Satellites in orbit

[Anilkumar]

a = v^2/r

 

The tangential speed has a lot to do with determining the centripetal force. Also "in our case" is very important. It doesn't have to be so — if the speed is wrong, you will not have circular motion.

 

I understand V= Tangential velocity. Which accelaration does 'a' denote?

[swansont]

I understand V= Tangential velocity. Which accelaration does 'a' denote?

 

a is the centripetal acceleration. That relation has to hold for uniform circular motion.

[Anilkumar]

a is the centripetal acceleration. That relation has to hold for uniform circular motion.

So what is keeping the satellite in orbit in the second model?

[swansont]

So what is keeping the satellite in orbit in the second model?

 

The two models are equivalent. Falling but missing the target is just a another way of explaining the same thing.

[Anilkumar]

The two models are equivalent. Falling but missing the target is just a another way of explaining the same thing.

 

I am not getting this falling part. Why would it fall at all? Tangential velocity is horizontal in direction?

 

Let us see if I can put it right.

 

"The centripetal force is pulling down the satellite at 9.8m/s^2 which is just sufficient enough, only to pull the satellite down by a distance equal to the length by which the surface of the earth goes down due to its curvature. So it does not come in contact & the distance between the satellite & surface remain unchanged"

 

Is this what happening, there?

[swansont]

I am not getting this falling part. Why would it fall at all? Tangential velocity is horizontal in direction?

 

Let us see if I can put it right.

 

"The centripetal force is pulling down the satellite at 9.8m/s^2 which is just sufficient enough, only to pull the satellite down by a distance equal to the length by which the surface of the earth goes down due to its curvature. So it does not come in contact & the distance between the satellite & surface remain unchanged"

 

Is this what happening, there?

 

Replace centripetal force with gravity and that's it. Gravity continually pulls the satellite down, but it's moving fast enough that it misses the earth.

[Anilkumar]

Replace centripetal force with gravity and that's it. Gravity continually pulls the satellite down, but it's moving fast enough that it misses the earth.

 

OK.

 

Then you said

a = v^2/r

The tangential speed has a lot to do with determining the centripetal force. Also "in our case" is very important. It doesn't have to be so — if the speed is wrong, you will not have circular motion.

 

How can Tangential velocity determine centripetal force? The centripetal force is here is nothing but Gravity. How can Tangential velocity determine Gravity?

[ewmon]

The equation a = v²/r concerns a tethered object, not a free-falling (ballistic) object.

 

The tangential speed has nothing to do with the centripetal (aka gravitational) force.

 

All orbits are elliptical to some extent (even though a circle is a special case of an ellipse). Perfectly circular orbits only occur when the attracting force varies as the inverse-fifth power of the distance between the objects.

[swansont]

The equation a = v²/r concerns a tethered object, not a free-falling (ballistic) object.

 

The tangential speed has nothing to do with the centripetal (aka gravitational) force.

 

No, that's not correct. For any object moving in a circle at constant speed, the net force must follow the centripetal force equation. In this example, the centripetal force is the gravitational force. They have to be equal in that case.

[Anilkumar]

No, that's not correct. For any object moving in a circle at constant speed, the net force must follow the centripetal force equation. In this example, the centripetal force is the gravitational force. They have to be equal in that case.

 

What & what have to be equal.

[swansont]

What & what have to be equal.

 

The gravitational force has to be equal to the centripetal force, because gravitational force is a centripetal force in this example. It is the net force acting toward the center.

[Anilkumar]

The gravitational force has to be equal to the centripetal force, because gravitational force is a centripetal force in this example. It is the net force acting toward the center.

 

But how equating gravitational/centripetal force alone would keep the satellite in orbit. The satellite is in orbit because the forces acting on it are in equilibrium & have kept it in orbit. Which is the other force that is countering the gravitational/centripetal force to keep it in the orbit in the second model.

 

You said

You really can't have and equilibrium between a force and a velocity

 

If Tangential velocity is not countering the gravitational/centripetal force then which other force is countering the gravitational/centripetal force.

 

[Edits: Words in italics have been added.]

Edited November 27, 2011 by Anilkumar

[michel123456]

 

from here.

 

Which is the other force that is countering the gravitational/centripetal force to keep it in the orbit in the second model.

 

Initial velocity.

[Anilkumar]

 

'a' which is the Gravitational/Centripetal force, pulls the satellite towards the earth & also gives it the near circular orbit.

 

'v' the Initial velocity, is spent solely in moving the satellite forward.

 

To keep the satellite into orbit we need another force which is equal & opposite in direction to 'a', so that it counter-balances 'a' & stops it from falling on Earth.

 

Where is it?

 

We can not say a vector component of 'v' will counter-balance 'a', because 'v' is perpendicular to 'a'. So its vector component in the opposite direction to 'a', which could counter-balance 'a' would be zero.

 

OR

 

Is it that, the satellite is given an initial force in such a direction that one vector component of that initial force counter-balances 'a' and second vector component of that initial force gives it a forward motion and its third vector component is zero.

 

And again,

 

For this to happen, there should be nill atmospheric resistance in the path of the satellite. Otherwise the initial force given to it gets reduced gradually & ultimately the satellite should fall on Earth.

[swansont]

 

'a' which is the Gravitational/Centripetal force, pulls the satellite towards the earth & also gives it the near circular orbit.

 

'v' the Initial velocity, is spent solely in moving the satellite forward.

 

To keep the satellite into orbit we need another force which is equal & opposite in direction to 'a', so that it counter-balances 'a' & stops it from falling on Earth.

 

No. Circular motion requires a net force. Motion in a circle represents an acceleration; the changing direction means the velocity is not constant. If the motion is circular at constant speed, the net force must conform to F= mv^2/r, no matter what that net force is. michel provided the vector diagrams.

 

In the second explanation, ones says that the satellite is falling toward earth, but the forward motion makes it continually miss.

 

 

For this to happen, there should be nill atmospheric resistance in the path of the satellite. Otherwise the initial force given to it gets reduced gradually & ultimately the satellite should fall on Earth.

 

Atmospheric drag is ignored here. In reality, that's precisely what happens to low-earth orbit satellites.

[Anilkumar]

Circular motion requires a net force.

I agree.

When a Satellite is orbiting the earth there is motion going on. So it certainly needs a net force.

 

Motion in a circle represents an acceleration; the changing direction means the velocity is not constant.

I have no doubts about it.

 

But when we consider the forces acting on the satellite, we see that there are only two forces acting on it.

 

'a'

 

and the force responsible for 'v'.

 

But taking into consideration only these two forces, we see that they are not sufficient to keep the satellite into orbit. There has to be another force that is keeping it in the orbit.

[michel123456]

I agree.

When a Satellite is orbiting the earth there is motion going on. So it certainly needs a net force.

 

 

I have no doubts about it.

 

But when we consider the forces acting on the satellite, we see that there are only two forces acting on it.

 

'a'

 

and the force responsible for 'v'.

 

But taking into consideration only these two forces, we see that they are not sufficient to keep the satellite into orbit. There has to be another force that is keeping it in the orbit.

 

I don't understand your argument.

If there were a rope, what force would extend the rope? The answer is v (most precisely a decomposed part of the v vector). If you cut the rope the object would fly away in a direction tangential to the circle at the moment when you cut it.

The rope is something that attracts the 2 bodies, in the case of astral bodies it is gravity.

So there are 2 "forces" in action: the one that wants the rotating body to fly away, and the other that wants the rotating body not to fly away.

I don't see anything missing.

Edited November 27, 2011 by michel123456

[Anilkumar]

I don't understand your argument.

If there were a rope, what force would extend the rope? The answer is v (most precisely a decomposed part of the v vector). If you cut the rope the object would fly away in a direction tangential to the circle at the moment when you cut it.

The rope is something that attracts the 2 bodies, in the case of astral bodies it is gravity.

So there are 2 "forces" in action: the one that wants the rotating body to fly away, and the other that wants the rotating body not to fly away.

I don't see anything missing.

 

I don't understand how two forces acting mutually perpendicular to each other, oppose each other. I have known that only forces which are acting in opposite directions or those which are at 180⁰ to each other and moving in opposite directions, oppose each other.

[michel123456]

I don't understand how two forces acting mutually perpendicular to each other, oppose each other. I have known that only forces which are acting in opposite directions or those which are at 180⁰ to each other and moving in opposite directions, oppose each other.

Ah. Now I see your objection. You don't accept vector decomposition of V because a is perpendicular, is that it?

[Anilkumar]

Ah. Now I see your objection. You don't accept vector decomposition of V because a is perpendicular, is that it?

 

In the Newton's Cannonball experiment the ball is thrown horizontall to the surface of the Earth i.e. perpendicular to the Gravity.

 

In this case, if you decompose 'v' which is horizontall to the surface of the Earth and so perpendicular to 'a', into its three components, the two of those components, other than the horizontal one, would be zero. None of them can counter-balance 'a'.

 

Unless the Cannonball/Satellite is given an Initial force in a direction such that the three components of that Initial force are such that,

 

one vector component of that initial force is equal & opposite to 'a' & counter-balances 'a';

 

and second vector component of that initial force gives it a forward motion;

 

and its third vector component is zero,

 

I don't see how an object would stay rotating in an orbit.

[michel123456]

Wait a moment. If you walk on the street on a straight line, and I come to push you from the side, will you change direction or not?

[Anilkumar]

Wait a moment. If you walk on the street on a straight line, and I come to push you from the side, will you change direction or not?

 

So you mean to say;

 

that the satellite is trying to move in a tangential direction;

 

and the Gravitational/Centripetal force is continously pulling it towards the Earth;

 

but the pull is not sufficient enough to make it fall on Earth.

 

It simply is able to change the direction of the satellite, to make it move in an orbit.

 

No other force is neither involved nor necessary.

 

And this is, what is called;

 

Circular motion.

 

Right?

[swansont]

I agree.

When a Satellite is orbiting the earth there is motion going on. So it certainly needs a net force.

 

 

I have no doubts about it.

 

But when we consider the forces acting on the satellite, we see that there are only two forces acting on it.

 

'a'

 

and the force responsible for 'v'.

 

But taking into consideration only these two forces, we see that they are not sufficient to keep the satellite into orbit. There has to be another force that is keeping it in the orbit.

 

Constant velocity never requires a force (from Newton's first law). You require a force to change velocity, and that gives rise to the acceleration in this example.

 

a and v are not forces. You can get forces that rely on them, via F=ma and F = mv^2/r

[michel123456]

So you mean to say;

 

that the satellite is trying to move in a tangential direction;

 

and the Gravitational/Centripetal force is continously pulling it towards the Earth;

 

but the pull is not sufficient enough to make it fall on Earth.

 

It simply is able to change the direction of the satellite, to make it move in an orbit.

 

No other force is neither involved nor necessary.

 

And this is, what is called;

 

Circular motion.

 

Right?

Orbital motion. Or orbit. Yes.

[ewmon]

a = v²/r

 

Another way of looking at it is this equation/concept supposedly computes the a (radial acceleration of a ballistic body) from the v and the r. With satellites, you don't compute the a, because the a derives from the force of gravity.

 

You can't say, for example, that you have a satellite 10,000 km from the center of the earth travelling tangentially at 25 km/sec, therefore the radial acceleration is blah, blah, blah. The radial acceleration is due only to gravity. Therefore, a satellite 10,000 km from the center of the earth has a radial acceleration due to gravity totally independent of its velocity.

 

Instead, the a (radial acceleration) and the v (total velocity) act in concert to produce the satellite's orbital motion.

 

To update the position, you double integrate the acceleration, ∫∫a, and you integrate the velocity, ∫v, and add them to the current position to produce the new position.

 

To update the velocity, you integrate the acceleration, ∫a, and add it to the existing velocity, v.

 

When you do this, you obtain the circular or elliptical orbit.

Edited November 27, 2011 by ewmon