Circular motion and gravitation

A revolution can be described by an angle

On a circle of radius r, an arc of length s corresponds to angle θ = s/r in radians. A radian is the angle subtending an arc equal to the radius. Since the circumference is 2πr, a full revolution corresponds to 2π rad, equivalent to 360°. Conversion is essential: s = rθ and v = rω assume radians.

If the body sweeps equal angles in equal times, angular velocity ω is constant and circular motion is uniform. Its magnitude is ω = Δθ/Δt, in rad/s. Tangential speed is v = rω, in m/s. Bodies on the same rigid disc have the same ω, but points farther from the axis have larger v.

Period T is the time for one revolution, in s, and frequency f is the number of revolutions per second, in Hz. Since one revolution measures 2π rad, ω = 2π/T = 2πf and v = 2πr/T. These relationships connect a time description to the geometry of the trajectory.

Constant speed is not constant velocity

Instantaneous velocity is tangent to the circle. As the body moves, that tangent changes direction. Even if the length of the velocity vector stays the same, the vector changes, so there is acceleration. Here “uniform” refers to the magnitude of velocity, not to the constancy of the whole vector.

Compare two nearby instants separated by a small angle Δθ. The triangle formed by the velocity vectors is similar to the triangle formed by the radii. Thus, for small intervals, |Δv|/v approaches Δs/r. Dividing by Δt and using Δs/Δt = v gives centripetal a = v²/r.

In the limit of ever closer instants, acceleration points towards the centre of the circle. We can also write centripetal a = ω²r. The unit of v²/r is (m²/s²)/m = m/s². Acceleration does not point along velocity: it is perpendicular to it in uniform circular motion.

Centripetal is the role of the resultant

By the second law, the sum of force components along the radial direction is mv²/r, directed towards the centre. “Centripetal force” names this role of the resultant; it is not a new interaction to add to the diagram. It can be provided by tension, gravity, friction, normal force or a combination.

On a flat horizontal bend, friction between tyre and road can provide the horizontal resultant towards the centre. For a body attached to a rope in idealised horizontal rotation, tension can provide the radial force. For a satellite, it is gravity. Always identify which bodies exert the real forces.

Choosing towards the centre as the positive radial direction helps write radial ΣF = mv²/r. If motion occurs in a vertical circle, weight may contribute towards the centre at one point and away from it at another. Tension need not have the same value throughout the revolution. If a calculation requires negative tension in a rope, the taut-rope model is no longer valid.

When speed also changes

In non-uniform circular motion there are two acceleration components: radial, with magnitude v²/r, which changes direction, and tangential, which describes the change in speed. The two are perpendicular. Their total magnitude can be obtained using Pythagoras if both are known.

In uniform circular motion, the resultant is perpendicular to velocity and does no work on the particle: kinetic energy remains constant. A force can therefore change motion without changing K. This connection clarifies the difference between changing a vector and changing only its magnitude.

If the constraint maintaining the curve disappears, the body does not automatically continue around the circle. At an idealised instant when significant forces cease to act, it follows the tangent direction with the velocity it had at that instant. The idea of a real force throwing it radially outwards does not belong in the diagram in an inertial reference frame.

The same interaction connects nearby and distant bodies

The law of universal gravitation gives F = GMm/r² for point masses, or spherically symmetric bodies external to one another. M and m are masses in kg, r is the distance between centres in m and G ≈ 6,67 × 10⁻¹¹ N·m²/kg². The force is attractive and each body exerts a force of the same magnitude on the other.

The distance is not height above the ground. For a body at height h above a spherical planet of radius R, r = R + h. The gravitational field has magnitude g(r) = GM/r². The approximation g ≈ constant used for small falls works because h is very small relative to R.

Doubling r reduces F to one quarter; doubling one mass doubles F. Gravity can be weak between small objects and still dominate astronomical motion because of the large masses. The expression no longer treats an extended body correctly if we are inside its mass distribution without considering that distribution.

A circular orbit is continuous falling

For a satellite of mass m in a circular orbit, neglecting other interactions and assuming much larger M, gravity provides the entire radial resultant: GMm/r² = mv²/r. The mass m cancels and we obtain v = √(GM/r). This does not mean gravity is absent; it is precisely gravity that curves the motion.

Since T = 2πr/v, we obtain T² = 4π²r³/(GM). For orbits around the same central mass, T² is proportional to r³. More distant bodies have lower circular orbital speeds and longer periods. The same v should not be imposed on every orbit without analysing the available force.

The feeling of weightlessness in a satellite occurs because astronauts and spacecraft are in free fall together. It is not necessary for g to be zero. Real orbits may be elliptical; then speed varies and circular-orbit expressions cannot be applied point by point without adaptation. Kepler’s laws describe this broader geometry, and the circular orbit is a special case.

Worked examples, without skipping steps

Follow the reasoning first. Then cover the solution and try to reconstruct it in your own words.

Now try it yourself

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Mistakes worth avoiding

Constant speed means zero acceleration.

Velocity direction can change, producing centripetal acceleration.

Centripetal is an additional interaction.

It is the role of the radial resultant of real forces.

A satellite is weightless because there is no gravity.

It is in free fall, with gravity providing orbital acceleration.

The r in gravitation is height above the ground.

It is the distance between centres in the spherical model.