analyse circular orbits in gravitational fields by relating the gravitational force to the centripetal acceleration it causes

Cambridge A‑Level Physics 9702 – Gravitational Fields & Circular Orbits

Learning objective

Analyse circular orbits in gravitational fields by relating the gravitational force to the centripetal acceleration it provides, and use the concepts of gravitational field strength, orbital energy and escape velocity.

13.1 Gravitational field

• Definition (syllabus wording): the gravitational field strength g at a point is the force that a test mass would experience per unit mass at that point.
• Mathematically

  \$\mathbf{g}=\frac{\mathbf{F}g}{m{\text{test}}}\qquad\text{units: N kg}^{-1}\;(= \text{m s}^{-2})\$

• Direction – always points radially toward the centre of the mass that creates the field.
• Field‑lines – drawn as arrows pointing inward; the density of lines is proportional to the magnitude of g.

  

Key result for a point (or spherically symmetric) mass M

At a distance r from the centre, the field strength is

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

with G = 6.674\times10^{-11}\;{\rm N\,m^{2}\,kg^{-2}}.

13.2 Gravitational force between point masses

Newton’s law of universal gravitation gives the attractive force between two point masses m₁ and m₂ separated by a centre‑to‑centre distance r:

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

• Magnitude:  \(F_g = \dfrac{GMm}{r^{2}}\) (inverse‑square dependence).
• Vector form shows the force is directed along the line joining the centres (central force).
• Because the work done depends only on the end points, the force is conservative; a potential energy function \(U=-GMm/r\) can be defined.

13.3 Deriving the gravitational field from the force

1. Place a test mass m at distance r from a point mass M.
2. From the law above, the force on the test mass is \(\displaystyle F_g=\frac{GMm}{r^{2}}\) toward M.
3. Divide by the test mass to obtain the field strength:

   \$\mathbf g = \frac{\mathbf F_g}{m}= \frac{GM}{r^{2}}\;\hat{\mathbf r}\$

This reproduces the result in 13.1, confirming that the field of a point mass falls off as \(1/r^{2}\).

13.4 Circular orbits in a gravitational field

13.4.1 Centripetal force for uniform circular motion

An object of mass m moving with speed v in a circle of radius r requires a centripetal force

\$F_c = \frac{mv^{2}}{r}\$

directed toward the centre of the circle.

13.4.2 Equating gravitational and centripetal forces

For a satellite of mass m orbiting a much larger body of mass M (planet, star, …) the gravitational attraction supplies the required centripetal force:

\$\frac{GMm}{r^{2}} = \frac{mv^{2}}{r}\$

• Cancel the satellite mass m and one factor of r.

Resulting orbital speed (circular‑orbit speed)

\$v = \sqrt{\frac{GM}{r}}\$

Note: the speed is independent of the satellite’s mass.

13.4.3 Orbital period – Kepler’s third law for circular orbits

The period \(T\) is the time for one full revolution. Using the circumference \(2\pi r\):

\$\$T = \frac{2\pi r}{v}=2\pi r\sqrt{\frac{r}{GM}}

= 2\pi\sqrt{\frac{r^{3}}{GM}}\$\$

Thus \(T^{2}\propto r^{3}\), the circular‑orbit form of Kepler’s third law.

13.4.4 Orbital energy

13.4.5 Escape velocity

The minimum speed needed to reach \(r\to\infty\) with zero kinetic energy is obtained by setting the total energy to zero:

\$\$\tfrac12 mv_{\text{esc}}^{2} - \frac{GMm}{r}=0

\;\;\Longrightarrow\;\;

v_{\text{esc}} = \sqrt{\frac{2GM}{r}}\$\$

Escape speed is \(\sqrt{2}\) times the circular‑orbit speed at the same radius.

Worked example – low‑Earth‑orbit satellite

1. Given: \(M_{\oplus}=5.97\times10^{24}\,\text{kg}\), altitude \(h=400\,\text{km}\).
   

   Earth radius \(R{\oplus}=6.37\times10^{6}\,\text{m}\) → orbital radius \(r=R{\oplus}+h=6.78\times10^{6}\,\text{m}\).

2. Orbital speed:

   \$\$v=\sqrt{\frac{GM_{\oplus}}{r}}

   =\sqrt{\frac{(6.674\times10^{-11})(5.97\times10^{24})}{6.78\times10^{6}}}

   \approx 7.67\times10^{3}\,\text{m s}^{-1}\$\$

3. Orbital period:

   \$\$T=2\pi\sqrt{\frac{r^{3}}{GM_{\oplus}}}

   \approx 5.5\times10^{3}\,\text{s}\approx 92\;\text{min}\$\$

4. Escape speed at this altitude:

   \$v_{\text{esc}}=\sqrt{2}\,v\approx1.09\times10^{4}\,\text{m s}^{-1}\$

Common misconceptions

• “Gravity depends on speed.” The gravitational force \(F_g\) depends only on the masses and separation, not on the orbital speed.
• “Larger radius → faster speed.” For a given central mass, \(v\propto r^{-1/2}\); increasing the radius actually reduces the speed.
• “Satellite mass is irrelevant.” The mass cancels in the expressions for \(v\) and \(T\), but it appears in kinetic energy, total orbital energy and the launch energy required.
• “All circular orbits have the same period.” Period varies as \(T\propto r^{3/2}\); only orbits with the same radius share a period.

Extension – elliptical orbits (vis‑viva equation)

For any bound orbit with semi‑major axis \(a\) and instantaneous distance \(r\) from the focus, the speed is

\$v = \sqrt{GM\!\left(\frac{2}{r}-\frac{1}{a}\right)}\$

• When \(r=a\) (a circular orbit) the equation reduces to \(v=\sqrt{GM/r}\).
• The total mechanical energy depends only on \(a\):

  \$E = -\frac{GMm}{2a}\$

Quick‑check questions

1. Derive the orbital period of a moon orbiting Jupiter, given \(M_J=1.90\times10^{27}\,\text{kg}\) and \(r=4.22\times10^{8}\,\text{m}\).
2. If the orbital radius of a satellite is doubled, by what factor does its orbital speed change? By what factor does its period change?
3. Explain why a low‑Earth‑orbit satellite must travel faster than a geostationary satellite.
4. Calculate the escape speed from the surface of the Moon (\(M{M}=7.35\times10^{22}\,\text{kg}\), \(R{M}=1.74\times10^{6}\,\text{m}\)).
5. Using the vis‑viva equation, find the speed of a spacecraft at periapsis (\(r_p=7.0\times10^{6}\,\text{m}\)) of an elliptical orbit around Earth with \(a=1.5\times10^{7}\,\text{m}\).

Summary

• The gravitational field of a point mass is \(\displaystyle g=GM/r^{2}\), directed toward the mass.
• Newton’s law of gravitation: \(\displaystyle \mathbf F_g = -GMm/r^{2}\,\hat{\mathbf r}\) (inverse‑square, central, conservative).
• For a circular orbit, equating gravity with the required centripetal force gives

  \(\displaystyle v=\sqrt{GM/r}\) and \(\displaystyle T=2\pi\sqrt{r^{3}/GM}\).

• Orbital energies: \(K=GMm/2r\), \(U=-GMm/r\), \(E=-GMm/2r\) (negative ⇒ bound).
• Escape velocity: \(\displaystyle v_{\text{esc}}=\sqrt{2GM/r}\), i.e. \(\sqrt{2}\) times the circular‑orbit speed.
• These relationships underpin the analysis of more complex motions (elliptical orbits, transfer manoeuvres, escape trajectories) required by the Cambridge 9702 syllabus.