Gravitational Fields

Use this page to revise the following concepts within gravitational fields:

Newton’s Law of Gravitation states that any two bodies in the universe attract each other with a force directly proportional to their masses and inversely proportional to the square of the distance between them described by the equation

\[F=G \frac{m_1m_2}{r^2}\]

Where \(m_1\) and \(m_2\) represent the different masses, \(r\) is the distance between the centre of mass of each object and \(G\) is the Universal Gravitational Constant.

\[ G = 6.67 \times 10^{-11} \mathrm{~Nm}^2\mathrm{~kg}^{-2} \]

A diagram demonstrating Newton’s Law of Universal Gravitation. It depicts the Sun (m1) and the Earth (m2). The distance between the centre of the two bodies is denoted by r. Arrows represent the force of the Sun pulled towards the Earth (F1) and the Earth towards the Sun (F2)

The gravitational field strength of any mass can be derived from equating equation \(F = mg\). Note we represent the letter g to represent the acceleration due to the Earth’s gravitational field or acceleration due to gravity.

\[ F_{m_1\text{ on }m_2} = m_2g \] \[ G\frac{m_1m_2}{r^2} = m_2g \] \[ G\frac{m_1}{r^2} = g \]

Where:

  • \(m_1\) is the mass of the object creating the field.
  • \(m_2\) is the mass of the object within the field.
  • \(r\) is the distance from the centre of mass of the object creating the field.
  • \(g\) is the gravitational field strength at that distance from the object creating the field.

Check your understanding of gravitational fields and forces by answering the following questions.

Worked Example

Show that the acceleration due to gravity on the surface of Earth is \(9.8 \mathrm{~ms}^{-2}\), correct to one decimal place, given:

\[ G = 6.67 \times 10^{-11} \mathrm{~Nm}^2\mathrm{~kg}^{-2} \]

\[ m_{\text{Earth}} = 5.98 \times 10^{24} \mathrm{~kg} \]

\[ r_{\text{Earth}} = 6.37 \times 10^6 \mathrm{~m} \]

\[ g = G\frac{m_E}{r^2} \]

\[ \begin{aligned} g &= \frac{(6.67 \times 10^{-11})(5.98 \times 10^{24})}{(6.37 \times 10^6)^2} \\ &\approx 9.83 \mathrm{~ms}^{-2} \\ &\approx 9.8 \mathrm{~ms}^{-2} \end{aligned} \]


Gravitational Fields and mass of objects within the field

From Newton’s Law of Gravitation, the field diagram to represent the field around any mass will always be a monopole that is attractive – that is, the field lines will have arrowheads pointing in towards the point mass.

The gravitational field strength is proportional to the mass of the object and inversely proportional to the square of the radius from the centre of mass.

\[g\propto{M}\]

\[g\propto\frac{1}{r^2}\]

Therefore, as the gravitational field strength is independent of the mass of any object within the field, each object will experience the same acceleration regardless of its own mass. This can be seen in the video below, where feathers and a bowling ball dropped from the same height in a vacuum fall at the same speed, despite their differing masses.

Inverse Square Law

As the gravitational field strength does decrease as the distance from the body creating the field increases by a squared factor as \(g\propto\frac{1}{r^2}\). The inverse square law is able to represent how the field decreases over an increasing distance. The image below depicts how the area affected by a point source is proportional to the square of the distance.

A diagram shows how field strength, E, decreases by the inverse of the distance from its source. At r=1 with a square-area of 1, E=1. At r=2 with a square-area of 4, E=1/4. At r=3 with a square-area of 9, E=1/9.

Check your understanding by filling in the blanks.