Relative Motion

Author(s): 
Larry Gladney and Dennis DeTurck

Note: The activities in this module make reference to the computer algebra system (CAS) Maple, and links are provided to download Maple files. Any other CAS can be used instead (e.g., Mathematica, Mathcad, etc.) as long as the user is familiar with that CAS system. Maple is not required for the use of the ideas in this module, but it is required for opening and executing the downloadable files.

Notes for Instructors

Larry Gladney is Associate Professor of Physics and Dennis DeTurck is Professor of Mathematics, both at the University of Pennsylvania.

Equations are used to describe the motion of objects. Usually the independent variable in these equations is t, for time. Depending on whether the motion is taking place along a line (one dimension), in a plane (two dimensions), or in space (three dimensions), we use one, two or three functions to specify the position of the object at any time. This module concerns the process of changing the point (and direction) of reference from which the motion is viewed -- this is important for solving physics problems, designing automatic pilots and other robotic devices, and video games.

 
Published July 2001
© 2001 by Larry Gladney and Dennis DeTurck

 

Relative Motion - Introduction

Author(s): 
Larry Gladney and Dennis DeTurck

Note: The activities in this module make reference to the computer algebra system (CAS) Maple, and links are provided to download Maple files. Any other CAS can be used instead (e.g., Mathematica, Mathcad, etc.) as long as the user is familiar with that CAS system. Maple is not required for the use of the ideas in this module, but it is required for opening and executing the downloadable files.

Notes for Instructors

Larry Gladney is Associate Professor of Physics and Dennis DeTurck is Professor of Mathematics, both at the University of Pennsylvania.

Equations are used to describe the motion of objects. Usually the independent variable in these equations is t, for time. Depending on whether the motion is taking place along a line (one dimension), in a plane (two dimensions), or in space (three dimensions), we use one, two or three functions to specify the position of the object at any time. This module concerns the process of changing the point (and direction) of reference from which the motion is viewed -- this is important for solving physics problems, designing automatic pilots and other robotic devices, and video games.

 
Published July 2001
© 2001 by Larry Gladney and Dennis DeTurck

 

Relative Motion - A Simple Example

Author(s): 
Larry Gladney and Dennis DeTurck

One-dimensional projectile motion and warmup exercises

Equations are used to describe the motion of objects. Usually, the independent variable in these equations is t, for time. Depending on whether the motion is taking place along a line (one dimension), in a plane (two dimensions) or in space (three dimensions), we use one, two or three functions to specify the position of the object at any time.

A simple example

An object that is projected straight up from the surface of the earth and then is subject only to gravity moves within a straight vertical line. Its motion can therefore be described by one function, y(t), which gives the height of the object above the earth for any time t. If we decide to measure time in seconds and height in feet, and if the object is projected upward at time = 0 with a velocity of 400 feet per second, then the function that gives the height of the object at time t is

y(t) = 400t - 16t2

Actually, this function only works for t between 0 and 25 seconds -- because 25 seconds after the object is projected upward, it hits the ground [ y(25) = 0].

Now suppose there are two objects. The first is projected upward when time t = 0 with initial velocity 400 feet per second -- so its height at time t is given by the function y(t) described above. The second is kept on the ground until time t = 5 seconds, and then it is projected upward with initial velocity 320 feet per second.


Problem 1: What function describes the motion of the second object for 0 < t < 25 seconds? Note that the function must be defined in pieces, since its value is zero for 0 < t < 5, and then it is given by some other formula for t > 5.


Problem 2: When does the second object hit the ground?


You can explore the motion of the two objects, both from an "external" point of view and from the point of view of one or the other of the objects in this Maple worksheet.


Problem 3: In the worksheet, the motion is viewed from the first object. You should also view the action from the point of view of the second object. How are the two related?


Relative Motion - Using relative velocities in one dimension

Author(s): 
Larry Gladney and Dennis DeTurck

A problem inspired by the driver's manual and the white Bronco

In the simple example in the preceding section, you have seen that the relative position of two objects moving in one dimension is simply the difference in their positions (measured in any reference system). Relative velocity is defined in the obvious way as the derivative of the relative position -- which is of course the difference of the velocities of the objects. Relative acceleration is defined the same way.

 

Driver's Ed redux

It is a basic principle of physics -- called Gallilean relativity -- that the laws of physics are the same when measured with respect to two reference systems that are moving at constant velocity with respect to one another. This can be especially useful when solving motion problems if one of the objects in the problem is moving with constant velocity relative to the "external" frame. A good example of this is provided by the following "classic" problem. You should solve it twice -- once from the point of view of an external stationary observer, and again from the point of view of the truck. Perhaps you can even solve it a third time, from the point of view of the car. At the end you will find an animation from the truck's point of view that reveals the answer to the problem.

 

 

Problem 4. The diagram above shows the passing ability of an automobile at low speed. From the data supplied in the figure, calculate the acceleration of the automobile during the pass and the time required for the pass. Assume constant acceleration. After finding the acceleration and the time required for the pass, use Maple to plot the position of both the automobile and the truck as a function of time. Using the graph, find the time corresponding to the point at which the automobile just passes the truck (i.e. the back of the car just passes the front of the truck).

( Follow this link to see an animated visual hint, and the answer.) [The video part of this no longer works. Ed.]

 


Relative Motion - Relative motion in two dimensions

Author(s): 
Larry Gladney and Dennis DeTurck

Circular motion and the motion of the planets

To consider motion in more than one dimension, we specify both the x and y coordinates of an object as functions of time. For example, for an object that is undergoing circular motion -- say its circular path has radius 5 and is centered at the origin, and it traverses a complete circle every 20 seconds -- we can express its motion as

x(t) = 5 cos (pt / 10); 

y(t) = 5 sin (pt / 10);

Here is an animation of the motion, produced by the Maple statement:

 animate([5*cos(Pi*t/10)+0.5*cos(s),5*sin(Pi*t/10)+0.5*sin(s),s=0..2*Pi],t=0..20);

Planetary Motion

To calculate relative motion in two dimensions, whether relative position, velocity, or acceleration, one proceeds as in one dimension, by subtracting the positions, velocities, or accelerations of the objects in question. Of course, you have to be careful to subtract x-coordinates from x-coordinates and y-coordinates from y-coordinates.

The most famous relative motion problem of all concerns the motion of the planets. Astronomers and astrologers realized that the planets moved against the backdrop of "fixed stars," but according to a pattern that seemed very mysterious -- most of the time planets move in one direction among the stars, but occasionally halt their motion and even regress a little from time to time. Ancient philosophers from Ptolemy to Aristarchus attempted to explain the peculiar motions of the planets -- mostly without complete success. Let's explore why this problem is so hard.

Consider the motion of Mars in the sky. We know now that the Earth circles the sun once a year in a roughly circular orbit that is about 93 million miles in radius. Mars is about 1.5 times as far from the sun as the earth, and takes about 2 Earth years to go once around the sun. For the moment, we assume that the orbits of the Earth and Mars are circular, with the sun at the center, and that the figures just given for the orbit of Mars are precise.

Then the Earth's orbit -- with distance measured in "astronomical units" (1 a.u. = 93,000,000 miles) and with time measured in years, is given by the equations:

xE  := cos(2*Pi*t);
yE  := sin(2*Pi*t);

These are the equations of a circle of radius 1 (note that xE2 + yE = 1) that is traversed once every 1 time unit.

Mars's orbit -- of radius 1.5 and period 2 -- is given by:

xM := 1.5*cos(Pi*t);
yM : = 1.5*sin(Pi*t);

As before, we can produce animations of these orbits together. Note that things are back where they started every 2 time units (2 years).

animate({[xE+0.1*cos(s),yE+0.1*sin(s),s=
0..2*Pi],
[xM+0.1*cos(s),yM+0.1*sin(s),s=0..2*Pi]},t=0..2);

For reference, here is the Maple worksheet.

This animation gives a picture relative to the sun -- the sun is at the center and remains there for the entire time. But until very recently, humans have been constrained to observe the heavens from the earth. So let's look at the motion of Mars relative to the earth:

xR := xM - xE;
yR := yM - yE;


Problem 5: Plot this curve using Maple, and animate it. See how Mars usually moves "around" the earth in one direction but occasionally exhibits "retrograde motion"? How often does this happen?

  Look here for the solution if you need help.


One good Web source for planetary data is NASA, in particular their Welcome to the planets site at CalTech's Jet Propulsion Lab.


Problem 6: This is a challenging problem! Of course, the orbits of the Earth and Mars are not circles, but ellipses with the sun at a focus. Look up more precise orbital data for the Earth and Mars (eccentricity of orbit) and repeat this analysis. Does the result change qualitatively in a substantial way?


Problem 7: Jupiter is about five times as far from the sun as the earth, and it takes about 12 earth years for Jupiter to go once around the sun. (You can use the NASA site for more precise information). Repeat the whole analysis for Jupiter. How often does Jupiter exhibit retrograde motion?


Problem 8: How about an inner planet? Find on the Web orbital data for Venus and repeat the analysis, and interpret the results. Are they consistent with what you know about observing Venus in the sky?


Problem 9: One last challenging problem! Look up data for Neptune and Pluto, including eccentricities etc... The interesting thing about these two planets is that sometimes Neptune is farther from the sun than Pluto, and sometimes the opposite is true. Repeat the above analysis from the point of view of an observer on Neptune.


Relative Motion - Adding rotation to the mix

Author(s): 
Larry Gladney and Dennis DeTurck

The motion of the moon and the earth

A new aspect of relative motion in two dimensions is that an observer can look in different directions -- instead of just forward and back (the distinction between positive and negative in one dimension). The effect of changing the direction in which the observer looks corresponds mathematically to a rotation of the plane. For example, if the observer at the origin turns 90 degrees clockwise, then what he or she sees seems to turn 90 degrees counterclockwise. So we need to derive what happens to the apparent positions of objects in the plane when the plane is rotated.

Here are the fomulas that describe clockwise rotation of the plane through the angle theta:

 
newx := t -> cos(theta)*oldx(t) + sin(theta)*oldy(t) 
 
newy := t -> -sin(theta)*oldx(t) + cos(theta)*oldy(t) 

To see that this really works, let's try looking at some stationary objects -- circles (originally) centered at the points (2,0) and (0,3), of radius 1/2. These circles are given by

 
[2+0.5*cos(s),0.5*sin(s),s=0..2*Pi]  and [0.5*cos(s),3+0.5*sin(s),s=0..2*Pi] 

respectively. The first and second coordinates of each of these will be our "oldx" and "oldy", so we can define:

 
oldx1:=2+0.5*cos(s); 
oldy1:=0.5*sin(s); 
 
oldx2:=0.5*cos(s); 
oldy2:=3+0.5*sin(s); 

Then we can animate what happens as theta increases (i.e., as the observer at the origin rotates) using the following definitions and animation. Notice that newx1, etc are defined as above.


newx1 := cos(theta)*oldx1 + sin(theta)*oldy1;
newy1 := -sin(theta)*oldx1 + cos(theta)*oldy1;

newx2 := cos(theta)*oldx2 + sin(theta)*oldy2;
newy2 := -sin(theta)*oldx2 + cos(theta)*oldy2;

animate({[newx1,newy1,s=0..2*Pi],[newx2,newy2,s=0..2*Pi]} ,theta=0..2*Pi);

Here is the Maple worksheet that implements this animation.


The motion of the earth, as seen from the moon

Let's apply the ideas of relative motion and rotation to two astronomical situations. The first is familiar to almost everyone, and the second, less well-known phenomenon is unusual and remarkable.

What's wrong with the name of this picture? It is called "Earthrise."

An astronomical fact that "everybody knows" -- and can observe -- is that the moon always presents the same face toward the earth. This is because the moon rotates as is moves around the earth, and its period of rotation is precisely the same as its period of revolution. We begin by producing an animation to illustrate this phenomenon from the point of view of the earth. The illustration will not be to scale, but will serve to indicate the motions of earth and moon from various vantage points.

Our illustration will have three components: the earth and the moon, which will be represented by circles, and the position of an object on the moon (so we can observe the moon's rotation), which will be represented by a line segment from the center of the moon through the position of the object on the moon's surface and extending somewhat into space (so as to be visible).

The earth is represented by:

 
earth:=[eposx+cos(s),eposy+sin(s),s=0..2*Pi]; 

For our first illustration, the earth will be stationary at the origin, so eposx = 0 and eposy = 0.

The moon is represented by

 
moon:=[mposx+0.5*cos(s),mposy+0.5*sin(s),s=0..2*Pi]; 

The moon will appear half as large as the earth -- as noted before, these pictures are not to scale. Also, since our first illustration will have the moon circling the stationary earth, we will have

 
mposx:=4*cos(t)   and     mposy:=4*sin(t) 

At this point, if we animate,

 
animate({earth,moon},t=0..2*Pi); 

we will see the moon circle the earth.

And here is the Maple worksheet that does this animation.


Problem 10: In real-life units, what are the units of t? (i.e., how often does the moon circle the earth?)

Now for the object. It is on the surface of the moon at some position -- let's say facing toward the earth. So relative to the standard polar coordinate system, its polar angle is 180 degrees (or Pi radians). The moon rotates in the same direction it revolves, and at the same angular rate. So the position of the object on the moon's surface is given by

oposx:=mposx+0.5*cos(t+Pi);  
oposy:=mposy+0.5*sin(t+Pi); 

To draw the line from the center of the moon through the object, we define

objectline:=[s*oposx+(1-s)*mposx, s*oposy+(1-s)*mposy, s=0..2]; 

(We use s=0..2 so that the line extends beyond the surface of the moon out into space.) Now we can animate,

 
animate({earth,moon,objectline},t=0..2*Pi); 

to see all three characters moving as they do from the point of view of the stationary earth. The Maple worksheet is here.


Problem 11 (think about this before we proceed to solve it): What does this look like from the point of view of the object on the moon?
To begin to answer this question, first we correct for the revolution of the moon around the earth -- to do this, as usual, we subtract the (originally varying) position of the center of the moon from the positions of everything under consideration:
mposnewx:=mposx-mposx; 
mposnewy:=mposy-mposy; 

(this puts the center of the moon at the origin).

 
eposnewx:=eposx-mposx; 
eposnewy:=eposy-mposy; 
 
oposnewx:=oposx-mposx; 
oposnewy:=oposy-mposy; 

Given these definitions for the positions of the earth, the moon and the object, we can define the following for the purposes of animation:

earthnew:=[eposnewx+cos(s),eposnewy+sin(s),s= 
 
0..2*Pi];moonnew:=[mposnewx+0.5*cos(s),mposnewy+0.5*sin(s),s=0..2*Pi]; 
 
objectlinenew:=[s*oposnewx+(1-s)*mposnewx, s*oposnewy+(1-s)*mposnewy, s=0..2]; 

Now, if we animate:

animate({earthnew,moonnew,objectnew},t=0..2*Pi);

we will see the moon centered at the origin and the earth will appear to circle the moon. Here is the Maple worksheet so far. But the object is still moving, since we have not yet compensated for the rotation of the moon. To do this, we will rotate eveything clockwise at precisely the same rate the moon was rotating counterclockwise --- this will have the effect of making the object stationary and we can see what happens to everything else.

So we define the positions of the earth, moon and object for the third time:

epos3x := cos(t)*eposnewx + sin(t)*eposnewy; 
epos3y := -sin(t)*eposnewx + cos(t)*eposnewy; 
 
mpos3x := cos(t)*mposnewx + sin(t)*mposnewy; 
mpos3y := -sin(t)*mposnewx + cos(t)*mposnewy; 
 
opos3x := cos(t)*oposnewx + sin(t)*oposnewy; 
opos3y := -sin(t)*oposnewx + cos(t)*oposnewy; 

and then we define the following for the purposes of animation:

 
earth3:=[epos3x+cos(s),epos3y+sin(s),s= 
 
0..2*Pi];moon3:=[mpos3x+0.5*cos(s),mpos3y+0.5*sin(s),s=0..2*Pi]; 
 
objectline3:=[s*opos3x+(1-s)*mpos3x, s*opos3y+(1-s)*mpos3y, s=0..2]; 

So now we can

animate({earth3,moon3,objectline3},t=0..2*Pi); 

Problem 12: Do the animation. (Here it is.) What do you see? Are you sure this is correct? Can you explain it?

Relative Motion - The peculiar day on Mercury

Author(s): 
Larry Gladney and Dennis DeTurck

Putting it all together

Another unusual celestial phenomenon is the apparent motion of the sun as seen from the surface of the planet Mercury. Mercury is the closest planet to the sun, and the relationship between its period of rotation (length of "day") and its period of revolution around the sun (length of "year") is very strange -- in fact, its day is almost as long as its year. It is this peculiar relationship, together with the fact that Mercury's orbit is quite elliptical, that causes the phenomenon we will study.

For the purposes of illustration, we will assume that the length of Mercury's day is 1/2 the length of its year. (This is fairly close to the correct ratio -- you get to make it more precise later.)

Mercury's orbit is an ellipse with the sun at the CENTER. (We will study the effect of varying the eccentricity of the ellipse -- you will also get to make this more realistic in two ways: first, by using the true eccentricity of the orbit, and second by putting the sun at a FOCUS of the elliptical orbit, in keeping with Kepler's first law of planetary motion.)

We start as we did with the earth and the moon. Everything is pretty easy from the point of view of the sun. The position of the (center of the) sun is:

 
sposx := 0; 
sposy := 0; 

Mercury orbits once every 2p time units (say). We'll let a and b be the lengths of the semi-axes of the ellipse, so the position of the center of Mercury is given by:

 
mposx := a*cos(t); 
mposy := b*sin(t); 

Now we put an object on the surface of Mercury that rotates along with the planet -- just as we did for the object on the moon in the preceding example. We'll take the radius of Mercury to be 1/2, and since the rotation period of Mercury is Pi, we get the position of the object to be:

oposx := mposx + 0.5 * cos(2*t); 
oposy := mposy + 0.5 * sin(2*t); 

Now we can define, for the purposes of animation,

sun := [sposx+cos(s), sposy+sin(s), s=0..2*Pi]; 
 
mercury := [mposx + 0.5*cos(s), mposy +0.5*sin(s), s=0..2*Pi]; 
 
objectline := [(1-s)*mposx + s*oposx,(1- s)*mposy + s*oposy, s=0..2]; 

and we can

animate({sun, mercury, objectline},t=0..2*Pi); 


Problem 13: Of course, this doesn't work (it just gives a picture of the stationary sun) -- we haven't assigned values to a and b yet. Make the orbit quite elliptical -- say b = 15 and a = 3 -- and then try the animation again! (Here is the worksheet.)


Problem 14: Why t=0..2*Pi?


So we have the motion of Mercury as seen from the sun. Now we need to bring Mercury to the center of the picture. As usual, we do this by subtracting the position of Mercury from that of everything:

spos2x := sposx - mposx; 
spos2y := sposy - mposy; 
 
mpos2x := mposx - mposx;   (of course, this is zero) 
mpos2y := mposy - mposy; 
 
opos2x := oposx - mposx; 
opos2y := oposy - mposy; 

and then

 
sun2 := [spos2x+cos(s), spos2y+sin(s), s=0..2*Pi]; 
 
mercury2 := [mpos2x + 0.5*cos(s), mpos2y +0.5*sin(s), s=0..2*Pi]; 
 
objectline2 := [(1-s)*mpos2x + s*opos2x, (1-s)*mpos2y + s*opos2y, s=0..2]; 

so we can animate:

animate({sun2, mercury2, objectline2}, t=0..2*Pi); 

This (the Maple worksheet so far) is almost what we want. The last thing we need to do is compensate for the rotation of Mercury. We do this the same way as we did for the rotation of the moon in the last example. The only tricky thing about this is the fact that the rotation is expressed in terms of trig functions of 2*t -- so our compensation must be expressed in the same way:

spos3x := cos(2*t)*spos2x + sin(2*t)*spos2y; 
spos3y := -sin(2*t)*spos2x + cos(2*t)*spos2y; 
 
mpos3x := cos(2*t)*mpos2x + sin(2*t)*mpos2y; 
mpos3y := -sin(2*t)*mpos2x + cos(2*t)*mpos2y; 
 
opos3x := cos(2*t)*opos2x + sin(2*t)*opos2y; 
opos3y := -sin(2*t)*opos2x + cos(2*t)*opos2y; 

and then

sun3 := [spos3x+cos(s), spos3y+sin(s), s=0..2*Pi]; 
 
mercury3 := [mpos3x + 0.5*cos(s), mpos3y +0.5*sin(s), s=0..2*Pi]; 
 
objectline3 := [(1-s)*mpos3x + s*opos3x, (1-s)*mpos3y + s*opos3y, s=0..2]; 

Then we animate (the check of course is that neither Mercury nor the object line should move):

animate({sun3, mercury3, objectline3}, t=0..2*Pi);

The motion of the sun in this animation seems somewhat curious. We can see the sun's path by plotting the path of the center of the sun (spos3x, spos3y):

plot([spos3x,spos3y,t=0..2*Pi]);


Problem 15: What does this mean about the apparent path of the sun through the sky as seen from a point on the surface of Mercury? Here is an animation against which you can check your intuition.


Problem 16: This analysis assumed that the sun is at the center of the elliptical orbit of Mercury. You can make this more realistic by moving the sun to a focus of the ellipse. How do you do this?


Problem 17: With both the "sun at the center" and the "sun at the focus" models, explore the effect of changing the eccentricity of the orbit (a and b).


Problem 18: Also explore the effect of changing the relationship between the periods of rotation and revolution of Mercury -- in particular, look up the exact values and use them.