Gravitation and orbital motion - TCE Physics (Tasmania)
“Apply Newton's law of universal gravitation and the field model to orbital motion.”
Every mass attracts every other mass with a force ; for a satellite this gravitational force supplies the centripetal force, which gives orbital speed and Kepler's third law .
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:
Here is the universal gravitational constant and 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 at a point is the force per unit mass placed there:
where is the mass creating the field. Near Earth's surface this evaluates to about , the same number as the free-fall acceleration in . 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.
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 at a point, the force on any mass placed there is simply , 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, , 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 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:
The satellite mass cancels, leaving the orbital speed:
A higher orbit means a slower speed. Substituting and rearranging gives Kepler's third law:
so is proportional to for all satellites of the same central body.
A satellite orbits Earth at a radius of measured from Earth's centre. Find the orbital speed and the orbital period. (Earth's mass .)
Step 1: Find the orbital speed
For a circular orbit, gravity supplies the centripetal force. Setting and cancelling the satellite mass gives . This formula shows that a satellite at a larger radius orbits more slowly because gravity is weaker there.
Step 2: Find the orbital period
The circumference of the orbit is , and the satellite covers this at constant speed , so the period is simply the circumference divided by the speed. This gives the time to complete one full orbit:
which is about minutes, consistent with real low Earth orbit satellites such as the International Space Station.
Final answer: The satellite has an orbital speed of and a period of .
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 in Kepler's third law gives an orbital radius of about .
A common error is to keep using for a satellite high above Earth. Field strength obeys the inverse-square law, so at large the field is much weaker. Always recompute 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 for field strength and weight, and use 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 marksA neutron star (pulsar) has a diameter of about and a mass of . Calculate the gravitational field strength at its surface.Show worked answer →
Use the field of a spherical mass: .
The radius is half the diameter: .
The surface gravity is about , around a hundred billion times Earth's. Markers want use of the radius (not the diameter) and the inverse-square form .
Practice5 marksA small moon orbits the asteroid 130 Elektra at with a period of 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 .
, :
Mass of Elektra from :
Elektra has a mass of about . Markers reward equating the moon's centripetal acceleration to the field strength, then solving .
Practice3 marksA Sun-like star HD137496 has a mass of . Its exoplanet 'b' orbits every 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: , which rearranges to .
.
Planet 'b' orbits at about . Markers want with the period in seconds.