Trajectory

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For other uses of "Trajectory", see Trajectory (disambiguation).

Trajectory is the path a moving object follows through space. The object might be a projectile or a satellite, for example. It thus includes the meaning of orbit - the path of a planet, an asteroid or a comet as it travels around a central mass. A trajectory can be described mathematically either by the geometry of the path, or as the position of the object over time.

$(f^k(x))_{k \in \mathbb{N}}$

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

Illustration showing the trajectory of a bullet fired at an uphill target.

1 Physics of trajectories
2 Examples
3 See also
4 External links

Physics of trajectories

A familiar example of a trajectory is the path of a projectile such as a thrown ball or rock. In a greatly simplified model the object moves only under the influence of a uniform homogenous gravitational force field. This can be a good approximation for a rock that is thrown for short distances for example, at the surface of the moon. In this simple approximation the trajectory takes the shape of a parabola. Generally, when determining trajectories it may be necessary to account for nonuniform gravitational forces, air resistance (drag and aerodynamics). This is the focus of the discipline of ballistics.

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 m, moving in a potential field V. Physically speaking, mass represents inertia, and the field V represents external forces, of a particular kind known as "conservative". That is, given V 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

$m \frac{\mathrm{d}^2 \vec{x}(t)}{\mathrm{d}t^2} = -\nabla V(\vec{x}(t))$ $\nabla V$

Examples

Uniform gravity, no drag or wind

A bouncing ball captured with a stroboscopic flash at 25 images per second. Note that the ball becomes significantly non-spherical after each bounce, especially after the first. That, along with spin and air-resistance, causes the curve swept out to deviate slightly from the expected perfect parabola

The case of uniform gravity, disregarding drag and wind, yields a trajectory which is a parabola. To model this, one chooses V = mgz, where g is the acceleration of gravity. This gives the equations of motion

$\frac{\mathrm{d}^2 x}{\mathrm{d}t^2} = \frac{\mathrm{d}^2 y}{\mathrm{d}t^2} = 0$ $\frac{\mathrm{d}^2 z}{\mathrm{d}t^2} = - g$

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.

$v_h\,$

Derivation

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

Let

$p(t)\;$ $t\;$ $v_h \;$ $v_v \;$

The path of the projectile is known to be a parabola so

$p(t) = ( A t, 0 , a t^2 + b t + c )\,$ $A,\,a,\,b,\,c$

$p'(t) = ( A , 0 , 2 a t + b ),\quad p''(t) = ( 0 , 0 , 2 a ).$

At t = 0

$p(0)= (0, 0, 0)\ p'(0)=(v_h,0,v_v),\ p''(0)=(0,0,-g)$

$A = v_h,\ a = -g/2,\ b = v_v,\ c = 0$

This yields the formula for a parabolic trajectory:

$p(t) = (v_h t,0,v_v t - g t^2/2)\,\qquad$

Range and height

The range R of the projectile is found when the z-component of p is zero, that is when

$0 = v_v t - g t^2/2 = t \left( v_v - g t/2\right)\,$ $R = 2 v_h v_v/g.\,$

From the symmetry of the parabola the maximum height occurs at the halfway point t = v_v / g at position

$p(v_v/g)=(v_h v_v/g,0,v_v^2/(2g))\,$

This can also be derived by finding when the z-component of p' is zero.

Angle of elevation

In terms of angle of elevation θ and initial speed v:

$v_h=v \cos \theta,\quad v_v=v \sin \theta \;$

giving the range as

$R= 2 v^2 \cos(\theta) \sin(\theta) / g = v^2 \sin(2\theta) / g\,.$

This equation can be rearranged to find the angle for a required range

${ \theta } = \frac 1 2 \sin^{-1} \left( { {g R} \over { v^2 } } \right)$

Note that the sine function is such that there are two solutions for θ for a given range d_h. Physically, this corresponds to a direct shot versus a mortar shot up and over obstacles to the target. The angle θ giving the maximum range can be found by considering the derivative or R with respect to θ and setting it to zero.

${\mathrm{d}R\over \mathrm{d}\theta}={2v^2\over g} \cos(2\theta)=0$ $2\theta=\pi/2=90^\circ$ ${\mathrm{d}H\over \mathrm{d}\theta}=v^2 \cos(\theta) /(2g)$ $y=-{g\sec^2\theta\over 2v_0^2}x^2+x\tan\theta+h$

where v₀ is the initial speed, h is the height the projectile is fired from, and g is the acceleration due to gravity).

Uphill/downhill in uniform gravity in a vacuum

The factual accuracy of this section is disputed.
Please see the relevant discussion on the talk page.(March 2008)

Given a hill angle α and launch angle θ as before, it can be shown that the range along the hill R_s forms a ratio with the original range R along the imaginary horizontal, such that:

$\frac{R_s} {R}=(1-\cot \theta \tan \alpha)\sec \alpha$

In this equation, downhill occurs when α is between 0 and -90 degrees. For this range of α we know: tan( − α) = − tanα and sec( − α) = secα. Thus for this range of α, R_s / R = (1 + tanθtanα)secα. Thus R_s / R is a positive value meaning the range downhill is always further than along level terrain. The lower level of terrain causes the projectile to remain in the air longer, allowing it to travel further horizontally before hitting the ground.

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 R_s / R = 1 (i.e. the slant range is equal to the level terrain range) and solving for the "critical angle" θ_cr:

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

Equation 11 may also be used to develop the "rifleman's rule" for small values of α and θ (i.e. close to horizontal firing, which is the case for many firearm situations). For small values, both tanα and tanθ 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:

$\frac{R_s} {R}=(1-0)\sec \alpha$

And solving for level terrain range, R

$R=R_s \cos \alpha \$

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."[1]

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 m in standard linear form at coordinates (x,y):

$y=mx+b \;$

Substituting the value of d_v = md_h into Equation 10:

$m x=-\frac{g}{2v^2{\cos}^2 \theta}x^2 + \frac{\sin \theta}{\cos \theta} x$ $x=\frac{2v^2\cos^2\theta}{g}\left(\frac{\sin \theta}{\cos \theta}-m\right)$

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

$y=mx=m \frac{2v^2\cos^2\theta}{g} \left(\frac{\sin \theta}{\cos \theta}-m\right)$

Now the slant range R_s is the distance of the intercept from the origin, which is just the hypotenuse of x and y:

$R_s=\sqrt{x^2+y^2}=\sqrt{\left(\frac{2v^2\cos^2\theta}{g}\left(\frac{\sin \theta}{\cos \theta}-m\right)\right)^2+\left(m \frac{2v^2\cos^2\theta}{g} \left(\frac{\sin \theta}{\cos \theta}-m\right)\right)^2}$

$=\frac{2v^2\cos^2\theta}{g} \sqrt{\left(\frac{\sin \theta}{\cos \theta}-m\right)^2+m^2 \left(\frac{\sin \theta}{\cos \theta}-m\right)^2}$ $=\frac{2v^2\cos^2\theta}{g} \left(\frac{\sin \theta}{\cos \theta}-m\right) \sqrt{1+m^2}$

Now α is defined as the angle of the hill, so by definition of tangent, m = tanα. This can be substituted into the equation for R_s:

$R_s=\frac{2v^2\cos^2\theta}{g} \left(\frac{\sin \theta}{\cos \theta}-\tan \alpha\right) \sqrt{1+\tan^2 \alpha}$ $\sec \alpha = \sqrt {1 + \tan^2 \alpha}$

$R_s=\frac{2v^2\cos\theta\sin\theta}{g}\left(1-\frac{\sin\theta}{\cos\theta}\tan\alpha\right)\sec\alpha$

Now the flat range R = v²sin2θ / g = 2v²sinθcosθ / g by the previously used trigonometric identity and sinθ / cosθ = tanθ so:

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

Orbiting objects

If instead of a uniform downwards gravitational force we consider two bodies orbiting with the mutual gravitation between them, we obtain Kepler's laws of planetary motion. The derivation of these was one of the major works of Isaac Newton and provided much of the motivation for the development of differential calculus.