# I am confused about the orbital velocity

1. Apr 8, 2010

### Misr

Hello,

We know that to calculate the orbital velocity we use the relation :

Vo=2* pi* r /time of one cycle

but if we want to calculate the orbital velocity when a body moves from the point a to the point b on a circular path we use the relation :

Vo = The line segment ab / time taken (as it is written im my book)

so i'm confused with this.
it should be: the arc ab /time taken

so Could u help?
Thanks

2. Apr 8, 2010

### Stonebridge

It's about the difference between speed (a scalar) and velocity (a vector)
If you calculate the quantity in your first example, the circumference of the circle divided by the time taken, you have calculated the speed of the object. (Or more correctly, the average speed if the value is not constant)

When an object moves from a to b on a circle, if you calculate the quantity arc length ab (aqb in diagram) divided by time, you have the speed of the object.
If you calculate the quantity chord length ab divided by time (apb in diagram) you have calculated the average velocity.
Velocity is displacement divided by time
Speed is distance divided by time
The distance travelled from a to b is the arc length aqb
The displacement from a to b is the chord length apb.

3. Apr 8, 2010

### sophiecentaur

In the limit, as the interval approaches zero, the magnitude of the two quantities is the same.
btw, is that a circle or an ellipse on post no 356? It may just be my glasses.

4. Apr 8, 2010

### sophiecentaur

In the limit, as the interval approaches zero, the magnitude of the two quantities is the same.
btw, is that a circle or an ellipse on post no 356? It may just be my glasses.

5. Apr 8, 2010

### sophiecentaur

Wow, deja vu!

6. Apr 8, 2010

### Stonebridge

It's more likely my glasses.

7. Apr 8, 2010

### Misr

the speed is constant in a circular motion .

but speed value is equal to velocity they should have the same value
still confused :(
Thanks so much

8. Apr 8, 2010

### Stonebridge

Often but not always.
Certainly in most physics problems like these, involving circular motion, the speed is constant.
Why should speed value and velocity be the same?
The answer I gave referred to "average velocity", not instantaneous velocity.
Unfortunately, we are not always precise with terminology. Sometimes people confuse, or are careless about, the terms speed and velocity.
They are not the same in physics.

If you run half way around a circle, diameter 200m, in 1 minute, you run a distance of Ï€d/2 = 314m
What was your average speed? Answer 314m/60s = 5.2m/s
But your end point is 200m from your start point. Your displacement is 200m.
What was your average velocity? Answer 200m/60s = 3.3 m/s
Distance and displacement measure different things; and average speed and average velocity measure different things.

When people refer to the Earth's orbital "velocity", they very often just mean the instantaneous speed at any point; the magnitude part of the velocity.

Last edited: Apr 8, 2010
9. Apr 9, 2010

### Misr

Last edited: Apr 9, 2010
10. Apr 9, 2010

### Stonebridge

What the proof does, is to say that the length of the arc and the length of the chord become nearer and nearer to equal as you make the arc smaller and smaller.
In the limit, when you take an infinitesimally small time interval, they become equal.
The calculus of the rate of change always uses this idea of taking a very small time interval, and looking at what happens as the interval gets smaller and smaller.

11. Apr 9, 2010

### sophiecentaur

The instantaneous speed is the same as the magnitude of the instantaneous velocity.

12. Apr 9, 2010

### Misr

Ahaa....I got it now
thanks so much

Yes
thanks very much ,Miss Sophie

13. Apr 9, 2010

### Misr

MMM
i tried it
where Vdt is ab arc

[PLAIN]http://img401.imageshack.us/img401/8510/limr.jpg [Broken]

suppose we have this circle

[PLAIN]http://img684.imageshack.us/img684/1360/circle1.jpg [Broken]

where ab is approx. 4 cm
ab arc = 9.4 cm

another circle after decreasing the angle we have

[PLAIN]http://img194.imageshack.us/img194/213/circle2i.jpg [Broken]

ab= 3cm
and ab arc = 3.14cm

so they approached each other when the angle decreases (or when time interval decreases)
till they both (the arc and the chord) equal zero when dt = zero
right or wrong??

Last edited by a moderator: May 4, 2017
14. Apr 9, 2010

### Stonebridge

Here are the actual values (to 3 significant figures) for your circle of radius 3cm
That's right
As the angle gets smaller the difference between the arc and the chord gets smaller

deg arc chord difference
90 4.712 4.243 0.470
80 4.189 3.857 0.332
70 3.665 3.441 0.224
60 3.142 3.000 0.142
50 2.618 2.536 0.082
40 2.094 2.052 0.042
30 1.571 1.553 0.018
20 1.047 1.042 0.005
10 0.524 0.523 0.001
00 0.000 0.000 0.000

15. Apr 9, 2010

### Misr

Then nothing wrong with what i have written?

16. Apr 9, 2010

### Stonebridge

What you wrote is fine.
The only mistake was the calculation of the arc for 90 degrees.
That's why I wrote out the table of values for you.

17. Apr 10, 2010

### Misr

Yeah right . Actually it is 3*pi/2
Thanks very much This really really helped me understand.