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A2 12–13

Circular motion and gravitational fields

A Level · syllabus topics 12 and 13 · theory on Paper 4; the practical skills behind it on Paper 5

These two topics belong together because the second is the most important application of the first. A satellite in orbit is nothing more than a body in circular motion whose centripetal force happens to be gravity.

🎯What you need to be able to do

  • Define the radian and express angular displacement in radians; understand and use angular speed, \( \omega = 2\pi/T \) and \( v = r\omega \).
  • Understand that a body in circular motion at constant speed has an acceleration directed towards the centre, and that a resultant force is needed to produce it.
  • Recall and use \( a = r\omega^{2} = v^{2}/r \), and \( F = mr\omega^{2} = mv^{2}/r \).
  • Understand that a gravitational field is a field of force, and define gravitational field strength as force per unit mass.
  • State and use Newton’s law of gravitation, \( F = Gm_1m_2/r^{2} \), and analyse circular orbits using it.
  • Derive and use \( g = GM/r^{2} \) for the field of a point mass, and understand why a uniform sphere behaves as a point mass at its centre for external points.
  • Define gravitational potential, use \( \phi = -GM/r \), and relate it to potential energy \( E_p = -GMm/r \).
  • Understand geostationary orbits and their uses.

📚The physics

Circular motion is accelerated motion even at constant speed, because the velocity is a vector and its direction changes. The acceleration points towards the centre and has magnitude \( v^{2}/r \). There is no such thing as a “centrifugal force” in an inertial frame; centripetal force is a role, filled by whatever real force is available — tension, friction, the normal contact force, gravity. Every question in this topic reduces to identifying which force is playing that role and setting it equal to \( mv^{2}/r \).

A body moving anticlockwise round a circle with velocity arrows tangent at two points A and B. Subtracting the two velocities gives a change in velocity pointing towards the centre. At a third point the acceleration v squared over r is drawn towards the centre.
\( \Delta v = v_B - v_A \) points to the centre, so the acceleration does too.

Newton’s law of gravitation is an inverse-square law between point masses, always attractive, with \( G = 6.67 \times 10^{-11} \) N m\(^{2}\) kg\(^{-2}\). Dividing by the mass being acted on gives the field strength \( g = GM/r^{2} \), which is also the free-fall acceleration at that point — the same quantity in two guises, N kg\(^{-1}\) and m s\(^{-2}\).

Orbits. For a satellite, gravity provides the centripetal force:

\[ \frac{GMm}{r^{2}} = \frac{mv^{2}}{r} \qquad\Longrightarrow\qquad T^{2} = \frac{4\pi^{2}r^{3}}{GM} \]

The satellite’s own mass cancels, which is why every satellite at a given radius has the same speed and period. Substituting \( v = 2\pi r/T \) gives Kepler’s third law, derived in three lines.

Potential is negative because the zero is defined at infinity and gravity is attractive: bringing a mass in from infinity releases energy, so the potential energy falls below zero. \( \phi = -GM/r \). Escape from a body requires kinetic energy at least equal to the depth of the well, \( \tfrac{1}{2}mv^{2} = GMm/r \), giving \( v_{\text{esc}} = \sqrt{2GM/r} \).

Two graphs against distance from the Earth's centre in Earth radii. Field strength falls from 9.81 newtons per kilogram at the surface as one over r squared; potential rises from minus 62.5 megajoules per kilogram towards zero as minus one over r. The geostationary radius, 6.6 Earth radii, is marked.
Outside the Earth: \( g = GM/r^{2} \) and \( \phi = -GM/r \), computed with the real values.

Geostationary orbits require three conditions together: period exactly 24 hours (strictly, one sidereal day), orbit in the plane of the equator, and travelling west to east. Only one radius satisfies the first, about \( 4.2 \times 10^{7} \) m from the Earth’s centre.

✏️Worked example

Find the radius of a geostationary orbit. Take \( M_{\text{Earth}} = 5.97 \times 10^{24} \) kg.

\( T = 24 \times 3600 = 8.64 \times 10^{4} \) s.

\( r^{3} = GMT^{2}/4\pi^{2} = (6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times (8.64 \times 10^{4})^{2})/(4\pi^{2}) \)

\[ r^{3} = 7.53 \times 10^{22}\ \text{m}^{3} \quad\Longrightarrow\quad r = 4.22 \times 10^{7}\ \text{m} \]
The Earth and the geostationary orbit drawn to scale, viewed from above the North Pole: the orbit radius is 4.22 times ten to the seven metres, about 6.6 Earth radii, and the height above the surface about 3.6 times ten to the seven metres. Both the Earth and the satellite move west to east.
The worked example’s answer to scale: \( r = 4.22 \times 10^{7} \) m is measured from the centre, not the surface.
The mark people actually lose here is answering with the height above the surface when the question asked for the orbital radius, or the reverse. \(r\) in every one of these formulae is measured from the centre of the Earth; the height above the surface is \(r\) minus \( 6.37 \times 10^{6} \) m, which is about \( 3.6 \times 10^{7} \) m. Read the question, then write down explicitly which of the two you have found. The other classic loss is using \( T = 24 \) without converting to seconds.

🔭See it happen

Whirl a bung on a string through a glass tube with a weight hanging below, and vary the radius: the tension is set by the hanging weight, so \( v^{2}/r \) is fixed and the period changes visibly with radius. It is one of the few A Level experiments where the relationship can be felt in the hand as well as measured.

🔗Go deeper — other people’s work

The links below are not mine. They are here because they are good, and they may move or disappear without warning.

  • PhET, Gravity and Orbits — turn gravity off mid-orbit and watch the satellite leave along the tangent.
  • The Physics Classroom, “Circular Motion and Satellite Motion”.
  • Isaac Physics, “Circular Motion” and “Gravitational Fields”.