What equations do I need for IB Physics circular motion and gravitation?

The central circular-motion equations are $ac=v^2/r=\omega^2r$ and $\Sigma F{\text{in}}=mv^2/r$. You should also know $v=\omega r$, $\omega=2\pi/T$, and $v=2\pi r/T$. For gravitation, use $F=GMm/r^2$ and $g=GM/r^2$. Circular-orbit questions combine gravity with circular motion to produce $v=\sqrt{GM/r}$ and $T^2=4\pi^2r^3/(GM)$. HL students must also handle gravitational potential, potential energy, orbital energy, and escape speed as specified in the current course.

Why is a satellite accelerating if its orbital speed is constant?

Acceleration is the rate of change of velocity, and velocity contains both magnitude and direction. In a circular orbit, the satellite's speed may remain constant while its direction changes continuously. Its velocity vector is tangent to the orbit at each point, whereas its acceleration points toward the centre. Gravity produces this centripetal acceleration by continually turning the velocity vector. If the inward gravitational force disappeared, the satellite would initially move along a straight tangent rather than continuing around the circle.

Is centripetal force a separate force on a free-body diagram?

No, centripetal force is not an additional physical interaction. It is the name given to the resultant force component directed toward the centre of circular motion. Depending on the situation, that resultant may be supplied by gravity, friction, tension, a normal force, or a combination of forces. A free-body diagram should contain only those real forces, followed by an equation such as $\Sigma F{\text{in}}=mv^2/r$. Adding a separate centripetal-force arrow usually counts the inward effect twice and leads to an incorrect equation.

Why must orbital radius be measured from the planet's centre?

Newton's gravitational equation uses the separation between the centres of mass of the two bodies. For a satellite at altitude $h$ above a spherical planet of radius $R$, this separation is $r=R+h$. Substituting altitude alone makes the distance far too small, especially for a low orbit, and produces an unrealistically large gravitational force or orbital speed. The same centre-to-centre radius must be used in $g=GM/r^2$, $v=\sqrt{GM/r}$, and the orbital-period equation. Writing $r=R+h$ explicitly is a simple way to secure the geometry before beginning the calculation.

Why does a satellite move more slowly in a higher circular orbit?

For a circular orbit, equating gravitational and centripetal forces gives $v=\sqrt{GM/r}$. Increasing orbital radius therefore reduces speed according to $v\propto r^{-1/2}$. The gravitational field is weaker farther from the central body, so less centripetal acceleration is required and supplied. Although energy must be added to transfer the satellite to the higher orbit, its final kinetic energy is lower while its gravitational potential energy is higher. Its period also increases because $T^2\propto r^3$ for orbits around the same central mass.

How should I revise circular motion and gravitation for IB exams?

Begin by learning the vector directions and practising free-body diagrams rather than memorising isolated formulas. Next, derive orbital speed and period from gravitational force and centripetal acceleration until the cancellations are familiar. Complete short proportional-reasoning questions alongside longer calculations involving altitude, field strength, and orbital energy. After each attempt, use a per-question worked or video solution to compare equation choice, algebra, units, and written explanation. Record recurring errors in categories such as incorrect radius, force identification, vector direction, or sign convention, then use targeted RevisionDojo Questionbank practice and Jojo AI feedback to address them.

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IB Physics Circular Motion & Gravitation Explained

• 31 Jul 2026•13

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IB Physics Circular Motion & Gravitation Explained

• 31 Jul 2026•13

Circular motion and gravitation in IB Physics centre on a small set of connected ideas: an object moving in a circle accelerates toward the centre, forces must provide that acceleration, and gravity can act as the centripetal force for an orbit. Exam questions usually test whether you can identify the inward resultant force, choose the correct radius, combine equations, and explain the direction of velocity, acceleration, force, or field strength.

This IB Physics circular motion & gravitation explained guide covers those ideas with an exam focus. It follows the current course, first assessed in 2025, in which circular motion appears within A.2 Forces and momentum and gravitational fields within D.1 Gravitational fields.

Where these ideas appear in the IB course

Circular motion and gravitation are taught in different parts of the current syllabus, but examination questions often connect them. The official IB Physics course page provides the current subject briefs, while the IB's Physics curriculum update confirms that the revised course was first assessed in May 2025.

At both SL and HL, students should be ready to work with circular motion, Newton's law of gravitation, gravitational field strength, and orbital motion. HL treatment extends gravitational fields to ideas including gravitational potential, potential energy, escape speed, and orbital energy. Check the level indicated by your teacher and current subject guide rather than relying on resources written for the pre-2025 course.

The present examination structure also matters. Paper 1A uses multiple-choice questions, Paper 1B uses data-based questions, and Paper 2 contains short-response and extended-response work. Consequently, the topic may be tested through quick proportional reasoning, calculations, graph interpretation, derivations, or written explanations.

Uniform circular motion

An object in uniform circular motion travels around a circle at constant speed. Its velocity is not constant because velocity includes direction, and that direction changes continuously. The object therefore has an acceleration even though its speed remains unchanged.

The acceleration is centripetal, meaning centre-seeking, and always points toward the centre of the circle:

$a_c=\frac{v^2}{r}=\omega^2r=\frac{4\pi^2r}{T^2}$

Here, $v$ is linear speed, $r$ is orbital radius, $\omega$ is angular speed, and $T$ is the period. Useful connections are

$v=\omega r,\qquad \omega=\frac{2\pi}{T},\qquad v=\frac{2\pi r}{T}.$

At any instant, velocity is tangent to the circle, while centripetal acceleration is directed radially inward. These vectors are perpendicular in uniform circular motion. That is why the inward resultant force changes the direction of motion without changing the object's kinetic energy.

Centripetal force is not an additional force

Applying Newton's second law gives

$F_{\text{resultant,inward}}=ma_c=\frac{mv^2}{r}=m\omega^2r.$

The phrase centripetal force describes the inward resultant of real forces. It is not a new force that should automatically be added to a free-body diagram.

Situation	Real force or resultant providing centripetal force
Satellite orbiting a planet	Gravitational force
Car turning on a level road	Friction from the road
Ball moving in a horizontal circle on a string	Inward component or resultant involving tension
Rider at the bottom of a vertical loop	Resultant of normal force upward and weight downward

For example, at the bottom of a vertical loop, inward is upward, so $N-mg=mv^2/r$ . At the top, inward is downward, and both weight and a downward normal force may contribute, giving $mg+N=mv^2/r$ when contact is maintained.

Newtonian gravitation and gravitational fields

Newton's law of universal gravitation gives the magnitude of the attraction between two point masses or spherically symmetric bodies:

$F=G\frac{Mm}{r^2}.$

The separation $r$ is measured between the centres of mass, not between the surfaces. The force on each body has the same magnitude and opposite direction, in accordance with Newton's third law.

Gravitational field strength is force per unit mass:

$g=\frac{F}{m}=\frac{GM}{r^2}.$

Its unit is $\text{N kg}^{-1}$ , which is dimensionally equivalent to $\text{m s}^{-2}$ . Field strength is a vector directed toward the mass creating the field. Near Earth's surface, $g$ is approximately constant over small height changes, but the inverse-square expression is required when the distance from Earth's centre changes significantly.

A frequent exam error is using altitude $h$ directly in the inverse-square equation. For a body at height $h$ above a planet of radius $R$ , the correct centre-to-centre distance is

$r=R+h.$

The OpenStax account of Newton's law of gravitation reinforces the same free-body-diagram and centre-to-centre approach.

Circular orbits: where the topics connect

For a satellite of mass $m$ in a circular orbit around a much larger mass $M$ , gravity supplies the centripetal force:

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

Cancelling $m$ and one factor of $r$ gives

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

This result shows that orbital speed does not depend on satellite mass. A larger circular orbit has a lower orbital speed because $v\propto r^{-1/2}$ .

Substitute $v=2\pi r/T$ into the force equation to obtain

$T^2=\frac{4\pi^2}{GM}r^3.$

Therefore, for satellites orbiting the same central mass, $T^2\propto r^3$ . This is the circular-orbit form of Kepler's third law. The standard derivation and its assumptions are also shown in OpenStax's treatment of satellite orbits.

A typical orbital calculation

Suppose a satellite orbits Earth at an altitude of $4.0\times10^5\ \text{m}$ . If Earth's radius is $6.37\times10^6\ \text{m}$ and $G$ , first calculate

$r=R+h=6.77\times10^6\ \text{m}.$

Then

$v=\sqrt{\frac{3.99\times10^{14}}{6.77\times10^6}}=7.68\times10^3\ \text{m s}^{-1}.$

This method earns credit because it states the physical relationship before substitution and uses distance from Earth's centre. Writing only a calculator result conceals the reasoning on which method marks may depend.

HL gravitational potential and orbital energy

At HL, gravitational potential at distance $r$ from an isolated spherical mass is

$V=-\frac{GM}{r},$

and the gravitational potential energy of mass $m$ is

$U=mV=-\frac{GMm}{r}.$

Potential is energy per unit mass, measured in $\text{J kg}^{-1}$ ; potential energy is measured in joules. Both are negative when zero is defined at infinity because work must be done against gravitational attraction to remove a bound object to infinity.

For a circular orbit,

$K=\frac{GMm}{2r},\qquad U=-\frac{GMm}{r},\qquad E=K+U=-\frac{GMm}{2r}.$

Moving a satellite to a larger circular orbit makes its total energy less negative, so energy must be supplied. Yet its final kinetic energy and orbital speed are lower. This apparently contradictory result is a common conceptual test: energy is added during transfer, but the satellite settles into a slower orbit farther from the central mass.

The escape speed from distance $r$ follows from setting the total energy at infinity to zero:

$v_e=\sqrt{\frac{2GM}{r}}.$

At the same radius, escape speed is $\sqrt{2}$ times the circular orbital speed. This Newtonian result assumes no propulsion after launch and neglects effects such as atmospheric resistance and planetary rotation.

How examiners phrase circular motion and gravitation questions

Recognising the command term helps determine what your answer must contain.

Wording	What a strong response does
State the direction	Names a precise direction, such as radially inward or tangent to the path
Explain why it accelerates	Links changing velocity direction to inward acceleration despite constant speed
Determine the force	Draws or interprets forces, then resolves the inward resultant
Show that / derive	Begins with accepted equations and presents connected algebraic steps
Compare two orbits	Uses a proportional relationship before inserting numbers
Sketch a graph	Shows the correct shape and physically meaningful intercepts or asymptotes

When asked to explain an orbit, avoid saying that gravity is “balanced by centripetal force.” Gravity is the centripetal force in the simplified model. If forces were balanced, the resultant force and acceleration would be zero, so the satellite would move in a straight line rather than a circle.

Common mistakes and an efficient solving method

The most frequent errors are predictable:

treating centripetal force as an extra force;
using diameter, altitude, or surface separation instead of orbital radius;
assuming constant speed means zero acceleration;
placing velocity toward the centre rather than tangent to the path;
retaining satellite mass in a derived orbital-speed expression;
confusing $G$ , the universal gravitational constant, with $g$ , local field strength;
using circular-orbit equations for an elliptical orbit without justification;
omitting the negative sign from HL potential or orbital energy.

Use the following sequence under exam conditions:

Draw the geometry and free-body diagram. Mark the centre and inward direction.
Convert to centre-to-centre radius. Write $r=R+h$ when necessary.
Identify the real inward forces. Do not add a separate centripetal-force arrow.
Write the governing relationship. Usually $\Sigma F_{\text{in}}=mv^2/r$ .

Turning theory into exam marks

Reading derivations is not enough. Practise selecting equations from unfamiliar wording, then compare every algebraic step with a worked method. RevisionDojo's A.2.4 Circular Motion topic page combines concept material with targeted practice, while the Circular Motion Questionbank lets you apply the method to exam-style questions.

For gravity and orbit questions, use the Topic D Fields Questionbank. The broader IB Physics Questionbank is useful for mixed practice, and the IB Physics data booklet resource helps you become familiar with the formulas available in an examination.

After attempting each question independently, watch or review the available per-question worked solution rather than merely checking the final number. A useful video-solution routine is to pause before each step, predict the next equation, and record why your own method differed. The IB Physics predicted papers also provide paper-level practice with model answers and video solutions where offered.

Conclusion

Circular motion and gravitation become manageable when every problem is reduced to geometry, real forces, and Newton's second law. Remember that velocity is tangential, acceleration is inward, gravity follows an inverse-square law, and orbital equations come from equating gravitational force with the required centripetal force.

For revision, combine short proportional questions with full derivations and orbit calculations. RevisionDojo's topic Questionbanks, worked video solutions, data booklet practice, and Jojo AI feedback can help you identify whether lost marks come from physics, algebra, or interpretation.

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