R^ what does it mean?

  • Thread starter Leonidas
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I was watching some Public television at 2:30 AM (it's when all the best programming is on)... and they were doing this bit on old astronomers and the mathematics behind it...

they didn't go into detail, and i understood everything that was spoken aloud... but they would flash equations in the backgrounds that I could ALMOST follow...

the main thing i didn't understand is they had an r with a ^ on top of it....

i assumed it had something to do with the radius... its not the average radius of an ellipse, is it?

anyways, if you feel like it, satisfy my curiosity...

also, if anyone wants to explain eccentricity of ellipses and how angular momentum is calculated, i'd be interested as long as they were in terms a high school calculus student could understand.

thanks! :shy:
 

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  • #2
SpaceTiger
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Do you mean:

[tex]\hat{r}[/tex]

That usually refers to a unit vector in the direction of something's radius. A unit vector is a vector of magnitude equal to one.
 
  • #3
dextercioby
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A more proper notation would be

[tex] \hat{e}_{r} [/tex]...

Daniel.
 
  • #4
SpaceTiger
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dextercioby said:
A more proper notation would be

[tex] \hat{e}_{r} [/tex]...

Do you have a reason for saying this? I have plenty of textbooks that use the [tex]\hat{r}[/tex] convention.
 
  • #5
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so in an equation about a heavenly body revolving around another heavenly body,

r^ basically says that it is defining the speed as 1 and the direction as around something else?


what does "in the direction of the radius" mean?
 
  • #6
dextercioby
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What about the other 2 unit vectors in the spherical coordinates ?

Daniel.
 
  • #7
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SpateTiger said:
[tex]\hat{r}[/tex]
dextercioby said:
[tex] \hat{e}_{r} [/tex]...



You're both wrong. It's [tex]\hat{i}_r[/tex] ... !
 
  • #8
dextercioby
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Actually the dot it's over the hat...:wink:

Daniel.
 
  • #9
SpaceTiger
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dextercioby said:
What about the other 2 unit vectors in the spherical coordinates ?

In the dynamics book I have by my side, they're denoted [tex]\hat{\theta}[/tex] and [tex]\hat{\phi}[/tex]...or is that supposed to be a response to the OP?
 
  • #10
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um... i still don't understand what it means.
 
  • #11
SpaceTiger
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Leonidas said:
um... i still don't understand what it means.

Yeah, sorry. :wink:

Imagine that you have an object around which you're basing your coordinate system. In orbital problems, it's often the sun. If you draw an arrow from the sun to any point in that coordinate system, that arrow is called the point's "radius vector". The earth's "radius vector" is an arrow pointing towards it from the sun. Now, [tex]\hat{r}[/tex] is a unit vector, so that means that it must have a length equal to one. So, to get the earth's unit radius vector, you just expand or contract the "radius vector" until it's equal to one. Its direction will be the same, but its length (or magnitude) will be different.
 
  • #12
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lol thanks, that makes a bit more sense... i think i'm in over my head here, though :-)

So when you say expand or contract the radius vector... what do you mean?

If you think its too hard to explain in terms i'll understand, i can wait till college :-)

-thanks!
 
  • #13
dextercioby
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Sorry,SpaceTiger,i was degenerating it into "conventions and likes vs. dislikes"...You know what i think,though...:smile:

Daniel.
 
  • #14
SpaceTiger
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Let me put it in terms of an equation, since this is hard to describe in words:

[tex]\hat{\bold r}=\frac{\vec{\bold r}}{|\vec{\bold r}|}[/tex]

What that means is that the unit vector is the radius vector (did you understand what I meant by that?) divided by its length. Unit vectors are generally used to specify a direction. If I want to specify a vector for something moving at speed v away from the sun, I just write:

[tex]\vec{\bold v}=v\hat{\bold r}[/tex]
 
  • #15
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I'm not sure I do understand what radius vector means...

I assumed it meant a measurement which coupled the length of the radius with the direction it's rotation.

"If I want to specify a vector for something moving at speed v away from the sun"

Away from the sun? do you mean around the sun?
 
  • #16
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Leonidas said:
So when you say expand or contract the radius vector... what do you mean?
Imagine you have the vector <2, 4>. The vector <1, 2> points in exactly the same direction as the first vector, but it is shorter. The vector <4, 8> points in the same direction as those other two vectors, but it is longer.

All three of these vectors are pointing the same way, but they differ in that some of them are longer or shorter than others. You could think of this as taking a vector, and expanding it (making it longer without changing the direction it's pointing in) or contracting it (making it shorter without changing the direction it's pointing in.)

If, after expanding or contracting the vector, the length of the vector was 1, you produced a unit vector.
 
  • #17
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alright thanks everyone! i'm not sure i understand yet, but I'll look into it a little bit more... maybe ask my math and physics teachers... it'll probably be easier to explain in person.

:-) thanks, and goodnight.
 
  • #18
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A vector in simple terms is basically an arrow. It points from one place to another. As spacetiger was saying, a radius vector is an arrow pointing from a certain place (the sun) to whereever you want it to (the earth). This is noted by the r with the hat on it. A unit vector is a vector which is one unit long. Say the distance frmo the earth to the sun is 5000 km (this is way wrong, just an example), then your vector will be 5000km long, but a unit vector is just 1m long. So what you do is you point your arrow in the direction of the earth (from the sun) and just squash it until its 1m long. However it still maintains its direction (from earth to sun), its just a lot shorter.
 
  • #19
SpaceTiger
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Have you ever worked with vectors before? This might be tough to explain in words if you haven't, but try this. If you have graph paper, draw a pair of coordinate axes on it. If not, draw a grid, like this.

Now, look at the point in that diagram. Do you know how to get its x and y coordinates? To get the coordinates, just count the number of squares over it is in the x and y directions. In this case, it's 7 squares in the x direction and 4 in the y direction, so its coordinates are (7,4).

Now try drawing an arrow to the point. This is called a "vector". The problem now is that we want to describe that vector. One way to do it is in terms of unit x and y vectors. They're called [tex]\hat{x}[/tex] and [tex]\hat{y}[/tex]. Graphically, you can describe them by drawing an arrow from (0,0) to (1,0) and an arrow from (0,0) to (0,1). The first is the x unit vector and the second is the y unit vector. It turns out that you can represent the arrow you drew first (the one to the point) by a linear combination of these vectors. That is:

[tex]\vec{\bold v}=7\hat{\bold x}+4\hat{\bold y}[/tex]

Try adding seven of the x unit vectors and four of y ones to verify. These are called "Cartesian coordinates". It turns out that you can also express this point in "polar" coordinates (or spherical, in three dimensions). To do that, instead of measuring x and y, you measure the distance from the center of the coordinate system (that's (0,0) in cartesian coordinates) and the angle from the x axis. For polar coordinates, the unit vectors are [tex]\hat{\bold r}[/tex] and [tex]\hat{\bold \theta}[/tex], an example of which can be shown here .
 
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  • #20
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Leonidas said:
Away from the sun? do you mean around the sun?

No, he meant away from the Sun in the same way that you'd say you were walking away from a building.
 
  • #21
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SpaceTiger said:
[tex]\hat{\bold v}=7\hat{\bold x}+4\hat{\bold y}[/tex]

You probably don't want to hat that [itex]v[/itex]... you'll make him think it's a unit vector! :tongue2:
 
  • #22
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Data said:
You probably don't want to hat that [itex]v[/itex]... you'll make him think it's a unit vector! :tongue2:

Ah yes indeed. Thanks. :wink:
 
  • #23
BobG
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Leonidas said:
I'm not sure I do understand what radius vector means...

I assumed it meant a measurement which coupled the length of the radius with the direction it's rotation.

"If I want to specify a vector for something moving at speed v away from the sun"

Away from the sun? do you mean around the sun?
No, he means directly away from the Sun (or directly toward the Sun).

For any coordinate system, you need a principle direction to measure from. In polar coordinates, you could measure the object's position from perihelion (in the case of the Sun) or perigee (in the case of the Earth). That's normally how you refer to an object's position.

If you're measuring the velocity of an orbiting object, it would be kind of complicated to analyze the velocity directly from perihelion. It's easier to use the position vector (radius vector) as your principle direction. You can then use the speed of the object and the angle of its motion relative to the radius. Or, you could break the velocity vector into two components: radial velocity (directly away from or toward the Sun) and tangential velocity (a component perpendicular to the radius). In fact, the equations to find the radial component and the tangential component of your velocity wind up being pretty simple.

Since your radius vector is referenced to the direction of perihelion, you can convert your velocity vector back to the same coordinate system your radius vector was identified in once you've finished working with it.
 
  • #24
BobG
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Leonidas said:
also, if anyone wants to explain eccentricity of ellipses and how angular momentum is calculated, i'd be interested as long as they were in terms a high school calculus student could understand.

thanks! :shy:
The short, general answer on eccentricity - an ellipse is kind of like a flattened circle. Eccentricity describes how flattened the circle is. Eccentricity can range from 0 to <1. If eccentricity is 0, you have a circle. The higher the eccentricity, the flatter the ellipse. If the ellipse equals 1, you no longer have an ellipse; you have a parabola and the object has reached an escape trajectory - the object's never coming back. Higher than 1, you have a hyperbolic trajectory which is also an escape trajectory.

There's a lot of ways to calculate the eccentricity. The simplest is a comparison of how far away the focus is from the geometric center of the ellipse (linear eccentricity, or c) to the semi-major axis, a. The eccentricity is just the linear eccentricity divided by the semi-major axis.

[tex]\frac{c}{a}=e[/tex]

If you only know the semi-major axis, a, and the semi-minor axis, b, you can also find the eccentricity, using:

[tex]\frac{\sqrt{a^2-b^2}}{a}[/tex]

For orbits, there's another way. If you know the position of the object, it's speed and direction of travel, and the gravitational potential of the object being orbited, you can compare the centrifugal force to the gravitational force to determine how eccentric the resulting orbit will be. This winds up being the best way, since, if you do this with vectors, not only will you have the shape of your ellipse, you'll also know where perihelion will be (and then you can use your perihelion as a reference point).
 
  • #25
BobG
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Leonidas said:
also, if anyone wants to explain eccentricity of ellipses and how angular momentum is calculated, i'd be interested as long as they were in terms a high school calculus student could understand.

thanks! :shy:
Angular momentum basically winds up being the amount of area swept out by an orbit per second (per Kepler's second law, that value remains constant).

If you know the semi-major axis, the semi-minor axis, and the time it takes to complete an orbit, the angular momentum is the area of the ellipse divided by the time it takes to complete one orbit.

There's also a few others ways to calculate it using cross products of the radius vector and velocity vector (observation) or by using your orbital elements (used to predict the object's location some time in the future).
 

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