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

Gravitational Field

13.1 Definition – a field of force

• A field of force is a region of space in which a test particle experiences a force even though there is no physical contact.
• The gravitational field is the specific field of force produced by mass.
• It is defined at a point in space as the gravitational force per unit test‑mass:

  \[

  \mathbf{g}(\mathbf{r}) \;=\; \frac{\mathbf{F}(\mathbf{r})}{m_{\text{test}}}

  \]

  where \(\mathbf{g}\) is a vector field (direction and magnitude vary with position).

• The scalar magnitude of the field is

  \[

  g = \frac{|\mathbf{F}|}{m_{\text{test}}}.

  \]

• Dimensions: \( \mathrm{LT^{-2}} \) (the same as acceleration).
  SI unit: metres per second squared (m s\(^{-2}\)).
• Direction: always toward the source mass; i.e. opposite to the outward radial unit vector \(\hat r\).
• Important reminder: the definition does not depend on the particular test mass chosen – any sufficiently small mass placed at the same point will experience the same \(\mathbf{g}\).

13.2 Gravitational field of a point mass

Newton’s law of universal gravitation gives the force on a test mass \(m\) due to a point source of mass \(M\) at a distance \(r\):

\[

\mathbf{F}= -\,\frac{GMm}{r^{2}}\;\hat r

\]

• The minus sign indicates that the force is attractive (directed toward the source).

Dividing the force by the test mass \(m\) (the definition of the field) gives the intermediate step:

\[

\mathbf{g}= \frac{\mathbf{F}}{m}= -\,\frac{GM}{r^{2}}\;\hat r

\]

• Vector form: \(\displaystyle \mathbf{g}= -\frac{GM}{r^{2}}\hat r\) – shows both magnitude and direction.
• Scalar magnitude: \(\displaystyle g = \frac{GM}{r^{2}}\).
• Gravitational constant: \(G = 6.674\times10^{-11}\;\text{N m}^{2}\text{kg}^{-2}\).

13.3 Field lines and the near‑Earth approximation

• Field lines are drawn tangent to \(\mathbf{g}\) at every point. For a point mass they are straight lines radiating inward toward the centre; the density of lines increases as \(r\) decreases, reflecting the inverse‑square dependence.
• Because the Earth’s radius \(R_{\oplus}=6.37\times10^{6}\,\text{m}\) is much larger than the height of everyday objects, the change in \(r\) over the region of interest is negligible. Hence the field can be treated as uniform near the surface:

  \[

  g{\text{near Earth}}\approx \frac{GM{\oplus}}{R_{\oplus}^{2}} = 9.81\;\text{m s}^{-2}.

  \]

  In this approximation the field lines are parallel and equally spaced, just as in a uniform electric field.

13.4 Using the gravitational field to find force

Once the field \(\mathbf{g}\) at a point is known, the gravitational force on any mass \(m\) placed at that point is obtained directly:

\[

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

\]

• This “shortcut’’ avoids recomputing the inverse‑square law each time a new mass is introduced.

Example – weight of a 2 kg object on Earth

1. Calculate the field at the Earth’s surface:

   \[

   g=\frac{GM{\oplus}}{R{\oplus}^{2}}

   =\frac{6.674\times10^{-11}\times5.97\times10^{24}}{(6.37\times10^{6})^{2}}

   \approx9.81\;\text{m s}^{-2}.

   \]

2. Force (weight) on a 2 kg mass:

   \[

   F = m g = 2\times9.81 = 19.6\;\text{N},

   \]

   directed vertically downward.

Comparison with other fields

Suggested diagram

Key points to remember

• The gravitational field is a vector field defined as force per unit test mass at a specific point in space.
• For a point mass: \(\displaystyle \mathbf g = -\frac{GM}{r^{2}}\hat r\) (inverse‑square law).
• Field lines point toward the source; near the Earth’s surface the field is effectively uniform.
• Knowing \(\mathbf g\) allows the immediate calculation of gravitational force: \(\mathbf F = m\mathbf g\).
• Remember the sign convention – the negative sign indicates that the field (and the force it produces) is directed toward the source mass.