\subsection{Gravitation and the Inverse-Square Field}

This subsection treats gravity from the local field viewpoint. A source mass creates a gravitational field $\vec{g}(\vec{r})$, and a test mass placed at position $\vec{r}$ experiences a gravitational force determined by that field.

\dfn{Gravitational field and the source--test-mass viewpoint}{Let $\vec{r}$ denote the position vector of a field point, and let a test mass $m$ be placed at that point. If $\vec{F}_g(\vec{r})$ denotes the gravitational force on the test mass due to some source mass distribution, then the \emph{gravitational field} at $\vec{r}$ is defined by
\[
\vec{g}(\vec{r})=\frac{\vec{F}_g(\vec{r})}{m}.
\]
Thus the field is the gravitational force per unit mass. Equivalently, any test mass $m$ placed at that point satisfies
\[
\vec{F}_g=m\vec{g}.
\]
The SI units of $\vec{g}$ are $\mathrm{N/kg}$, which are equivalent to $\mathrm{m/s^2}$. The test mass is assumed small enough that it does not significantly change the source field.}

\thm{Newton's law of gravitation and the inverse-square field}{Let $G$ denote the universal gravitational constant. Let a point mass or a spherically symmetric source have total mass $M$ and center at the origin. Let $\vec{r}$ denote the position vector of a field point, let $r=|\vec{r}|$ denote its distance from the center, and let $\hat{r}=\vec{r}/r$ denote the outward radial unit vector. If the source is spherically symmetric, assume the field point lies outside the source. If a test mass $m$ is placed at that point, then the gravitational force on the test mass is
\[
\vec{F}_g(\vec{r})=-\frac{GMm}{r^2}\hat{r}.
\]
Therefore the gravitational field produced by the source is
\[
\vec{g}(\vec{r})=-\frac{GM}{r^2}\hat{r}.
\]
Its magnitude is
\[
g(r)=\frac{GM}{r^2}.
\]
The negative sign shows that the field points toward the source, so gravity is attractive.}

\pf{Connecting the force law to the field law}{Let $M$ denote the source mass, let $m$ denote the test mass, and let $r$ denote the separation between their centers. Newton's law of gravitation states that the magnitude of the gravitational force is
\[
F_g=\frac{GMm}{r^2}.
\]
Because gravity is attractive, the force on the test mass points toward the source. Since $\hat{r}$ points radially outward, the force direction is $-\hat{r}$. Therefore,
\[
\vec{F}_g(\vec{r})=-\frac{GMm}{r^2}\hat{r}.
\]
Now apply the definition of gravitational field:
\[
\vec{g}(\vec{r})=\frac{\vec{F}_g(\vec{r})}{m}=-\frac{GM}{r^2}\hat{r}.
\]
Thus the field depends on the source mass $M$ and the location $\vec{r}$, but not on the test mass $m$.}

\cor{Near-Earth constant-$g$ approximation}{Let Earth have mass $M_E$ and radius $R_E$. Let a point above Earth's surface have altitude $h$, and let $r=R_E+h$ denote its distance from Earth's center. Then the gravitational field is
\[
\vec{g}(\vec{r})=-\frac{GM_E}{(R_E+h)^2}\hat{r}.
\]
If $h\ll R_E$, then $R_E+h\approx R_E$, so the field magnitude is approximately constant:
\[
g(r)\approx \frac{GM_E}{R_E^2}=g.
\]
Over a small region near Earth's surface, the radial direction changes very little. If the positive $y$-axis is chosen upward, then locally
\[
\vec{g}\approx -g\hat{\jmath},
\]
which is the constant-acceleration free-fall model used near Earth's surface.}

\qs{Worked example}{Let Earth have mass $M_E=5.97\times 10^{24}\,\mathrm{kg}$, let Earth have radius $R_E=6.37\times 10^6\,\mathrm{m}$, and let the gravitational constant be $G=6.67\times 10^{-11}\,\mathrm{N\,m^2/kg^2}$. A spacecraft has mass $m=500\,\mathrm{kg}$ and is at altitude $h=4.00\times 10^5\,\mathrm{m}$ above Earth's surface. Let $r=R_E+h$ denote the spacecraft's distance from Earth's center, let $\hat{r}$ denote the outward radial unit vector, and let $g_0=9.80\,\mathrm{m/s^2}$ denote the near-surface gravitational field magnitude.

Find the gravitational field magnitude $g(r)$ at the spacecraft, the gravitational force $\vec{F}_g$ on the spacecraft, and the percent by which the near-surface value $g_0$ overestimates the true field magnitude at that altitude.}

\sol First compute the distance from Earth's center:
\[
r=R_E+h=6.37\times 10^6\,\mathrm{m}+4.00\times 10^5\,\mathrm{m}=6.77\times 10^6\,\mathrm{m}.
\]
The inverse-square field magnitude is
\[
g(r)=\frac{GM_E}{r^2}.
\]
Substitute the given values:
\[
g(r)=\frac{\left(6.67\times 10^{-11}\,\mathrm{N\,m^2/kg^2}\right)\left(5.97\times 10^{24}\,\mathrm{kg}\right)}{\left(6.77\times 10^6\,\mathrm{m}\right)^2}.
\]
This gives
\[
g(r)\approx 8.69\,\mathrm{m/s^2}.
\]
So the gravitational field at the spacecraft has magnitude $8.69\,\mathrm{m/s^2}$ and points toward Earth's center.

Now use $\vec{F}_g=m\vec{g}$. Since $\vec{g}=-(8.69\,\mathrm{m/s^2})\hat{r}$ at that location,
\[
\vec{F}_g=m\vec{g}=\left(500\,\mathrm{kg}\right)\left[-(8.69\,\mathrm{m/s^2})\hat{r}\right].
\]
Therefore,
\[
\vec{F}_g\approx -4.34\times 10^3\hat{r}\,\mathrm{N}.
\]
Its magnitude is
\[
F_g\approx 4.34\times 10^3\,\mathrm{N},
\]
and the negative sign means the force points toward Earth's center.

Finally, compare this with the near-surface value $g_0=9.80\,\mathrm{m/s^2}$. The amount of overestimate is
\[
g_0-g(r)=9.80-8.69=1.11\,\mathrm{m/s^2}.
\]
Thus the percent overestimate is
\[
\frac{g_0-g(r)}{g(r)}\times 100\% = \frac{1.11}{8.69}\times 100\% \approx 12.8\%.
\]
So at an altitude of $4.00\times 10^5\,\mathrm{m}$, the true gravitational field magnitude is about $8.69\,\mathrm{m/s^2}$, the spacecraft's gravitational force is about $4.34\times 10^3\,\mathrm{N}$ toward Earth, and the near-surface constant-$g$ model overestimates the field by about $12.8\%$.