Skip to main content

Gravitation and orbital motion - TCE Physics (Tasmania)

Syllabus dot point

“Apply Newton's law of universal gravitation and the field model to orbital motion.”

TCEPhysicsUnit 3: Gravity and Electromagnetism9 min read

Quick answer

Every mass attracts every other mass with a force F=Gm1m2r2F = G\dfrac{m_1 m_2}{r^2}; for a satellite this gravitational force supplies the centripetal force, which gives orbital speed v=GMrv = \sqrt{\dfrac{GM}{r}} and Kepler's third law T2∝r3T^2 \propto r^3.

What this dot point is asking

This dot point links gravity to circular motion, showing how the same inverse-square force that makes an apple fall also keeps the Moon and artificial satellites in orbit.

Newton's law of universal gravitation

Newton proposed that any two point masses attract each other along the line joining them with a force proportional to the product of their masses and inversely proportional to the square of their separation:

F=Gm1m2r2F = G\frac{m_1 m_2}{r^2}

Here G=6.67×10−11 N m2 kg−2G = 6.67 \times 10^{-11}\ \text{N m}^2\,\text{kg}^{-2} is the universal gravitational constant and rr is the distance between the centres of the masses. The inverse-square dependence is crucial: doubling the separation reduces the force to a quarter.

The gravitational field

It is often more useful to describe gravity as a field. The gravitational field strength gg at a point is the force per unit mass placed there:

g=Fm=GMr2g = \frac{F}{m} = \frac{GM}{r^2}

where MM is the mass creating the field. Near Earth's surface this evaluates to about 9.8 N kg−19.8\ \text{N kg}^{-1}, the same number as the free-fall acceleration in m s−2\text{m s}^{-2}. Field strength is a vector pointing toward the mass that creates it, and it falls off as the inverse square of distance from the centre.

Gravitational field strength

The gravitational field strength at a point is the gravitational force experienced per unit mass placed at that point, measured in newtons per kilogram.

The field model is powerful because it separates the source of the field from the object that responds to it. Once you know gg at a point, the force on any mass placed there is simply F=mgF = mg, with no need to revisit the inverse-square calculation. This is the same logic used for electric fields, and it lets you treat field strength as a property of the space itself, mapped out by field lines that point toward the source mass.

Comparing gravity with Coulomb's law

Newton's law of gravitation and Coulomb's law share the same inverse-square form, F∝1r2F \propto \dfrac{1}{r^2}, with a product of source quantities on top. The differences are instructive: gravity acts on mass, which is always positive, so it is always attractive, while the electric force acts on charge of either sign and can attract or repel. Gravity is also vastly weaker: between two protons the Coulomb force exceeds the gravitational force by about 103610^{36} times. Recognising this shared structure lets you carry over the same ratio techniques between the two topics.

Orbital motion

For a satellite in a stable circular orbit, gravity is the only force acting and it points toward the central body, so gravity provides exactly the centripetal force needed:

GMmr2=mv2r\frac{GMm}{r^2} = \frac{m v^2}{r}

The satellite mass mm cancels, leaving the orbital speed:

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

A higher orbit means a slower speed. Substituting v=2πrTv = \dfrac{2\pi r}{T} and rearranging gives Kepler's third law:

T2=4π2GM r3T^2 = \frac{4\pi^2}{GM}\,r^3

so T2T^2 is proportional to r3r^3 for all satellites of the same central body.

Worked example: period of a low Earth satellite

A satellite orbits Earth at a radius of r=6.6×106 mr = 6.6 \times 10^6\ \text{m} measured from Earth's centre. Find the orbital speed and the orbital period. (Earth's mass M=6.0×1024 kgM = 6.0 \times 10^{24}\ \text{kg}.)

Step 1: Find the orbital speed

For a circular orbit, gravity supplies the centripetal force. Setting GMmr2=mv2r\dfrac{GMm}{r^2} = \dfrac{mv^2}{r} and cancelling the satellite mass mm gives v=GMrv = \sqrt{\dfrac{GM}{r}}. This formula shows that a satellite at a larger radius orbits more slowly because gravity is weaker there.

v=GMr=(6.67×10−11)(6.0×1024)6.6×106=7.8×103 m s−1.v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{(6.67\times10^{-11})(6.0\times10^{24})}{6.6\times10^{6}}} = 7.8\times10^{3}\ \text{m s}^{-1}.

Step 2: Find the orbital period

The circumference of the orbit is 2πr2\pi r, and the satellite covers this at constant speed vv, so the period is simply the circumference divided by the speed. This gives the time to complete one full orbit:

T=2πrv=2π(6.6×106)7.8×103=5.3×103 s,T = \frac{2\pi r}{v} = \frac{2\pi (6.6\times10^{6})}{7.8\times10^{3}} = 5.3\times10^{3}\ \text{s},

which is about 8989 minutes, consistent with real low Earth orbit satellites such as the International Space Station.

Final answer: The satellite has an orbital speed of 7.8×103 m s−17.8\times10^{3}\ \text{m s}^{-1} and a period of 5.3×103 s5.3\times10^{3}\ \text{s}.

A geostationary satellite is a special case: it has a period of exactly one day so it stays above the same point on the equator. Setting T=86400 sT = 86400\ \text{s} in Kepler's third law gives an orbital radius of about 4.2×107 m4.2 \times 10^7\ \text{m}.

Do not use surface g for distant orbits

A common error is to keep using g=9.8 N kg−1g = 9.8\ \text{N kg}^{-1} for a satellite high above Earth. Field strength obeys the inverse-square law, so at large rr the field is much weaker. Always recompute g=GMr2g = \dfrac{GM}{r^2} using the orbital radius measured from Earth's centre, not the surface value.

When tackling orbital questions, decide first whether you need the field equation or the orbit equation. Use g=GMr2g = \dfrac{GM}{r^2} for field strength and weight, and use GMmr2=mv2r\dfrac{GMm}{r^2} = \dfrac{mv^2}{r} whenever the body is actually orbiting, since that is what links gravity to circular motion.

Exam-style questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice2 marks
A neutron star (pulsar) has a diameter of about 30 km30\ \text{km} and a mass of 2.79×1030 kg2.79 \times 10^{30}\ \text{kg}. Calculate the gravitational field strength at its surface.
Show worked answer →

Use the field of a spherical mass: g=GMr2g = \dfrac{GM}{r^2}.

The radius is half the diameter: r=15 km=1.5×104 mr = 15\ \text{km} = 1.5 \times 10^4\ \text{m}.

g=(6.67×10−11)(2.79×1030)(1.5×104)2=1.861×10202.25×108=8.27×1011 N kg−1.g = \frac{(6.67\times10^{-11})(2.79\times10^{30})}{(1.5\times10^4)^2} = \frac{1.861\times10^{20}}{2.25\times10^8} = 8.27 \times 10^{11}\ \text{N kg}^{-1}.

The surface gravity is about 8×1011 N kg−18 \times 10^{11}\ \text{N kg}^{-1}, around a hundred billion times Earth's. Markers want use of the radius (not the diameter) and the inverse-square form g=GMr2g = \dfrac{GM}{r^2}.

Practice5 marks
A small moon orbits the asteroid 130 Elektra at 501 km501\ \text{km} with a period of 1.191.19 Earth days. Calculate Elektra's gravitational field strength near this moon and hence the mass of Elektra.
Show worked answer →

For the orbiting moon, gravity supplies the centripetal acceleration, so the field strength at the moon equals its centripetal acceleration g=4π2rT2g = \dfrac{4\pi^2 r}{T^2}.

T=1.19×86400=1.028×105 sT = 1.19 \times 86400 = 1.028 \times 10^5\ \text{s}, r=5.01×105 mr = 5.01 \times 10^5\ \text{m}:

g=4π2(5.01×105)(1.028×105)2=1.978×1071.057×1010=1.87×10−3 m s−2.g = \frac{4\pi^2 (5.01\times10^5)}{(1.028\times10^5)^2} = \frac{1.978\times10^7}{1.057\times10^{10}} = 1.87 \times 10^{-3}\ \text{m s}^{-2}.

Mass of Elektra from g=GMr2g = \dfrac{GM}{r^2}:

M=gr2G=(1.87×10−3)(5.01×105)26.67×10−11=7.04×1018 kg.M = \frac{g r^2}{G} = \frac{(1.87\times10^{-3})(5.01\times10^5)^2}{6.67\times10^{-11}} = 7.04 \times 10^{18}\ \text{kg}.

Elektra has a mass of about 7×1018 kg7 \times 10^{18}\ \text{kg}. Markers reward equating the moon's centripetal acceleration to the field strength, then solving M=gr2GM = \dfrac{g r^2}{G}.

Practice3 marks
A Sun-like star HD137496 has a mass of 2.1×1030 kg2.1 \times 10^{30}\ \text{kg}. Its exoplanet 'b' orbits every 1.621.62 Earth days. Calculate the orbital radius of 'b'.
Show worked answer →

Combine gravity as the centripetal force with the period to get Kepler's third law: GMr2=4π2rT2\dfrac{GM}{r^2} = \dfrac{4\pi^2 r}{T^2}, which rearranges to r3=GMT24π2r^3 = \dfrac{GM T^2}{4\pi^2}.

T=1.62×86400=1.400×105 sT = 1.62 \times 86400 = 1.400 \times 10^5\ \text{s}.

r3=(6.67×10−11)(2.1×1030)(1.400×105)24π2=2.745×103039.48=6.95×1028.r^3 = \frac{(6.67\times10^{-11})(2.1\times10^{30})(1.400\times10^5)^2}{4\pi^2} = \frac{2.745\times10^{30}}{39.48} = 6.95 \times 10^{28}.

r=(6.95×1028)1/3=4.11×109 m.r = (6.95\times10^{28})^{1/3} = 4.11 \times 10^9\ \text{m}.

Planet 'b' orbits at about 4.1×109 m4.1 \times 10^9\ \text{m}. Markers want r3=GMT24π2r^3 = \dfrac{GM T^2}{4\pi^2} with the period in seconds.

Practise this

Sources & how we know this

ExamExplained