Apply Newton's law of universal gravitation F = G m1 m2 / r^2 and the gravitational field strength g = G M / r^2 to calculate gravitational force, field strength and acceleration at points in a radial gravitational field

What this dot point is asking

QCAA wants you to apply Newton's law of universal gravitation to calculate the force between two masses, determine the gravitational field strength at a point in a radial field, and explain how the inverse-square dependence on distance shapes planetary, lunar and satellite phenomena. You need both the conceptual fluency (action at a distance vs field model, why $g$ varies with altitude) and the numerical fluency to compute $F$ and $g$ at any distance.

The answer

Newton's law of universal gravitation

Every pair of point masses (or spherically symmetric masses) attracts each other with a force directed along the line joining their centres:

where:

• IMATH_5 N m$^2$ / kg$^2$ is the universal gravitational constant.
• IMATH_8 and $m_2$ are the two masses in kilograms.
• IMATH_10 is the distance between their centres (not between their surfaces).

The force is mutual: each mass exerts the same magnitude of force on the other (Newton's third law).

The inverse-square law

Force is inversely proportional to the square of the distance. Doubling $r$ reduces $F$ to one quarter. Halving $r$ quadruples $F$. This rapid fall-off explains why Earth's gravity dominates near the surface but becomes negligible far from the planet, while still extending to infinity in principle.

Gravitational field strength

The gravitational field strength $g$ at a point is the gravitational force per unit mass on a test mass placed there:

where $M$ is the mass of the source body and $r$ is the distance from its centre. The units are N/kg, which are numerically equal to m/s$^2$.

At Earth's surface ($r = R_E = 6.37 \times 10^6$ m, $M_E = 5.97 \times 10^{24}$ kg):

$g = \frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{(6.37 \times 10^6)^2} = 9.81 \text{ N/kg}$.

Acceleration due to gravity

For a freely falling object of mass $m$ in a gravitational field $g$, the acceleration is $a = g$ (independent of $m$, because $F = m g$ and $a = F / m$). All objects fall with the same acceleration in a given gravitational field, in the absence of air resistance. This is why $g$ appears in projectile-motion equations as the constant downward acceleration.

Field model versus action at a distance

Two equivalent descriptions:

• Action at a distance. The two masses pull on each other directly across empty space.
• Field model. Each mass creates a gravitational field around itself, and any other mass placed in that field experiences a force. The field model is preferred for QCE Physics because it generalises cleanly to electric and magnetic fields in Topic 2.

The field of a point or spherical mass is radial, pointing toward the source mass, and uniform in magnitude on any sphere centred on the source. The field-line picture has lines pointing inward at every point on a sphere, getting denser closer to the mass.

Try it: Universal gravitation calculator. Plug in any two masses and separation, with Earth-Moon and Sun-Earth presets.

Worked example: Earth-Moon force

The Earth ($M_E = 5.97 \times 10^{24}$ kg) and Moon ($M_m = 7.35 \times 10^{22}$ kg) are separated by $r = 3.84 \times 10^8$ m. The force between them is:

$F = \frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 7.35 \times 10^{22}}{(3.84 \times 10^8)^2}$

Numerator: $6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 7.35 \times 10^{22} = 2.93 \times 10^{37}$.

Denominator: $(3.84 \times 10^8)^2 = 1.47 \times 10^{17}$.

$F = 1.99 \times 10^{20}$ N.

This is the force that holds the Moon in orbit around the Earth, and the gravitational pull that drives the tides.

How this appears in IA1 and IA2

IA1 data test. Expect a table of $g$ at various altitudes and a question asking you to extract $M$ of the planet, or a single-line problem asking you to compute $g$ at a given altitude and discuss the apparent weight of an astronaut. Markers focus on candidates who substitute altitude $h$ for $r$ instead of $R_E + h$.

IA2 student experiment. A common IA2 measures local $g$ by timing a simple pendulum at different lengths and linearising $T^2$ against $L$ (slope $= 4 \pi^2 / g$). The Unit 3 universal-gravitation theory provides the framework in the justification section: the value of $g$ extracted should agree with $G M_E / R_E^2$ at the school's altitude.

Common traps

Using altitude instead of distance from the centre. $r$ in the formulas is always measured from the centre of the source body. For a satellite at altitude $h$ above Earth, $r = R_E + h$.

Forgetting to square $r$. The denominator is $r^2$, not $r$. Halving the distance multiplies the force by four, not two.

Confusing $g$ and $G$. $G$ is a universal constant ($6.67 \times 10^{-11}$ N m$^2$/kg$^2$). $g$ depends on the source mass and your distance from it, and is a field strength in N/kg (or an acceleration in m/s$^2$).

Assuming $g = 9.8$ m/s$^2$ everywhere. This is only true near Earth's surface. At higher altitudes or on other planets, recompute $g = G M / r^2$.

Treating gravity as having a cut-off altitude. Gravity extends to infinity, just very weakly. The "edge" of Earth's gravity that students sometimes invoke is fictional.

Conflating gravity with apparent weight. Apparent weightlessness in orbit is caused by free fall, not by the absence of gravity. The astronaut in low Earth orbit still feels nearly the surface gravitational field.

In one sentence

Newton's law of universal gravitation, $F = G m_1 m_2 / r^2$, gives the attractive force between any two masses, and the resulting field strength $g = G M / r^2$ falls off as the inverse square of the distance from the centre of the source, with $g = 9.81$ N/kg at Earth's surface and smaller values at higher altitudes.