Theory Question - Uniform Circular Motion

[tahayassen]

-Question Removed-

 

[Doc Al]

What was your reasoning? Show how you determined that answer.

 

[TaxOnFear]

You teacher was right, the correct answer is false.

If the radius of an object in orbit increases, the speed of said object must decrease.

 

[tahayassen]

v=2*pi*radius/T

As the radius increases, the speed also increases.

 

[Vanadium 50]

And what about T?

 

[tahayassen]

It stays constant in order to maintain uniform circular motion.

 

[Doc Al]

v=2*pi*radius/T

As the radius increases, the speed also increases.

You assume that T is constant. Rethink that.

 

[tahayassen]

You assume that T is constant. Rethink that.

Shouldn't angular velocity always be constant no matter what the radius is?

 

[Doc Al]

It stays constant in order to maintain uniform circular motion.

No it doesn't.

Try getting back to basics. What force holds the satellite in orbit? Apply Newton's 2nd law.

 

[Doc Al]

Shouldn't angular velocity always be constant no matter what the radius is?

Your formula was for linear velocity, not angular velocity. And no, the angular velocity is not constant as the radius changes.

 

[tahayassen]

No it doesn't.

Try getting back to basics. What force holds the satellite in orbit? Apply Newton's 2nd law.

(mv^2)/r = (Gmm)/r^2

v^2 = Gm/r

I'm confused. What do I do here?

Your formula was for linear velocity, not angular velocity. And no, the angular velocity is not constant as the radius changes.

Yes, but 2PI/T is the formula for angular velocity, correct? In order to maintain uniform circular motion, how can the angular velocity not be constant?

 

[Doc Al]

(mv^2)/r = (Gmm)/r^2

v^2 = Gm/r

Excellent. Now realize that Gm = (v^2)r must be constant. That's what you need.

Yes, but 2PI/T is the formula for angular velocity, correct?

Yes.

In order to maintain uniform circular motion, how can the angular velocity not be constant?

For a given radius the angular velocity will be constant, but not if you change the radius.

 

[Doc Al]

edit: From this formula, I can see that as the radius increases, the velocity goes down, but this doesn't account for uniform circular motion.

Sure it does. The formulas are derived for the case of uniform circular motion.

 

[tahayassen]

For a given radius the angular velocity will be constant, but not if you change the radius.

Even if you change the radius, the angular velocity will stay the same.

angular velocity=2PI/T

You can change the radius as much as you'd like, but the angular velocity will stay the same.

I can see why v^2 = Gm/r would work with large bodies that involve large gravitational forces, but what if we assumed gravity is negligible? Then, I guess we can use:

v=2*PI*radius/T

And from that we can see, since angular velcoity is constant (2PI/T), as you increase the radius, the velocity also increases.

edit: Actually, it's impossible for large bodies of mass to maintain uniform circular motion after the radius is changed. If we plug in the v=2*PI*radius/T into the equation, then we can see that:

v^2 = Gm/r
4(PI^2)(r^2)/T^2 = Gm/r
4(PI^2)(r^3)/T^2 = Gm

Gm has to stay constant. 4PI^2 has to stay constant. In order to maintain uniform circular motion, T has to stay constant, therefore it is impossible to change the radius AND maintain uniform circular motion at the same time.

 

[Doc Al]

Even if you change the radius, the angular velocity will stay the same.

angular velocity=2PI/T

You can change the radius as much as you'd like, but the angular velocity will stay the same.

Why do you think this? What's your reasoning?

I can see why v^2 = Gm/r would work with large bodies that involve large gravitational forces, but what if we assumed gravity is negligible?

This is a question about satellites in orbit. You can't very well ignore gravity.

Then, I guess we can use:

v=2*PI*radius/T

You can always use that for uniform circular motion. But it doesn't tell you much.

And from that we can see, since angular velcoity is constant (2PI/T), as you increase the radius, the velocity also increases.

You merely assume that angular velocity is constant. Why?

 

[tahayassen]

angular velocity = 2PI/T

Because angular velocity comes from multiplying 2 times PI divided by T, the radius has no effect on the angular velocity.

Assume the radius is 5 m and the period is 10 seconds.

angular velocity 1 = 2PI/10

Assume the radius is 10 m and the period is 10 seconds.

angular velocity 2 = 2PI/10

angular velocity 1 = angular velocity 2

 

[tahayassen]

You can also think of it logically.

To maintain the same angular velocity, if you increase the radius of anything, it will need to cover the larger amount of arc length over the same amount of time. Therefore, the velocity must increase, to maintain the same amount of time taken.

 

[Doc Al]

angular velocity = 2PI/T

Because angular velocity comes from multiplying 2 times PI divided by T, the radius has no effect on the angular velocity.

Assume the radius is 5 m and the period is 10 seconds.

angular velocity 1 = 2PI/10

Assume the radius is 10 m and the period is 10 seconds.

angular velocity 2 = 2PI/10

angular velocity 1 = angular velocity 2

For a given orbital radius, the angular velocity is constant. That's what uniform circular motion means. But when you change the radius, the angular velocity changes to a new value. So you cannot assume that angular velocity remains constant for different radii.

 

[tahayassen]

For a given orbital radius, the angular velocity is constant. That's what uniform circular motion means. But when you change the radius, the angular velocity changes to a new value. So you cannot assume that angular velocity remains constant for different radii.

Alright, so you agree that when there is uniform circular motion, the angular velocity is constant?

 

[Doc Al]

You can also think of it logically.

To maintain the same angular velocity, if you increase the radius of anything, it will need to cover the larger amount of arc length over the same amount of time. Therefore, the velocity must increase, to maintain the same amount of time taken.

Sure, if angular velocity remains the same. But it doesn't!

(By the way, for uniform circular motion both angular and linear velocity are constant.)

 

[Doc Al]

Alright, so you agree that when there is uniform circular motion, the angular velocity is constant?

That's the definition of uniform circular motion. But all that means is that for a given radius, the velocity will be constant. (As opposed to having a changing speed.) Of course, for a satellite that constant velocity depends on the radius of the orbit!

 

[tahayassen]

I see the issue. The question doesn't state that the uniform circular motion is maintained nor is the angular velocity maintained. What is the significance of stating that the orbit before the radius is changed is in uniform circular motion?

 

[tahayassen]

Actually, I've change my mind (again :[).

For a given orbital radius, the angular velocity is constant. That's what uniform circular motion means. But when you change the radius, the angular velocity changes to a new value. So you cannot assume that angular velocity remains constant for different radii.

So in other words, you are saying that linear speed is constant? But angular speed is not constant?

v = 2*PI*r/T

If linear speed is the same, then when you increase the radius, the period also increases.

 

[Doc Al]

So in other words, you are saying that linear speed is constant? But angular speed is not constant?

v = 2*PI*r/T

If linear speed is the same, then when you increase the radius, the period also increases.

For a given orbit (radius), both linear speed and angular speed are constant. (At least in uniform circular motion.)

Of course, if you change to a new orbit, you'll get different speeds.

 

[JHamm]

Look, for uniform circular motion there must be a centripetal acceleration of \frac{v^2}{r} which means there must be a force of m \frac{v^2}{r} with m being the mass of the moving body. In an orbiting satellite this force is \frac{GMm}{r^2} with M being the mass of the central body, which means that
v^2 = \frac{GM}{r}

And you can see that the velocity decreases with radius, that's really all there is to this problem.

If you're still not convinced take your equation of angular velocity being \frac{2 \pi}{T} and from (one of?) Kepler's law(s) you know that as radius increases orbit time also increases and the length of the orbital path also increases so your velocity will definitely be less at further radii.

 

[tahayassen]

Alright, I think I understand. This is how my teacher told me the answer was:

She assumed that centripetal acceleration stayed the same.

a1 = a2
(v1)^2/r1 = (v2)^2/r2

She said that as r2 increases, v2 must decrease. Does that method work too?

 

[Doc Al]

Alright, I think I understand. This is how my teacher told me the answer was:

She assumed that centripetal acceleration stayed the same.

a1 = a2
(v1)^2/r1 = (v2)^2/r2

She said that as r2 increases, v2 must decrease. Does that method work too?

No, that method is wrong. You cannot assume that centripetal acceleration stays the same!

The acceleration is given by F/m, where F is the force of gravity. F obviously depends on distance, so as the radius changes so does the centripetal acceleration.

The correct analysis was given in posts #11 and #12.

 

[Doc Al]

More on this:

a1 = a2
(v1)^2/r1 = (v2)^2/r2

She said that as r2 increases, v2 must decrease. Does that method work too?

If that equation were true it would imply that as r increased so must v. Just the opposite of what she said.