2.6 Gravitational Force

Learning objectives

Locate the interaction

Every pair of masses attracts. Each gravitational force acts along the line joining the systems' centers of mass, toward the other system.

Calculate force magnitude

$|\vec F_g|=G\frac{m_1m_2}{r^2}$

Predict proportional changes

Change	New gravitational-force magnitude
Double either mass	$2\|\vec F_g\|$
Double both masses	$4\|\vec F_g\|$
Double center-to-center distance $r$	$\|\vec F_g\|/4$

Move from force to field

A gravitational field describes the force effect at each position without requiring a test object to remain there. For source mass $M$ , $|\vec g|=\dfrac{|\vec F_g|}{m}=G\dfrac{M}{r^2}$ . Field strength has units $\text{N/kg}$ ; if gravity is the only force, its numerical value equals free-fall acceleration in $\text{m/s}^2$ .

Distinguish mass and weight

The gravitational force from an astronomical body on a relatively small nearby object is its weight: $\text{Weight}=F_g=mg$ . Mass is measured in kilograms; weight is a force measured in newtons. Here $r$ is center-to-center distance, not the gap between surfaces.

Check whether change is negligible

Gravitational force can be treated as constant over a motion when the change in center-to-center distance is small enough that the resulting change in gravitational force is negligible for the analysis.

Use the near-Earth approximation

$g\approx 10\ \text{N/kg}\approx 10\ \text{m/s}^2\quad\text{near Earth's surface}$

Apply it locally

Example: for an object moving through a classroom-scale height near Earth's surface, its distance from Earth's center changes by a negligible fraction. Use one nearly constant $g$ , so a fixed mass has nearly constant weight $F_g=mg$ throughout that motion.

Know when it fails

This is an approximation, not a universal rule. If the change in center-to-center distance is large enough that $G M/r^2$ changes appreciably, calculate the changing field or force instead of using one constant $g$ .

Name the measured force

Your apparent weight is the magnitude of the normal force exerted on you by a supporting surface. Your gravitational force is $F_g=mg$ ; these are different forces and need not have equal magnitudes.

Write the vertical force sum

$\sum F_y=N-mg=ma_y\quad(\text{upward positive})$

Compare acceleration cases

Vertical contact case	Apparent weight $N$
$a_y>0$	$N>mg$ : you feel heavier
$a_y=0$	$N=mg$
$a_y<0$ while contact remains	$0<N<mg$ : you feel lighter
Free fall	$N=0$ : weightless

Explain weightlessness

Example: you and a freely falling elevator accelerate together under gravity alone. The floor no longer needs to push on you, so $N=0$ and your apparent weight is zero—even though gravity still acts and $F_g=mg$ is not zero.

Apply the boundary

Weightlessness means zero support force, not zero gravity. The equivalence principle says that, using only local observations, an observer in a noninertial frame cannot distinguish apparent-weight effects from those produced by a gravitational field.

Measure resistance to change

Inertial mass measures how strongly an object's motion resists changing during an interaction. For the same net force, a larger inertial mass produces a smaller acceleration.

Measure gravitational interaction

Gravitational mass determines how strongly a system participates in gravitational attraction: it appears in the gravitational-force relationship between masses.

Separate the two roles

Role	Revealed by	Relationship
Inertial mass $m_i$	Response to net force	$\vec F_{\text{net}}=m_i\vec a$
Gravitational mass $m_g$	Strength of gravitational interaction	$\|\vec F_g\|\propto m_g$

Interpret equivalence

Experiments verify that inertial and gravitational mass are equivalent: for a given object, their measured values are equal. The roles remain conceptually distinct—one describes response to force, the other gravitational interaction—so equivalence is an empirical result, not merely a definition.

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