I am confused about the orbital velocity

[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 I am my book)

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

so Could u help?
Thanks

 

[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 traveled from a to b is the arc length aqb
The displacement from a to b is the chord length apb.

 

[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.

 

[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.

 

[sophiecentaur]

Wow, deja vu!

 

[Stonebridge]

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.

It's more likely my glasses.

 

[Misr]

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 traveled from a to b is the arc length aqb
The displacement from a to b is the chord length apb.

(Or more correctly, the average speed if the value is not constant)

the speed is constant in a circular motion .

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.

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

 

[Stonebridge]

the speed is constant in a circular motion .

Often but not always.
Certainly in most physics problems like these, involving circular motion, the speed is constant.

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

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.

Edit to add:
If you run half way around a circle, diameter 200m, in 1 minute, you run a distance of πd/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.

 

[Misr]

now its quite obvious

i'll tell u my real problem
http://dev.physicslab.org/Document....Motion_CentripetalAcceleration.xml#disclaimer

I can't understand why do we suppose the arc length equal to its chord??

dv/v = c/r

we substitute for c(chord length) by s (arc length)

 

[Stonebridge]

now its quite obvious

i'll tell u my real problem
http://dev.physicslab.org/Document....Motion_CentripetalAcceleration.xml#disclaimer

I can't understand why do we suppose the arc length equal to its chord??

dv/v = c/r

we substitute for c(chord length) by s (arc length)

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.

 

[sophiecentaur]

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

 

[Misr]

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.

Ahaa...I got it now
thanks so much

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

Yes
thanks very much ,Miss Sophie

 

[Misr]

MMM
i tried it
I don't know anything about limits but is there anything wrong about this equation?
where Vdt is ab arc

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



suppose we have this circle

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

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

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??

 

[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

 

[Misr]

Then nothing wrong with what i have written?

 

[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.

 

[Misr]

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.

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