KAHIBARO
Table of Contents
Outside a Spherical Body
A very important result in gravitation is that a spherically symmetric body acts, for points outside it, as if all of its mass were concentrated at its center. This means that a planet, star, or uniform sphere can often be treated like a point mass when we are outside it.
If a spherical body has total mass $M$, and a point is at distance $r$ from its center, with $r \ge R$ where $R$ is the radius of the body, then the gravitational field strength is
$$
g(r) = \frac{GM}{r^2}
$$
and it points toward the center of the sphere.
This is the same formula as for a point mass. The direction is radial, meaning along the line joining the point to the center.
For any point outside a spherically symmetric mass distribution,
$$
\vec g(r) = -\frac{GM}{r^2}\hat r
$$
The minus sign shows that the field points inward, toward the center.
This result is why we can use simple gravitational formulas for Earth, the Moon, and planets when studying motion of satellites and falling objects far from the surface details.
At the Surface of a Spherical Body
At the surface of a spherical body, the distance from the center is $r = R$. So the gravitational field strength at the surface is
$$
g = \frac{GM}{R^2}
$$
This quantity is especially important for planets and moons. For Earth, this gives the familiar value near
$$
g \approx 9.8 \, \text{m/s}^2
$$
if we use Earth's mass and radius.
The field depends on two things only, the total mass and the radius. A more massive body gives a stronger field, while a larger radius weakens the field at the surface.
Inside a Uniform Spherical Body
For a uniform sphere, the gravitational field inside the body behaves differently. If you are at a distance $r$ from the center, with $r < R$, only the mass enclosed within radius $r$ contributes to the net field in the simple spherical way.
If the density is uniform, then the enclosed mass is
$$
M_r = M\frac{r^3}{R^3}
$$
Substituting this into the field formula gives
$$
g(r) = \frac{GM_r}{r^2}
= \frac{G}{r^2}\left(M\frac{r^3}{R^3}\right)
= \frac{GM}{R^3}r
$$
So inside a uniform sphere, the field increases linearly with distance from the center.
At the very center, $r=0$, so the field is zero. As you move outward, the field becomes stronger, reaching its maximum value at the surface.
Inside a uniform spherical body,
$$
g(r) = \frac{GM}{R^3}r \qquad (r<R)
$$
So the gravitational field is zero at the center and grows linearly up to the surface.
This result is very different from the outside formula, where the field decreases as $1/r^2$.
Why the Field Is Zero at the Center
At the center of a spherical body, mass surrounds you equally in all directions. The gravitational pulls from opposite sides cancel exactly. Because of this perfect symmetry, the net gravitational field at the center is zero.
More generally, anywhere inside a uniform sphere, the outer layers do not increase the net field the way one might first expect. Spherical symmetry causes their contributions to cancel in the right way, leaving only the effect of the mass inside radius $r$.
Piecewise Formula
For a uniform spherical body of radius $R$ and total mass $M$, the gravitational field strength can be written as
$$
g(r) =
\begin{cases}
\dfrac{GM}{R^3}r, & r<R \\
\dfrac{GM}{r^2}, & r\ge R
\end{cases}
$$
The direction is always toward the center.
This formula shows a smooth transition at the surface, since at $r=R$ both expressions give
$$
g(R)=\frac{GM}{R^2}
$$
Comparison of Inside and Outside Behavior
The behavior of the field is summarized below.
| Region | Formula for magnitude of field | Behavior |
|---|---|---|
| Center, $r=0$ | $0$ | Zero field |
| Inside uniform sphere, $r<R$ | $\dfrac{GM}{R^3}r$ | Increases linearly with $r$ |
| Surface, $r=R$ | $\dfrac{GM}{R^2}$ | Maximum value for uniform sphere |
| Outside sphere, $r>R$ | $\dfrac{GM}{r^2}$ | Decreases as $1/r^2$ |
Graph of Field Strength
If we plot $g$ versus $r$ for a uniform spherical body, the graph starts at zero, rises in a straight line inside the sphere, then curves downward outside according to the inverse square law.
Spherical Shell
A useful special case is a thin spherical shell of mass. For points outside the shell, the field is again
$$
g = \frac{GM}{r^2}
$$
as if all the mass were at the center. But for any point inside the shell, the net gravitational field is zero.
Inside a thin spherical shell,
$$
\vec g = 0
$$
everywhere.
This is a remarkable consequence of spherical symmetry.
Visualizing the Radial Field
The field lines around a spherical body point directly inward toward the center. Farther from the body, the lines spread out, showing that the field gets weaker with distance.
Physical Meaning
These results help explain why the gravitational field near a planet is almost radial, why the field weakens with altitude, and why the internal structure of a sphere matters only when we are inside it. Outside, only the total mass matters. Inside a uniform body, the enclosed mass changes with position, so the field changes in a different way.
For many practical problems, Earth is treated as a sphere, so near or above its surface we use
$$
g(r)=\frac{GM_E}{r^2}
$$
where $M_E$ is Earth's mass. This becomes especially useful in later studies of orbital motion and satellites.
Key results for a spherical body:
$$
\vec g(r) = -\frac{GM}{r^2}\hat r \qquad (r \ge R)
$$
$$
\vec g(r) = -\frac{GM}{R^3}r\,\hat r \qquad (r < R,\ \text{uniform sphere})
$$
$$
\vec g = 0 \qquad (\text{inside a thin spherical shell})
$$
Example
Suppose a uniform planet has mass $M$ and radius $R$. At a point halfway from the center to the surface, $r = R/2$. Then
$$
g\left(\frac{R}{2}\right)=\frac{GM}{R^3}\left(\frac{R}{2}\right)=\frac{GM}{2R^2}
$$
So the field there is half the surface value.
At a point twice the radius from the center, $r=2R$, the field is
$$
g(2R)=\frac{GM}{(2R)^2}=\frac{GM}{4R^2}
$$
so it is one quarter of the surface value.
This nicely shows the contrast between the linear increase inside and inverse square decrease outside.