation for a test mass $m$ placed at distance $r$ from a source mass $M$ : $F = \dfrac{GMm}{r^2}$ , directed from $m$ toward $M$ (attractive, always toward the source).

Step 2. Apply the definition of gravitational field intensity, $E_g = F/m$ , to this force. Substitute the expression from Step 1:

E_g = \dfrac{1}{m}\left(\dfrac{GMm}{r^2}\right)

Step 3. Cancel the test mass $m$ from numerator and denominator — it appears in both, so it drops out entirely, leaving a result that depends only on $G$ , $M$ and $r$ :

E_g = \dfrac{GM}{r^2}

Step 4. Interpret the result. Because $m$ cancelled, $E_g$ is a property of the source mass $M$ and the point's distance $r$ alone — every test mass placed at that same point, however large or small, measures the identical value of $E_g$ . Setting $r$ equal to the Earth's radius $R$ reproduces $g = GM/R^2$ , confirming that field intensity at the surface is exactly the familiar acceleration due to gravity.

Worked example

Find the gravitational field intensity at a point $5.0\times10^5\ \text{m}$ from the centre of a planet of mass $4.0\times10^{22}\ \text{kg}$ , and state the force felt by a $3.0\ \text{kg}$ mass placed there. Take $G = 6.67\times10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2$ .

Step 1. Identify the quantities: $M = 4.0\times10^{22}\ \text{kg}$ , $r = 5.0\times10^5\ \text{m}$ , and the test mass whose force we want is $m = 3.0\ \text{kg}$ . Since $r$ is measured from the centre and we are outside the planet, use $E_g = GM/r^2$ directly.

Step 2. Compute the numerator: $GM = (6.67\times10^{-11})(4.0\times10^{22}) = 2.668\times10^{12}$ .

Step 3. Compute the denominator and divide: $r^2 = (5.0\times10^5)^2 = 2.5\times10^{11}$ , so

E_g = \dfrac{2.668\times10^{12}}{2.5\times10^{11}} = 10.67\ \text{N/kg}

Step 4. Find the force on the $3.0\ \text{kg}$ mass using $F = mE_g$ (the definition rearranged, exactly the reverse of the derivation above): $F = 3.0\times10.67 = 32.0\ \text{N}$ , directed toward the planet's centre. Its acceleration if released would be $F/m = E_g = 10.67\ \text{m/s}^2$ — the same for any mass, confirming field intensity is the local " $g$ " at that point.

Field intensity: $\mathbf{10.67\ N/kg}$ ; force on the 3.0 kg test mass: $\mathbf{32.0\ N}$ .

Common traps

Believing field intensity depends on the test mass $m$ used to measure it — it does not; $m$ cancels in $E_g = F/m$ , and $E_g = GM/r^2$ involves only the source mass and distance.
Treating field intensity as a scalar — it is a vector directed toward the source, so fields from two masses at a point must be added by vector (not arithmetic) addition unless they lie along the same line.
Thinking $\text{N/kg}$ and $\text{m/s}^2$ are different kinds of quantities — they are dimensionally identical ( $[LT^{-2}]$ ) and interchangeable here.
Using $E_g = GM/r^2$ with the full planetary mass for a point inside a uniform sphere — only the mass enclosed within radius $r$ contributes there, the same rule used for the depth variation of $g$ .

Ask the tutor: Want to see how gravitational field intensity leads into gravitational potential and potential energy, or practice combining fields from two masses at once? Ask the AI tutor about Gravitational field intensity.

Why does the test mass cancel out of the field intensity formula?

Because gravitational force is proportional to the test mass ( $F = GMm/r^2$ ), and field intensity is defined by dividing that force by the same test mass ( $E_g = F/m$ ). The $m$ in the numerator and the $m$ in the denominator are literally the same number, so they cancel algebraically, leaving $E_g = GM/r^2$ — a quantity built only from the source mass and the distance. This cancellation is why fields are useful: the same $E_g$ value applies no matter what test mass you use to probe the point.

Is gravitational field intensity really the same thing as g, or just similar?

They are the same physical quantity, not merely similar. At the Earth's surface, $r$ equals the Earth's radius $R$ , and $E_g = GM/R^2$ is the exact expression already derived for $g$ in the variation-of- $g$ topic. Whether you call it "field intensity" or "acceleration due to gravity" depends only on whether you are thinking of it as force-per-unit-mass or as an acceleration — the units $\text{N/kg}$ and $\text{m/s}^2$ are dimensionally identical, so the numbers and the physics match exactly.

How do I combine the gravitational fields from two different masses at one point?

Because field intensity is a vector, you add the individual fields using vector addition, not by simply adding magnitudes. Compute $\vec E_{g1} = GM_1/r_1^2$ pointing toward mass $M_1$ and $\vec E_{g2} = GM_2/r_2^2$ pointing toward mass $M_2$ , then add them as vectors (using components, or the parallelogram rule if the directions are known geometrically). Only when both fields point along exactly the same line can you add or subtract their magnitudes directly.

Gravitational field intensity

The core idea

Worked example

Common traps

Before this

Where this leads