understand that a gravitational field is an example of a field of force and define gravitational field as force per unit mass

Published by Patrick Mutisya · 14 days ago

Cambridge A-Level Physics 9702 – Gravitational Field

Gravitational Field

1. What is a field of force?

A field of force is a region of space in which a force can be experienced by a test particle placed at any point in that region. The magnitude and direction of the force at each point are described by a vector quantity called the field.

  • It allows us to describe interactions without direct contact.

  • Examples include electric fields, magnetic fields, and gravitational fields.

2. Definition of the gravitational field

The gravitational field, denoted by \$\mathbf{g}\$, is defined as the gravitational force \$\mathbf{F}\$ experienced by a unit mass placed at a point in space:

\$\mathbf{g} = \frac{\mathbf{F}}{m}\$

Thus, the gravitational field has the dimensions of acceleration and its SI unit is metres per second squared (m s\(^{-2}\)).

3. Expression for the gravitational field of a point mass

For a point mass \$M\$, the magnitude of the gravitational field at a distance \$r\$ from the centre of the mass is given by Newton’s law of gravitation:

\$g = \frac{GM}{r^{2}}\$

where:

  • \$G = 6.674 \times 10^{-11}\ \text{N m}^{2}\text{kg}^{-2}\$ is the universal gravitational constant.

  • \$M\$ is the mass creating the field.

  • \$r\$ is the radial distance from the mass centre.

4. Direction of the gravitational field

The gravitational field vector points radially inward toward the mass that creates the field. It is therefore a central field.

5. Comparison with other fields

QuantitySymbolUnitExpression
Gravitational field\$\mathbf{g}\$m s\(^{-2}\)\$\displaystyle \mathbf{g} = \frac{\mathbf{F}}{m} = -\frac{GM}{r^{2}}\hat{r}\$
Electric field\$\mathbf{E}\$V m\(^{-1}\) (or N C\(^{-1}\))\$\displaystyle \mathbf{E} = \frac{\mathbf{F}}{q} = \frac{kQ}{r^{2}}\hat{r}\$
Magnetic field\$\mathbf{B}\$T (tesla)\$\displaystyle \mathbf{F}=q\mathbf{v}\times\mathbf{B}\$ (field defined via force on moving charge)

6. Using the gravitational field to find force

Once the gravitational field \$\mathbf{g}\$ at a point is known, the force on any mass \$m\$ placed at that point is simply:

\$\mathbf{F}=m\mathbf{g}\$

This relationship mirrors the way electric force is obtained from the electric field.

7. Example calculation

  1. Find the gravitational field at the surface of the Earth (\$R{\oplus}=6.37\times10^{6}\ \text{m}\$, \$M{\oplus}=5.97\times10^{24}\ \text{kg}\$).

  2. Use \$g = \dfrac{GM{\oplus}}{R{\oplus}^{2}}\$.

  3. Substituting the values gives \$g \approx 9.81\ \text{m s}^{-2}\$.

  4. For a 2 kg mass placed on the surface, the weight (gravitational force) is \$F = mg = 2 \times 9.81 = 19.62\ \text{N}\$.

8. Suggested diagram

Suggested diagram: Vector representation of the gravitational field \$\mathbf{g}\$ around a spherical mass \$M\$, showing radial arrows decreasing in magnitude with distance \$r\$.

9. Key points to remember

  • The gravitational field is a vector field representing force per unit mass.

  • Its magnitude for a point mass follows an inverse‑square law: \$g = GM/r^{2}\$.

  • Direction is always toward the source mass.

  • Knowing \$\mathbf{g}\$ allows immediate calculation of the gravitational force on any mass via \$\mathbf{F}=m\mathbf{g}\$.