= Solution
The <volume> $V$ is the three-dimensional space occupied by the body. Its <mass> $m$ measures its inertia: in <Newtonian mechanics> the resultant <force> is $m\mathbf a$. Its <gravitational weight> is the <force> exerted on it by <gravity>,
$$
\boxed{\mathbf W=m\mathbf g.}
$$
For a uniform body, $m=\rho V$, where $\rho$ is its <mass density>; more generally $m=\int\rho\,dV$. <Volume> can change with <temperature> or <pressure> while the body's <mass> remains fixed in the classical closed-system model. Equal volumes of different materials need not have equal <masses>.
\b[The same <mass> need not have the same <gravitational weight>.] The local <gravitational acceleration> changes between planets and with position on Earth. Outside a spherical Earth, $g(r)=GM_\oplus/r^2$, so increasing altitude reduces it. Earth's rotation and shape also affect terrestrial weighing: a scale at rest on the rotating Earth responds to the effective acceleration including the centrifugal contribution, rather than simply to a spherical gravitational field.
Distinguish <gravitational weight> from <apparent weight>. A supporting scale measures a contact <force>. In a vertically accelerating cabin, upward <acceleration> $b$ gives $N=m(g+b)$ while contact is maintained. During common <free fall>, $N=0$ although $mg$ remains nonzero. Immersion in a <fluid> can similarly reduce the support needed through <buoyancy>, without removing the gravitational attraction. These distinctions explain why a change in a balance reading is not automatically a change of <mass>.
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