Velocity and Acceleration in Space, multivariable and vector calculus lesson

A car rounding a bend at a steady 50 kilometers per hour is still accelerating, because its direction changes. A thrown ball, a satellite and a car all follow the same rules once position is a vector function of time. Two derivatives carry the kinematics, and a few clearly stated assumptions turn them into projectile paths and orbits.

Key terms

Velocity

Definition Let $\mathbf{r}(t)$ be the position at time $t$ of a particle moving in space, with $\mathbf{r}$ differentiable on an open interval. The velocity is

\mathbf{v}(t) = \mathbf{r}'(t) = \lim_{h \to 0} \frac{\mathbf{r}(t + h) - \mathbf{r}(t)}{h}.

When $\mathbf{v}(t) \ne \mathbf{0}$ , it points along the tangent line in the direction of motion.

In plain words Velocity is the displacement per unit time over a vanishingly short interval. It is a vector, so it carries both a rate and a direction.

Example For $\mathbf{r}(t) = \langle 2t, \ t^2, \ \tfrac{1}{3}t^3 \rangle$ , measured in meters with $t$ in seconds,

\mathbf{v}(t) = \langle 2, \ 2t, \ t^2 \rangle,

so at $t = 1$ the particle moves in the direction $\langle 2, 2, 1 \rangle$ .

Speed

Definition The speed is the scalar $v(t) = \lvert \mathbf{v}(t) \rvert \ge 0$ . If $s(t)$ is arc length measured from a fixed time, then $ds/dt = v(t)$ , and the distance traveled from $t = a$ to $t = b$ is $\int_a^b v(t) \, dt$ for a $C^1$ path.

In plain words Speed is what a speedometer shows. It has no direction, and it is never negative.

Example For the particle above,

v(t) = \sqrt{4 + 4t^2 + t^4} = t^2 + 2,

so at $t = 1$ the speed is $3$ meters per second. From $t = 0$ to $t = 3$ it travels $\int_0^3 (t^2 + 2)\, dt = 15$ meters.

Acceleration

Definition If $\mathbf{v}$ is differentiable, the acceleration is $\mathbf{a}(t) = \mathbf{v}'(t) = \mathbf{r}''(t)$ . For a particle of constant mass $m$ under a total force $\mathbf{F}$ , Newton’s second law states $\mathbf{F} = m\mathbf{a}$ .

In plain words Acceleration measures how the velocity vector changes. Changing speed and changing direction both count.

Example Uniform circular motion $\mathbf{r}(t) = \langle R\cos \omega t, R\sin \omega t \rangle$ with $R > 0$ and $\omega > 0$ has constant speed $R\omega$ , yet

\mathbf{a}(t) = -\omega^2 \mathbf{r}(t).

The acceleration points to the center and has size $R\omega^2$ . With $R = 5$ meters and $\omega = 2$ radians per second, the speed is $10$ meters per second and the acceleration has size $20$ meters per second squared.

Tangential and normal acceleration

Definition Let $\mathbf{r}$ be $C^2$ with $\mathbf{v}(t) \ne \mathbf{0}$ , unit tangent $\mathbf{T}$ and curvature $\kappa$ . Then

\begin{gathered} \mathbf{a} = a_T \mathbf{T} + a_N \mathbf{N}, \\[0.4em] a_T = \frac{dv}{dt}, \qquad a_N = \kappa v^2, \end{gathered}

where the $\mathbf{N}$ term is present only when $\kappa \ne 0$ (if $\kappa = 0$ then $\mathbf{a} = a_T \mathbf{T}$ ). In practice,

a_T = \frac{\mathbf{v} \cdot \mathbf{a}}{\lvert \mathbf{v} \rvert}, \qquad a_N = \frac{\lvert \mathbf{v} \times \mathbf{a} \rvert}{\lvert \mathbf{v} \rvert},

and $a_T^2 + a_N^2 = \lvert \mathbf{a} \rvert^2$ . The acceleration has no binormal component.

In plain words The tangential part speeds you up or slows you down. The normal part is the sideways push you feel in a turn, and it grows with the square of the speed.

Example For $\mathbf{r}(t) = \langle 2t, t^2, \tfrac{1}{3}t^3 \rangle$ at $t = 1$ , $\mathbf{v} = \langle 2, 2, 1 \rangle$ and $\mathbf{a} = \langle 0, 2, 2 \rangle$ . Then

\begin{gathered} a_T = \frac{0 + 4 + 2}{3} = 2, \\[0.4em] a_N = \frac{\lvert \langle 2, -4, 4 \rangle \rvert}{3} = \frac{6}{3} = 2. \end{gathered}

As a check, $a_T = \frac{d}{dt}(t^2 + 2) = 2t = 2$ at $t = 1$ , and $a_T^2 + a_N^2 = 8 = \lvert \mathbf{a} \rvert^2$ . The curvature there is $\kappa = a_N / v^2 = \tfrac{2}{9}$ .

Projectile motion

Definition Model a projectile as a point mass moving under gravity alone, with constant acceleration $g$ straight down, no air resistance and a flat horizontal ground at height $0$ . If it starts at height $h \ge 0$ with speed $v_0$ at angle $\alpha$ above the horizontal, integrating $\mathbf{a} = \langle 0, -g \rangle$ twice gives the position $\mathbf{r}(t) = \langle x(t), y(t) \rangle$ with

\begin{aligned} x(t) &= v_0 t\cos\alpha, \\[0.4em] y(t) &= h + v_0 t\sin\alpha - \tfrac{1}{2} g t^2, \end{aligned}

valid from launch until it hits the ground. When $h = 0$ and $0 < \alpha < \pi/2$ , the flight time $T_f$ , the range $D$ and the maximum height $H$ are

\begin{gathered} T_f = \frac{2v_0\sin\alpha}{g}, \quad D = \frac{v_0^2 \sin 2\alpha}{g}, \\[0.4em] H = \frac{(v_0\sin\alpha)^2}{2g}. \end{gathered}

In plain words Horizontally nothing pushes, so the horizontal velocity stays constant. Vertically gravity pulls down at a steady rate, so the height is a downward parabola in time.

Example A ball leaves the ground at $20$ meters per second at $30^\circ$ , with $g = 9.8$ meters per second squared. The flight time is $\frac{2 \cdot 20 \cdot 0.5}{9.8} \approx 2.04$ seconds. The range and the highest point are

\begin{gathered} \frac{400 \sin 60^\circ}{9.8} \approx 35.3 \text{ m}, \\[0.4em] \frac{10^2}{2 \cdot 9.8} \approx 5.10 \text{ m.} \end{gathered}

Central forces

Definition A force on a particle is central, with center at the origin, if at every time $t$ it is parallel to the position vector, that is $\mathbf{F}(t) = c(t)\,\mathbf{r}(t)$ for some scalar function $c$ . A common case is a force depending only on position, $\mathbf{F} = f(\lvert \mathbf{r} \rvert)\, \mathbf{r}/\lvert \mathbf{r} \rvert$ for $\mathbf{r} \ne \mathbf{0}$ .

In plain words A central force always pulls straight toward a fixed point or pushes straight away from it, never sideways.

Example In uniform circular motion, $\mathbf{F} = m\mathbf{a} = -m\omega^2 \mathbf{r}$ , so the force is central with $c = -m\omega^2$ . A string tied to a post supplies such a force. A projectile thrown with some horizontal velocity is not under a central force about any point $P$ . Its force always points straight down, so it is parallel to the line from $P$ to the projectile only while the projectile is directly above or below $P$ , and that happens at most at one instant.

Angular momentum

Definition The angular momentum about the origin of a particle of constant mass $m$ is $\mathbf{L} = \mathbf{r} \times m\mathbf{v}$ . If $\mathbf{r}$ is $C^2$ , then since $\mathbf{v} \times \mathbf{v} = \mathbf{0}$ ,

\frac{d\mathbf{L}}{dt} = \mathbf{r} \times m\mathbf{a} = \mathbf{r} \times \mathbf{F},

the torque. For a central force the torque is $\mathbf{0}$ , so $\mathbf{L}$ is constant. If $\mathbf{L} \ne \mathbf{0}$ , the particle then stays in the plane through the origin perpendicular to $\mathbf{L}$ , and the area swept by the segment from the origin to the particle grows at the constant rate

\frac{dA}{dt} = \frac{\lvert \mathbf{L} \rvert}{2m}.

In plain words Angular momentum measures how much a particle circulates around a point. A force aimed at that point cannot change it, which keeps the motion flat and sweeping area at a steady pace.

Example A particle of mass $2$ kilograms at $\mathbf{r} = \langle 1, 0, 0 \rangle$ with velocity $\langle 0, 3, 4 \rangle$ has

\begin{aligned} \mathbf{L} &= 2\, \langle 1, 0, 0 \rangle \times \langle 0, 3, 4 \rangle \\[0.4em] &= \langle 0, -8, 6 \rangle. \end{aligned}

If only a central force acts, the motion stays in the plane $-8y + 6z = 0$ , which contains both $\mathbf{r}$ and $\mathbf{v}$ .

Kepler’s laws with inverse-square-force assumptions

Definition Treat the planet and the sun as point masses. Assume a planet of mass $m$ moves under one force only, the gravity of a sun of mass $M$ held fixed at the origin,

\mathbf{F} = -\frac{GMm}{\lvert \mathbf{r} \rvert^3}\, \mathbf{r}, \qquad \mathbf{r} \ne \mathbf{0},

with $G$ the gravitational constant, and assume $\mathbf{L} \ne \mathbf{0}$ . Then the following hold.

The orbit is a conic section with a focus at the origin. If the orbit is bounded, it is an ellipse.
The segment from the sun to the planet sweeps out equal areas in equal times. This one holds for every central force.
For an elliptical orbit with semimajor axis $a$ and period $T$ , the square of the period is proportional to the cube of the semimajor axis,

T^2 = \frac{4\pi^2}{GM}\, a^3.

Real planets also pull on the sun and on each other, so these laws describe an idealized model. If the sun is allowed to move and only the two bodies interact, the same laws hold for the relative position, with $GM$ replaced by $G(M + m)$ in the third law.

In plain words An inverse-square pull toward a fixed point forces bounded orbits to be ellipses, makes the planet move faster near the sun so that equal times sweep equal areas, and ties the size of an orbit to how long it takes.

Example For a circular orbit of radius $a$ , gravity supplies the normal acceleration, so $GM/a^2 = v^2/a$ and $v = \sqrt{GM/a}$ . The period is

T = \frac{2\pi a}{v} = 2\pi\sqrt{\frac{a^3}{GM}},

which agrees with the third law. The same model fits a satellite around the Earth, with the Earth as the fixed center. With $GM = 3.986 \times 10^{14}$ cubic meters per second squared for the Earth and $a = 6.771 \times 10^{6}$ meters, about $400$ kilometers above the surface, this gives $v \approx 7670$ meters per second and $T \approx 5545$ seconds, about $92$ minutes. Doubling $a$ multiplies the period by $2^{3/2} \approx 2.83$ .

Check yourself

Find the tangential and normal components of acceleration for $\mathbf{r}(t) = \langle \cos t, \sin t, t \rangle$ .

Show the answer

The velocity is $\langle -\sin t, \cos t, 1 \rangle$ , with constant speed $\sqrt{2}$ , and the acceleration is $\langle -\cos t, -\sin t, 0 \rangle$ . Since the speed is constant, $a_T = 0$ . All the acceleration is normal, so $a_N = \lvert \mathbf{a} \rvert = 1$ .

A stone is thrown horizontally at $15$ meters per second from the top of a $20$ meter cliff. Ignoring air resistance and taking $g = 9.8$ meters per second squared, how far from the base of the cliff does it land?

Show the answer

The height is $20 - 4.9t^2$ , which is $0$ at $t = \sqrt{40/9.8} \approx 2.02$ seconds. The horizontal velocity stays $15$ , so it lands about $15 \cdot 2.02 \approx 30.3$ meters from the base.

A particle of mass $1$ moves along $\mathbf{r}(t) = \langle \cos t, 2\sin t, 0 \rangle$ . Show that the force on it is central and that its angular momentum is constant. Then use the area rate to find the area of the ellipse it traces.

Show the answer

The acceleration is $\langle -\cos t, -2\sin t, 0 \rangle = -\mathbf{r}$ , so the force is central. The angular momentum is

\begin{aligned} \mathbf{L} &= \langle \cos t, 2\sin t, 0 \rangle \\ &\qquad \times \langle -\sin t, 2\cos t, 0 \rangle \\[0.4em] &= \langle 0, 0, 2\cos^2 t + 2\sin^2 t \rangle \\[0.4em] &= \langle 0, 0, 2 \rangle, \end{aligned}

which is constant. The area rate is $\lvert \mathbf{L} \rvert / 2 = 1$ , and one trip takes $2\pi$ , so the area is $2\pi$ , matching $\pi \cdot 1 \cdot 2$ .

Velocity and Acceleration in Space

Key terms

Velocity

Speed

Acceleration

Tangential and normal acceleration

Projectile motion

Central forces

Angular momentum

Kepler’s laws with inverse-square-force assumptions

Check yourself

How this lesson connects