§ 13.5 — Gravitational Field and Force

The Concept of a Gravitational Field

A gravitational field is the region of space around any mass where gravitational effects are experienced. Rather than thinking of one object directly pulling another across empty space, we can imagine that each mass creates a field that permeates space, and other masses respond to that field.

This field concept, generalized by Einstein in general relativity, proves mathematically elegant and physically powerful. It separates the problem into two parts: (1) how a mass creates a field, and (2) how other masses respond to that field.

Gravitational Field Intensity

The gravitational field \( \vec{g}(\vec{r}) \) is defined as the gravitational force per unit mass at a point:

\[ \vec{g}(\vec{r}) = \frac{\vec{F}}{m} = -G\frac{M}{r^2}\hat{r} \]

where \( M \) is the source mass creating the field, and \( \hat{r} \) is the unit vector pointing outward from \( M \). The negative sign indicates the field points inward (toward the mass).

Units: The gravitational field has units of acceleration (m/s² or N/kg).

At Earth's surface, the gravitational field has magnitude:

\[ g = G\frac{M_E}{R_E^2} = \frac{(6.674 \times 10^{-11})(5.972 \times 10^{24})}{(6.371 \times 10^6)^2} \approx 9.81 \, \text{m/s}^2 \]

This is the familiar acceleration experienced by falling objects.

Gravitational Potential

Related to the gravitational field is the gravitational potential \( \Phi(r) \):

\[ \Phi(r) = -G\frac{M}{r} \]

The relationship between potential and field is analogous to the relationship between electric potential and electric field:

\[ \vec{g} = -\nabla\Phi = -\frac{d\Phi}{dr}\hat{r} \]

The gradient of potential gives the field. In one dimension:

\[ g_r = -\frac{d\Phi}{dr} = -\frac{d}{dr}\left(-G\frac{M}{r}\right) = -G\frac{M}{r^2} \]

The potential is particularly useful because potentials from different masses superpose (add algebraically), whereas forces require vector addition. For a uniform spherical shell of mass \( M \) at radius \( R \):

\[ \Phi(r) = \begin{cases} -\frac{GM}{R} & \text{if } r \geq R \\ -\frac{GM}{R} & \text{if } r < R \end{cases} \]

Remarkably, the potential inside a uniform spherical shell is constant!

Equipotential Surfaces

Surfaces of constant potential are called equipotential surfaces. For a spherically symmetric mass, equipotentials are concentric spheres. Since \( \vec{g} = -\nabla\Phi \), the field is always perpendicular to equipotential surfaces.

This geometry has important consequences: no work is required to move a mass along an equipotential surface (since the force is always perpendicular to the displacement).

Superposition of Fields and Potentials

For multiple masses \( M_i \) at positions \( \vec{r}_i \), the net gravitational field at point \( \vec{r} \) is:

\[ \vec{g}(\vec{r}) = -G\sum_i \frac{M_i}{|\vec{r} - \vec{r}_i|^2}\hat{r}_i \]

Alternatively, calculate the potential first (scalar addition):

\[ \Phi(\vec{r}) = -G\sum_i \frac{M_i}{|\vec{r} - \vec{r}_i|} \]

Then find the field: \( \vec{g} = -\nabla\Phi \). This approach often simplifies calculations for symmetric configurations.

A Parallel Problem: Gravitational Field of a Spherical Shell Segment

Consider a hemispherical shell of uniform density with total mass \( M = 1000 \, \text{kg} \) and radius \( R = 2.0 \, \text{m} \). Find the gravitational field at the center of the flat face (at the geometric center of the original complete sphere).

Solution:

By symmetry, we need only consider the component perpendicular to the flat face. Consider a small mass element \( dm \) on the shell at angle \( \theta \) from the axis perpendicular to the flat face.

For the hemisphere, integrate from \( \theta = 0 \) to \( \theta = \pi/2 \):

The surface area element in spherical coordinates is \( dA = R^2\sin\theta \, d\theta \, d\phi \).

For a uniform shell, \( \sigma = M/(2\pi R^2) \) (mass per unit area of hemisphere).

The mass element is \( dm = \sigma dA = \sigma R^2\sin\theta \, d\theta \, d\phi \).

The distance from any point on the shell to the center is \( r = R \).

The component of gravitational field along the axis (perpendicular to the flat face) is:

\[ g_z = -\int_0^{2\pi}\int_0^{\pi/2} G\frac{dm}{R^2}\cos\theta \]

where \( \cos\theta \) is the component along the axis.

\[ g_z = -G\sigma \int_0^{2\pi}d\phi \int_0^{\pi/2}\sin\theta\cos\theta \, d\theta \]

\[ g_z = -2\pi G\sigma \int_0^{\pi/2}\sin\theta\cos\theta \, d\theta \]

Using \( \sin\theta\cos\theta = \frac{1}{2}\sin(2\theta) \):

\[ g_z = -\pi G\sigma \int_0^{\pi/2}\sin(2\theta) \, d\theta = -\pi G\sigma \left[-\frac{1}{2}\cos(2\theta)\right]_0^{\pi/2} \]

\[ g_z = -\pi G\sigma \left[-\frac{1}{2}(-1 - 1)\right] = -\pi G\sigma \]

Substituting \( \sigma = M/(2\pi R^2) \):

\[ g_z = -\pi G \frac{M}{2\pi R^2} = -\frac{GM}{2R^2} \]

\[ g_z = -\frac{(6.674 \times 10^{-11})(1000)}{2(2.0)^2} = -8.34 \times 10^{-11} \, \text{m/s}^2 \]

The negative sign indicates the field points toward the hemisphere (into the flat face). The magnitude is exactly half what it would be for a complete spherical shell of the same radius.

Real-World Applications

Understanding gravitational fields enables precise calculations in gravitational surveying (detecting subsurface density variations), modeling planetary interiors, and predicting perturbations in spacecraft trajectories. The field approach also forms the foundation for understanding tidal forces, which we'll explore in the final section.

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