Consider a particle moving along the perimeter of a circle at a uniform rate, such that it makes one complete revolution every hour. To describe the motion mathematically, a vector is constructed from the centre of the circle to the particle. The vector then makes one complete revolution every hour. In other words, the vector behaves exactly like the large hand on a wristwatch, an arrow of fixed length that makes one complete revolution every hour. The motion of the point of the vector is an example of uniform circular motion, and the period T of the motion is equal to one hour (T = 1 h). The arrow sweeps out an angle of 2π radians (one complete circle) per hour. This rate is called the angular frequency and is written ω = 2π h−1. Quite generally, for uniform circular motion at any rate,
These definitions and relations are the same as they are for harmonic motion, discussed above.
Consider a coordinate system, as shown in , with the circle centred at the origin. At any instant of time, the position of the particle may be specified by giving the radius r of the circle and the angle θ between the position vector and the x-axis. Although r is constant, θ increases uniformly with time t, such that θ = ωt, or dθ/dt = ω, where ω is the angular frequency in equation (26). Contrary to the case of the wristwatch, however, ω is positive by convention when the rotation is in the counterclockwise sense. The vector r has x and y components given by
One meaning of equations (27) and (28) is that, when a particle undergoes uniform circular motion, its x and y components each undergo simple harmonic motion. They are, however, not in phase with one another: at the instant when x has its maximum amplitude (say, at θ = 0), y has zero amplitude, and vice versa.
In a short time, Δt, the particle moves rΔθ along the circumference of the circle, as shown in . The average speed of the particle is thus given by
The average velocity of the particle is a vector given by
This operation of vector subtraction is indicated in . It yields a vector that is nearly perpendicular to r(t) and r(t + Δt). Indeed, the instantaneous velocity, found by allowing Δt to shrink to zero, is a vector v that is perpendicular to r at every instant and whose magnitude is
The relationship between r and v is shown in . It means that the particle’s instantaneous velocity is always tangent to the circle.
Notice that, just as the position vector r may be described in terms of the components x and y given by equations (27) and (28), the velocity vector v may be described in terms of its projections on the x and y axes, given by
Imagine a new coordinate system, in which a vector of length ωr extends from the origin and points at all times in the same direction as v. This construction is shown in . Each time the particle sweeps out a complete circle, this vector also sweeps out a complete circle. In fact, its point is executing uniform circular motion at the same angular frequency as the particle itself. Because vectors have magnitude and direction, but not position in space, the vector that has been constructed is the velocity v. The velocity of the particle is itself undergoing uniform circular motion at angular frequency ω.
Although the speed of the particle is constant, the particle is nevertheless accelerated, because its velocity is constantly changing direction. The acceleration a is given by
Since v is a vector of length rω undergoing uniform circular motion, equations (29) and (30) may be repeated, as illustrated in , giving
Thus, one may conclude that the instantaneous acceleration is always perpendicular to v and its magnitude is
Since v is perpendicular to r, and a is perpendicular to v, the vector a is rotated 180° with respect to r. In other words, the acceleration is parallel to r but in the opposite direction. The same conclusion may be reached by realizing that a has x and y components given by
similar to equations (32) and (33). When equations (38) and (39) are compared with equations (27) and (28) for x and y, it is clear that the components of a are just those of r multiplied by −ω2, so that a = −ω2 r. This acceleration is called the centripetal acceleration, meaning that it is inward, pointing along the radius vector toward the centre of the circle. It is sometimes useful to express the centripetal acceleration in terms of the speed v. Using v = ωr, one can write
Circular orbits
The detailed behaviour of real orbits is the concern of celestial mechanics (see the article celestial mechanics). This section treats only the idealized, uniform circular orbit of a planet such as Earth about a central body such as the Sun. In fact, Earth’s orbit about the Sun is not quite exactly uniformly circular, but it is a close enough approximation for the purposes of this discussion.
A body in uniform circular motion undergoes at all times a centripetal acceleration given by equation (40). According to Newton’s second law, a force is required to produce this acceleration. In the case of an orbiting planet, the force is gravity. The situation is illustrated in . The gravitational attraction of the Sun is an inward (centripetal) force acting on Earth. This force produces the centripetal acceleration of the orbital motion.
Before these ideas are expressed quantitatively, an understanding of why a force is needed to maintain a body in an orbit of constant speed is useful. The reason is that, at each instant, the velocity of the planet is tangent to the orbit. In the absence of gravity, the planet would obey the law of inertia (Newton’s first law) and fly off in a straight line in the direction of the velocity at constant speed. The force of gravity serves to overcome the inertial tendency of the planet, thereby keeping it in orbit.
The gravitational force between two bodies such as the Sun and Earth is given by
where M S and M E are the masses of the Sun and Earth, respectively, r is the distance between their centres, and G is a universal constant equal to 6.674 × 10−11 Nm2/kg2 (Newton metres squared per kilogram squared). The force acts along the direction connecting the two bodies (i.e., along the radius vector of the uniform circular motion), and the minus sign signifies that the force is attractive, acting to pull Earth toward the Sun.
To an observer on the surface of Earth, the planet appears to be at rest at (approximately) a constant distance from the Sun. It would appear to the observer, therefore, that any force (such as the Sun’s gravity) acting on Earth must be balanced by an equal and opposite force that keeps Earth in equilibrium. In other words, if gravity is trying to pull Earth into the Sun, some opposing force must be present to prevent that from happening. In reality, no such force exists. Earth is in freely accelerated motion caused by an unbalanced force. The apparent force, known in mechanics as a pseudoforce, is due to the fact that the observer is actually in accelerated motion. In the case of orbital motion, the outward pseudoforce that balances gravity is called the centrifugal force.
For a uniform circular orbit, gravity produces an inward acceleration given by equation (40), a = −v 2/r. The pseudoforce f needed to balance this acceleration is just equal to the mass of Earth times an equal and opposite acceleration, or f = M E v 2/r. The earthbound observer then believes that there is no net force acting on the planet—i.e., that F + f = 0, where F is the force of gravity given by equation (41). Combining these equations yields a relation between the speed v of a planet and its distance r from the Sun:
It should be noted that the speed does not depend on the mass of the planet. This occurs for exactly the same reason that all bodies fall toward Earth with the same acceleration and that the period of a pendulum is independent of its mass. An orbiting planet is in fact a freely falling body.
Equation (42) is a special case (for circular orbits) of Kepler’s third law, which is discussed in the article celestial mechanics. Using the fact that v = 2πr/T, where 2πr is the circumference of the orbit and T is the time to make a complete orbit (i.e., T is one year in the life of the planet), it is easy to show that T 2 = (4π2/GM S )r 3. This relation also may be applied to satellites in circular orbit around Earth (in which case, M E must be substituted for M S ) or in orbit around any other central body.
Angular momentum and torque
A particle of mass m and velocity v has linear momentum p = m v. The particle may also have angular momentum L with respect to a given point in space. If r is the vector from the point to the particle, then
Notice that angular momentum is always a vector perpendicular to the plane defined by the vectors r and p (or v). For example, if the particle (or a planet) is in a circular orbit, its angular momentum with respect to the centre of the circle is perpendicular to the plane of the orbit and in the direction given by the vector cross product right-hand rule, as shown in . Moreover, since in the case of a circular orbit, r is perpendicular to p (or v), the magnitude of L is simply
The significance of angular momentum arises from its derivative with respect to time,
where p has been replaced by m v and the constant m has been factored out. Using the product rule of differential calculus,
In the first term on the right-hand side of equation (46), d r/dt is simply the velocity v, leaving v × v. Since the cross product of any vector with itself is always zero, that term drops out, leaving
Here, d v/dt is the acceleration a of the particle. Thus, if equation (47) is multiplied by m, the left-hand side becomes d L/dt, as in equation (45), and the right-hand side may be written r × m a. Since, according to Newton’s second law, m a is equal to F, the net force acting on the particle, the result is
Equation (48) means that any change in the angular momentum of a particle must be produced by a force that is not acting along the same direction as r. One particularly important application is the solar system. Each planet is held in its orbit by its gravitational attraction to the Sun, a force that acts along the vector from the Sun to the planet. Thus, the force of gravity cannot change the angular momentum of any planet with respect to the Sun. Therefore, each planet has constant angular momentum with respect to the Sun. This conclusion is correct even though the real orbits of the planets are not circles but ellipses.
The quantity r × F is called the torque τ. Torque may be thought of as a kind of twisting force, the kind needed to tighten a bolt or to set a body into rotation. Using this definition, equation (48) may be rewritten
Equation (49) means that if there is no torque acting on a particle, its angular momentum is constant, or conserved. Suppose, however, that some agent applies a force F a to the particle resulting in a torque equal to r × F a . According to Newton’s third law, the particle must apply a force −F a to the agent. Thus, there is a torque equal to −r × F a acting on the agent. The torque on the particle causes its angular momentum to change at a rate given by d L/dt = r × F a . However, the angular momentum L a of the agent is changing at the rate d L a /dt = −r × F a . Therefore, d L/dt + d L a /dt = 0, meaning that the total angular momentum of particle plus agent is constant, or conserved. This principle may be generalized to include all interactions between bodies of any kind, acting by way of forces of any kind. Total angular momentum is always conserved. The law of conservation of angular momentum is one of the most important principles in all of physics.
