# Trajectory

A trajectory is an imagined trace of positions followed by an object moving through space. Some common examples of trajectories: (i) the path taken by a falling body, and (ii) the orbit of a spacecraft. A particular trajectory may be described mathematically either by the geometry of the entire trajectory (i.e. the set of all positions taken by the object), or as the position of the object as function of time.

A familiar example is a projectile launched under the influence of only a uniform gravitational force field. A rock thrown on the practically airless surface of the Moon is a good approximation. In this case, the trajectory takes the shape of a parabola, provided the rock is not thrown too far. More generally, the precise trajectory of a projectile requires taking into account nonuniform gravitational forces and other forces such as drag and wind. This is the focus of the discipline of ballistics. A projectile, such as a baseball, when thrown through the air, is influenced by both gravity and aerodynamics.

More generally, trajectory refers to the ordered set of intermediate states assumed by a dynamical system as a result of time evolution.

The word trajectory is also often used metaphorically, for instance, to describe an individual's career.

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## Physics of trajectories

One of the remarkable achievements of Newtonian mechanics was the derivation of the laws of Kepler, in the case of the gravitational field of a single point mass (representing the Sun). The trajectory is a conic section, like an ellipse or a parabola. This agrees with the observed orbits of planets and comets, to a reasonably good approximation. Although if a comet passes close to the Sun, then it is also influenced by other forces, such as the solar wind and radiation pressure, which modify the orbit, and cause the comet to eject material into space.

Newton's theory later developed into the branch of theoretical physics known as classical mechanics. It employs the mathematics of differential calculus (which was, in fact, also initiated by Newton, in his youth). Over the centuries, countless scientists contributed to the development of these two disciplines. Classical mechanics became a most prominent demonstration of the power of rational thought, i.e. reason, in science as well as technology. It helps to understand and predict an enormous range of phenomena. Trajectories are but one example.

Consider a particle of mass [itex]m[itex], moving in a potential field [itex]V[itex]. Physically speaking, mass represents inertia, and the field [itex]V[itex] represents external forces, of a particular kind known as "conservative". That is, given [itex]V[itex] at every relevant position, there is a way to infer the associated force that would act at that position, say from gravity. Not all forces can be expressed in this way, however.

The motion of the particle is described by the second-order differential equation

[itex] m \frac{d^2 \vec{x}(t)}{dt^2} = -\nabla V(\vec{x}(t)) [itex] with [itex]\vec{x} = (x, y, z)[itex]

On the right-hand side, the force is given in terms of [itex]\nabla V[itex], the gradient of the potential, taken at positions along the trajectory. This is the mathematical form of Newton's second law of motion: mass times acceleration equals force, for such situations.

### Example: Uniform gravity, no drag or wind

The case of uniform gravity, disregarding drag and wind, yields a trajectory which is a parabola. To model this, one chooses [itex]V = m g z[itex], where [itex]g[itex] (gee) is the so-called acceleration of gravity. This gives the equations of motion

[itex] \frac{d^2 x}{dt^2} = \frac{d^2 y}{dt^2} = 0[itex]
[itex] \frac{d^2 z}{dt^2} = - g[itex]

Simplifications are made for the sake of studying the basics. The actual situation, at least on the surface of Earth, is considerably more complicated than this example would suggest, when it comes to computing actual trajectories. By deliberately introducing such simplifications, into the study of the given situation, one does, in fact, approach the problem in a way that has proved exceedingly useful in physics.

The present example is one of those originally investigated by Galileo Galilei. To neglect the action of the atmosphere, in shaping a trajectory, would (at best) have been considered a futile hypothesis by practical minded investigators, all through the Middle Ages in Europe. Nevertheless, by anticipating the existence of the vacuum, later to be demonstrated on Earth by his collaborator Evangelista Torricelli, Galileo was able to initiate the future science of mechanics. And in a near vacuum, as it turns out for instance on the Moon, his simplified parabolic trajectory proves essentially correct.

Relative to a flat terrain, let the initial horizontal speed be [itex]v_h[itex], and the initial vertical speed be [itex]v_v[itex]. It will be shown that, the range is [itex]2v_h v_v/g[itex], and the maximum altitude is [itex]{v_v^2}/2g[itex]. The maximum range, for a given total initial speed [itex]v[itex], is obtained when [itex]v_h=v_v[itex], i.e. the initial angle is 45 degrees. This range is [itex]v^2/g[itex], and the maximum altitude at the maximum range is a quarter of that.

#### Derivation

The equations of motion may be used to calculate the characteristics of the trajectory.

Let:

[itex]t \;[itex] be the time into the flight of the projectile
[itex]d_h(t) \;[itex] be the horizonal displacement at time t
[itex]d_v(t) = z \;[itex] be the vertical displacement at time t
[itex]v_h \;[itex] be the horizonal velocity (which is constant)
[itex]v_v \;[itex] be the initial vertical velocity upwards
[itex]v \;[itex] be the initial speed
[itex]v_v(t) \;[itex] be the vertical velocity at time t

Along the horizontal dimension, [itex]v_h[itex] is a constant and thus by the equations of motion,

[itex]d_h(t)=v_h t \;[itex] (Equation 1)

The vertical distance, or altitude follows the equations of motion for constant negative acceleration [itex]g[itex]:

[itex]d_v=v_vt- {{gt^2} \over 2}[itex] (Equation 2)
[itex]v_v = \frac{d}{dt} d_v(t) = v_v-gt[itex] (Equation 3: velocity equation which is the derivative of equation 2)

The range of the projectile occurs when [itex]d_v(t)[itex] is zero again and intercepts the ground. This occurs when [itex]d_v[itex] in equation 2 is zero:

[itex]0=v_vt- {{gt^2} \over 2 }[itex]

Solving this for time [itex]t[itex] gives the time of the projectile's flight:

[itex]t={2v_v \over g }[itex] (Equation 4: "hang time" of projectile)

The maximum range occurs when equation 4 is substituted into equation 1:

[itex]d_h(t) = { v_h t } = { v_h ({2 v_v \over g})} = { {2 v_h v_v} \over g } [itex] (Equation 5: range of projectile)

The maximum altitude for a given trajectory occurs when the vertical velocity is zero. Thus set equation 3 to zero:

[itex]0=v_v-gt \;[itex]

Solving for [itex]t[itex]

[itex]t= { {v_v} \over g } [itex]

This can be substituted into equation 2 to give the maximum altitude:

[itex]d_v(max)=v_v ( {v_v \over g}) - \frac 1 2 g( { v_v \over g } )^2 = {{v_v}^2 \over 2g } [itex] (Equation 6: maximum altitude of projectile)

Thus, not surprisingly, for a given initial speed the attained altitude is highest if the initial velocity was straight up. This altitude is twice the attained altitude when the range is maximized.

#### Derviation in polar coordinates

In terms of angle of elevation [itex]\theta[itex] and initial speed [itex]v[itex]:

[itex]v = \sqrt{ {v_h}^2 + {v_v}^2 } [itex]
[itex]v_h=v \cos \theta \;[itex]
[itex]v_v=v \sin \theta \;[itex]

Substituting into Equation 1 gives:

[itex]d_h=v_h t = vt \cos \theta \;[itex] (Equation 1a)

Substituting into Equation 2 gives:

[itex]d_v=v_vt- {{gt^2} \over 2} = ( v \sin \theta ) t - { { g t^2 } \over 2 } [itex] (Equation 2a)

Taking the derivative gives the vertical velocity:

[itex]v_v=\frac{d}{dt} d_v = \frac{d}{dt} (v \sin \theta) t - {{g t^2} \over 2} = v \sin \theta -gt [itex] (Equation 3a: vertical velocity)

Hang time calculated above in equation 4 may be expressed in terms of angle of elevation:

[itex]t={2v_v \over g} = {{2 v \sin \theta } \over g }[itex] (Equation 4a)

Equation 4a may be substituted into Equation 1a to get the horizontal distance or range:

[itex]d_h=v_h t = { vt \cos \theta } =v ({{2 v \sin \theta} \over g}) \cos \theta = { { 2 v^2 \sin \theta \cos \theta } \over g}[itex]

Now using the trigonometric identity for [itex]2 \sin \theta \cos \theta = sin 2 \theta[itex]:

[itex]d_h={{ v^2 \sin 2 \theta } \over g}[itex] (Equation 5a: range of projectile)

This may be solved for angle [itex]\theta[itex] to give the "angle" equation to hit a target at range [itex]d_h[itex]:

[itex] { \theta } = \frac 1 2 \sin^{-1} ( { {g d_h} \over { v^2 } } ) [itex] (Equation 7: angle of projectile launch)

Note that the sine function is such that there are two solutions for [itex]\theta[itex] for a given range [itex]d_h[itex]. Physically, this corresponds to a direct shot versus a mortar shot up and over obstacles to the target.

The maximum altitude for a given range may be determined by setting the vertical velocity to zero in equation 3a and solving for [itex]t[itex]:

[itex]0= v \sin \theta -gt \;[itex]
[itex]t= \frac{v \sin \theta}{g} [itex] (rearrange and solve for [itex]t[itex])

Now substitute into the vertical height equation 2a:

[itex]d_v = (v \sin \theta) t - \frac{g t^2}{2} = (v \sin \theta)(\frac {v \sin \theta}{g}) - \frac {g (\frac{v \sin \theta}{g})^2}{2} = \frac{v^2 {\sin}^2 \theta}{g} - \frac{v^2 {\sin}^2 \theta}{2g} = \frac{v^2 {\sin}^2 \theta}{2g}[itex] (Equation 6a: max altitude for a given launch angle)

#### Maximum range

Given the above range and altitude equations, the maximum range and altitude may be determined. Both equations for the range, equations 5 and 5a may be used to determine the maximum range by setting their derivatives to zero. For equation 5, range of the projectile is a function of [itex]v_h[itex] and [itex]v_v[itex] such that [itex]v_v^2+v_h^2=v^2[itex] where v is the total initial velocity and is constant. Thus, the range may be expressed as a function of [itex]v_v[itex] by solving for [itex]v_h[itex]:

[itex]v_h=\sqrt{v^2-v_v^2}[itex] (Equation 8)

And substituting [itex]v_h[itex] into equation 5:

[itex]d_h={ {2 v_h v_v} \over g } = { {2v_v \sqrt{v^2-v_v^2} } \over g} [itex]

The maximum [itex]d_h[itex] may be determined by calculating the derivative and setting it to zero. The derivative is calculated as follows:

[itex]\frac{d}{dv_v} d_h={1 \over g} \frac{d}{dv_v} {2v_v \sqrt{v^2-v_v^2} } [itex]
[itex]={1 \over g} (2 \sqrt{v^2-{v_v}^2} +2v_v \frac{d}{dv_v} \sqrt{v^2-v_v^2} [itex] (application of product rule)
[itex]={1 \over g} (2 \sqrt{v^2-{v_v}^2} +2v_v \frac{d \sqrt{v^2-v_v^2} }{d{v_v}^2} \frac{d{v_v}^2}{dv_v}) [itex] (application of chain rule)
[itex]={1 \over g} (2 \sqrt{v^2-{v_v}^2} +2v_v (\frac{-1} {2 \sqrt{v^2-{v_v}^2}}) 2v_v) [itex] (derivative of square root)
[itex]={1 \over g} (2 \sqrt{v^2-{v_v}^2} - \frac{2{v_v}^2}{\sqrt{v^2-{v_v}^2}}) [itex] (simplify second term)

Set to zero and solve for [itex]v[itex]:

[itex]0={1 \over g} (2 \sqrt{v^2-{v_v}^2} - \frac{2{v_v}^2}{\sqrt{v^2-{v_v}^2}}) [itex]
[itex]\sqrt{v^2-{v_v}^2}=\frac {{v_v}^2}{\sqrt{v^2-{v_v}^2}}[itex]
[itex]v^2-{v_v}^2=v_v^2[itex]
[itex]v^2=2v_v^2[itex] (Equation 9)

Thus maximum range occurs when [itex]v^2[itex] is [itex]2v_v^2[itex] and this can be substituted back into equation 8:

[itex]v_h=\sqrt{v^2-v_v^2}=\sqrt{v^2-v_v^2}=\sqrt{2v_v^2-v_v^2}=v_v[itex]

Thus the maximum range occurs when [itex]v_h=v_v[itex].

The actual maximum range may now be calculated by substituting [itex]v_h=v_v[itex] and equation 9 into equation 5:

[itex]d_h = { {2 v_h v_v} \over g } = { {2 v_v v_v} \over g } = { {2 v_v^2} \over g } = {v^2 \over g} [itex]

#### Maximum range in polar coordinates

The same conclusion may be drawn by starting with equation 5a.

[itex]\frac{d}{d\theta} d_h = \frac{d}{d\theta} {{ v^2 \sin 2 \theta } \over g} [itex]
[itex]= \frac{v^2}{g} \frac{d}{d2\theta} \sin 2 \theta \frac{d2\theta}{d\theta} [itex] (application of chain rule)
[itex]= \frac{v^2}{g} (\cos 2 \theta) 2 \quad [itex]

Set to zero and solve for [itex]\theta[itex]:

[itex]0= \frac{v^2}{g} (\cos 2 \theta) 2 [itex]

Now cosine is zero at [itex]\pi \over 2[itex]:

[itex]2 \theta=\frac \pi 2[itex] (also directly clear from equation 5a, it gives the maximum possible sine value of 1)
[itex]{\theta}_{\max} = \frac \pi 4 [itex] radians

Thus the maximum range occurs when the angle is 45 degrees.

The actual maximum range may now be calculated by substituting 45 degrees into equation 5a:

[itex]d_h={{ v^2 \sin 2 \theta } \over g}={{v^2 \sin 2 ( \frac{\pi}{4} ) } \over g} = {v^2 \over g}[itex]

#### Maximum altitude at maximum range

Equations 6 and 6a may be used to calculate the maximum altitude at the maximum range. Equation 9 may be substituted into equation 6:

[itex]d_v=\frac{v_v^2}{2g}=\frac{(\frac{v^2}{2})}{2g}=\frac{v^2}{4g}[itex]

Likewise 45 degrees may be substituted into equation 6a:

[itex]d_v=\frac{v^2 {\sin}^2 \theta}{2g}=\frac{v^2(\frac{1}{\sqrt 2})^2}{2g}=\frac{v^2}{4g} [itex]

#### As a parabola

Equations 1 and 2 are parametric equations that describe a parabola. They may be rearranged into the more familiar quadratic form by solving equation 1 for [itex]t[itex] and substituting into equation 2:

[itex]t=\frac{d_h}{v_h}[itex] (rearrange equation 1 for [itex]t[itex])

Substituting this into equation 2:

[itex]d_v=v_vt- \frac{gt^2}{2}=v_v(\frac{d_h}{v_h})-\frac{g(\frac{d_h}{v_h})^2}{2}=-\frac{g}{2{v_h}^2}(d_h)^2 + \frac{v_v}{v_h}(d_h)[itex]

This is now in the form

where

[itex]y=d_v, x=d_h, a=-g/{2{v_h}^2}, b=v_v/v_h, c=0[itex].

This is the form of a parabola and thus the trajectory is a parabola.

Likewise equations 1a and 2a can be rearranged into quadratic form. Equation 1a may be rearranged to:

[itex]t=\frac{d_h}{v \cos \theta}[itex]

And this may be substituted into equation 2a:

[itex]d_v=vt \sin \theta - \frac{gt^2}{2}=v \sin \theta \frac{d_h}{v \cos \theta} - \frac{g {(\frac{d_h}{v \cos \theta})}^2}{2} [itex]

Now [itex]\tan \theta=\sin \theta / \cos \theta[itex], so:

[itex]d_v=-\frac{g}{2v^2{\cos}^2 \theta}d_h^2 + d_h \tan \theta [itex] (Equation 10)

This is again now in the form [itex]y=ax^2+bx+c[itex] where [itex]y=d_v[itex], [itex]x=d_h[itex], [itex]a=-g/{2v^2{\cos}^2 \theta}[itex], [itex]b=\sin \theta / \cos \theta[itex] and [itex]c=0[itex] demonstrating that this is a parabola.

The quadratic formula gives the location of the intersection of the parabola and the x-axis. This is where the projectile trajectory starts and ends and thus may be used directly to calculate the range. In terms of rectinlinear coordinate systems:

[itex]

d_h=\frac{-b \pm \sqrt {b^2-4ac\ }}{2a}=\frac{-(v_v/v_h) \pm \sqrt{(v_v/v_h)^2 - 0} }{2(-g/2v_h^2)} = \frac{-v_v/v_h \pm v_v/v_h}{-g/v_h^2}=0, \frac{2{v_v}{v_h}}{g} [itex]

This is the same result as equation 5 above.

In polar coordinates and using the trigonometric identity [itex] 2 \sin \theta \cos \theta = \sin 2 \theta[itex], the intersections are:

[itex]

d_h=\frac{-b \pm \sqrt {b^2-4ac\ }}{2a}=\frac{-(\sin \theta / \cos \theta) \pm \sqrt{(\sin \theta / \cos \theta)^2 - 0 }}{2(-g/2v^2 \cos^2 \theta)} =0, \frac{2 \sin \theta / \cos \theta}{g/v^2 \cos^2 \theta} = 0, \frac{2 v^2 \sin \theta \cos \theta}{g} = 0, \frac{v^2 \sin 2 \theta}{g} [itex]

This is the same result as in equation 5a above.

Similarly, the vertex of the parabola is the maximum altitude for a given range.

### Uphill/downhill in uniform gravity in a vacuum

Given a hill angle [itex]\alpha[itex] and launch angle [itex]\theta[itex] as before, it can be shown that the range along the hill [itex]R_s[itex] forms a ratio with the original range [itex]R[itex] along the imaginary horizontal, such that:

[itex]\frac{R_s} {R}=(1-\tan \theta \tan \alpha)\sec \alpha [itex] (Equation 11)

In this equation, downhill occurs when [itex]\alpha[itex] is between 0 and -90 degrees. For this range of [itex]\alpha[itex] we know: [itex]\tan(-\alpha)=-\tan \alpha[itex] and [itex]\sec ( - \alpha ) = \sec \alpha[itex]. Thus for this range of [itex]\alpha[itex], [itex]R_s/R=(1+\tan \theta \tan \alpha)\sec \alpha [itex]. Thus [itex]R_s/R[itex] is a positive value meaning the range downhill is always further than along level terrain. This makes perfect sense as it is expected that gravity will assist the projectile, giving it greater range.

While the same equation applies to projectiles fired uphill, the interpretation is more complex as sometimes the uphill range may be shorter or longer than the equivalent range along level terrain. Equation 11 may be set to [itex]R_s/R=1[itex] (i.e. the slant range is equal to the level terrain range) and solving for the "critical angle" [itex]\theta_{cr}[itex]:

[itex]1=(1-\tan \theta \tan \alpha)\sec \alpha \quad \; [itex]
[itex]\theta_{cr}=\arctan((1-\csc \alpha)\cot \alpha) \quad \; [itex]

Equation 11 may also be used to develop the "rifleman's rule" for small values of [itex]\alpha[itex] and [itex]\theta[itex] (i.e. close to horizontal firing, which is the case for many firearm situations). For small values, both [itex]\tan \alpha[itex] and [itex]\tan \theta[itex] have a small value and thus when multiplied together (as in equation 11), the result is almost zero. Thus equation 11 may be approximated as:

[itex]\frac{R_s} {R}=(1-0)\sec \alpha [itex]

And solving for level terrain range, [itex]R[itex]

[itex]R=R_s \cos \alpha \ [itex] "Rifleman's rule"

Thus if the shooter attempts to hit the level distance R, s/he will actually hit the slant target. "In other words, pretend that the inclined target is at a horizontal distance equal to the slant range distance multiplied by the cosine of the inclination angle, and aim as if the target were really at that horizontal position." (http://www.snipertools.com/article4.htm)

#### Derivation based on equations of a parabola

The intersect of the projectile trajectory with a hill may most easily be derived using the trajectory in parabolic form in cartesian coordinates (Equation 10) intersecting the hill of slope [itex]m[itex] in standard linear form at coordinates [itex](x,y)[itex]:

[itex]y=mx+b \;[itex] (Equation 12) where in this case, [itex]y=d_v[itex], [itex]x=d_h[itex] and [itex]b=0[itex]

Substituting the value of [itex]d_v=m d_h[itex] into Equation 10:

[itex]m x=-\frac{g}{2v^2{\cos}^2 \theta}x^2 + \frac{\sin \theta}{\cos \theta} x[itex]
[itex]x=\frac{2v^2\cos^2\theta}{g}(\frac{\sin \theta}{\cos \theta}-m)[itex] (Solving above x)

This value of x may be substituted back into the linear equation 12 to get the corresponding y coordinate at the intercept:

[itex]y=mx=m \frac{2v^2\cos^2\theta}{g} (\frac{\sin \theta}{\cos \theta}-m)[itex]

Now the slant range [itex]R_s[itex] is the distance of the intercept from the origin, which is just the hypoteneuse of x and y:

[itex]R_s=\sqrt{x^2+y^2}=\sqrt{(\frac{2v^2\cos^2\theta}{g}(\frac{\sin \theta}{\cos \theta}-m))^2+(m \frac{2v^2\cos^2\theta}{g} (\frac{\sin \theta}{\cos \theta}-m))^2}[itex]
[itex]=\frac{2v^2\cos^2\theta}{g} \sqrt{(\frac{\sin \theta}{\cos \theta}-m)^2+m^2(\frac{\sin \theta}{\cos \theta}-m)^2}[itex]
[itex]=\frac{2v^2\cos^2\theta}{g} (\frac{\sin \theta}{\cos \theta}-m) \sqrt{1+m^2}[itex]

Now [itex]\alpha[itex] is defined as the angle of the hill, so by definition of tangent, [itex]m=\tan \alpha[itex]. This can be substituted into the equation for [itex]R_s[itex]:

[itex]R_s=\frac{2v^2\cos^2\theta}{g} (\frac{\sin \theta}{\cos \theta}-\tan \alpha) \sqrt{1+\tan^2 \alpha}[itex]

Now this can be refactored and the trigonometric identity for [itex]\sec \alpha = \sqrt {1 + \tan^2 \alpha}[itex] may be used:

[itex]R_s=\frac{2v^2\cos\theta\sin\theta}{g}(1-\frac{\sin\theta}{\cos\theta}\tan\alpha)\sec\alpha[itex]

Now the flat range [itex]R=v^2\cos 2 \theta / g = 2v^2\cos\theta\sin\theta[itex] by the previously used trigonometric identity and [itex]\sin\theta/\cos\theta=tan \theta[itex] so:

[itex]R_s=R(1-\tan\theta\tan\alpha)\sec\alpha \;[itex]
[itex]\frac{R_s}{R}=(1-\tan\theta\tan\alpha)\sec\alpha[itex]

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