Satellites in orbit

[Anilkumar]

Hello everybody,

 

The satellites revolve in orbit.

Are they in orbit due to;

 

 

• The equilibrium between the centripetal force provided by the Gravity & the Tangential velocity of the satellite? OR

 

• Due to the 'catch me if you can' game played between the Curvature of the earth's surface & the Escape velocity of the satellite.

 

Thank you

Edited November 25, 2011 by Anilkumar

[logearav]

They are due to balancing of the Centripetal force and Electrostatic force of attraction

[Anilkumar]

They are due to balancing of the Centripetal force and Electrostatic force of attraction

 

No, Electrostatic force does not come into play there.

[ewmon]

The equilibrium between the centripetal force provided by the Gravity & the Tangential velocity of the satellite

Correct.

[Anilkumar]

Correct.

 

But the other option;

• The 'catch me if you can' game played between the Curvature of the earth's surface & the Escape velocity of the satellite i.e. when the satellite falls down to contact the surface of the earth, the surface of the earth is not reachable due to its curvature, it too goes down by the same ammount.

is also proposed as the reason behind the phenomenon. So then which one is correct.

Edited November 25, 2011 by Anilkumar

[CaptainPanic]

But the other option;

• The 'catch me if you can' game played between the Curvature of the earth's surface & the Escape velocity of the satellite i.e. when the satellite falls down to contact the surface of the earth, the surface of the earth is not reachable due to its curvature, it too goes down by the same ammount.

is also proposed as the reason behind the phenomenon. So then which one is correct.

I am not familiar with the "catch me if you can" model... but it sounds to me that it is ultimately just another way to describe exactly the same balance of forces (gravity and the centrifugal force).

 

I'm a chemical engineer, so not exactly the expert when it comes to orbital mechanics, but I always thought this picture (answers.com) to be correct. The centrifugal force makes the satellite want to leave orbit (and go higher), while the gravity acts as a centripetal force and balances the centrifugal force.

 

Anyway, I think that the catch me if you can model describes the orbit from a point of view of acceleration, while the other one takes the approach of forces.

 

But since F=m*a, and m is a constant, you can describe it using the F or the a, and it shouldn't matter much.

[Anilkumar]

I am not familiar with the "catch me if you can" model... but it sounds to me that it is ultimately just another way to describe exactly the same balance of forces (gravity and the centrifugal force).

 

I'm a chemical engineer, so not exactly the expert when it comes to orbital mechanics, but I always thought this picture (answers.com) to be correct. The centrifugal force makes the satellite want to leave orbit (and go higher), while the gravity acts as a centripetal force and balances the centrifugal force.

 

Anyway, I think that the catch me if you can model describes the orbit from a point of view of acceleration, while the other one takes the approach of forces.

 

But since F=m*a, and m is a constant, you can describe it using the F or the a, and it shouldn't matter much.

There is some sense in what you say. Even I too had the gut feeling that both may be different versions of the same thing.

Anyway, I think that the catch me if you can model describes the orbit from a point of view of acceleration, while the other one takes the approach of forces. <BR sab="810"><BR sab="811">But since F=m*a, and m is a constant, you can describe it using the F or the a, and it shouldn't matter much.

This is good mathematical thinking.

But even if both hold good; the first i.e. the 'centrepetal force & tangential velocity' model gives a clear cut explanation. But even I feel the second 'catch me if you can' model is some what vague. I fail to understand the phenomenon thoroughly from the POV of 'catch me if you can' model. I would like to check if my understanding of the second model is correct by putting forward the reverse model of the second 'catch me if you can' model.

"The earth is trying to grab the satellite with its gravitational hand, but the satellite which is fast enough not to get caught, escapes by moving forward before the gravitational hand reaches it. It would get caught if its speed is slower than the escape velocity"

Does this hold good?

Thank you CaptainPanic for your interest.

[Klaynos]

I'd say it's a combination of the ideas from both.

 

You need to gravitation and the tangential velocity to give you your curved trajectory. But you also need that curve to be such that the object orbits and doesn't just hit the deck but continually misses.

[Anilkumar]

I'd say it's a combination of the ideas from both.

 

You need to gravitation and the tangential velocity to give you your curved trajectory. But you also need that curve to be such that the object orbits and doesn't just hit the deck but continually misses.

 

Does this mean that the 'catch me if you can' model explains the phenomenon without taking into consideration Gravitation & Tangential velocity. Then what does the curve depend on?

[michel123456]

the 'catch me if you can' model looks exactly the same as Newton's canonball.

 

http://galileoandeinstein.physics.virginia.edu/more_stuff/Applets/newt/newtmtn.html

Edited November 25, 2011 by michel123456

[Klaynos]

Does this mean that the 'catch me if you can' model explains the phenomenon without taking into consideration Gravitation & Tangential velocity. Then what does the curve depend on?

 

No, I just don't think it's a very good explanation. Escape velocity is the wrong term to use.

[Ophiolite]

The 'catch me if you can' model is a wau of explaining the physics without mentioning the physics. I recall it in the following version. Imagine a tall tower with a canon on top. Fire the canon and the ball will follow a curved path towards the ground, landing some distance from the tower. Apply a larger charge, or a lighter ball and it will travel further. The larger charge has given it a higher velocity, so that it can move further horziontally in the time it takes to fall vertically the same distance. But if the distance is substantial because the velocity is high, there will come a time in which the distance fallen vertically matches the curvature of the Earth and the ball will now be in orbit. (Obviously in this simple mind experiment ari resitance is ignored.)

[Anilkumar]

the 'catch me if you can' model looks exactly the same as Newton's canonball.

 

http://galileoandein...wt/newtmtn.html

 

No question of looking exactly the same. It is the Newton's canonball.

 

No, I just don't think it's a very good explanation. Escape velocity is the wrong term to use.

 

But it is the escape velocity that puts the satellite into orbit. If not, what does the curve depend on?

 

But if the distance is substantial because the velocity is high, there will come a time in which the distance fallen vertically matches the curvature of the Earth . . .

 

Why does question of the canon ball falling down arise here, when the canon ball is moving horizontally tangential to Earth's curvature? We shouldn't bring gravity & its effects in this POV, I suppose. We need to explain on the basis of ACCELERATION of the ball, alone.

[Klaynos]

Escape velocity has a very specific meaning. If it was moving at the escape velocity it would escape.

[Anilkumar]

Escape velocity has a very specific meaning. If it was moving at the escape velocity it would escape.

 

Here, we will consider it as the velocity just needed to prevent it from falling towards the Earth. [Does it have any other scientific term designated to it?]

[swansont]

Hello everybody,

 

The satellites revolve in orbit.

Are they in orbit due to;

 

 

• The equilibrium between the centripetal force provided by the Gravity & the Tangential velocity of the satellite? OR

 

• Due to the 'catch me if you can' game played between the Curvature of the earth's surface & the Escape velocity of the satellite.

 

Thank you

 

These both describe the same thing. Sort of. You really can't have and equilibrium between a force and a velocity, and as Klaynos has pointed out, escape velocity is the wrong term to use. But one is an attempt to be more technical and the other is more of an analogy, and both approaches can be valid.

[Anilkumar]

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

 

Then how do we explain it from the POV of the first model?

[Klaynos]

Here, we will consider it as the velocity just needed to prevent it from falling towards the Earth. [Does it have any other scientific term designated to it?]

 

Yes, it has a well defined meaning. For the sake of understanding it is best not to change them.

 

http://en.wikipedia.org/wiki/Escape_velocity

[Anilkumar]

Yes, it has a well defined meaning. For the sake of understanding it is best not to change them.

 

http://en.wikipedia....Escape_velocity

OK let's not use the term escape velocity. We shall use the term 'ORBITING VELOCITY' i.e. the velocity needed to just prevent the satellite from falling down & escape away.

[swansont]

Then how do we explain it from the POV of the first model?

 

The tangential velocity is such that the gravitational force is equal to the centripetal force (assuming we're talking about circular orbits)

[ewmon]

Having been a rocket scientist for 10 years, I choose the force of gravity and tangential velocity answer although the total velocity must be considered.

 

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

IN terms of a mathematical solution, this "equilibrium" is a matter of the distance from earth converted through its 1/R² relationship to obtain a force resulting in an acceleration (a=F/m) that is double-integrated to produce a change in position combined with integrating the satellite's velocity ("total" not just "tangential") to produce a change in position. If the new position is still outside the earth's surface (and atmosphere), you're still orbiting. The total velocity is then updated with the incremental velocity obtained by integrating the acceleration (v=at). I've done this graphically on paper by hand.

 

I've never heard of a catch me if you can game, and I don't know what it means. (I've even seen the movie and read about Frank Abagnale Jr and his company, including the interview with Australia's Radio National.) The surface of the earth is a spherical ellipsoid, and the escape velocity vector is normal to the surface. It all seems fairly static and non-specific (ie, not relative to a specific situation).

[Anilkumar]

The tangential velocity is such that the gravitational force is equal to the centripetal force (assuming we're talking about circular orbits)

The gravitational force itself is the centripetal force. The question of equating them does not arise. They are one and the same in our case. And tangential velocity has no role in determining either forces.

[michel123456]

the 'catch me if you can' model looks exactly the same as Newton's canonball.

 

http://galileoandeinstein.physics.virginia.edu/more_stuff/Applets/newt/newtmtn.html

 

It is funny that in the little animation the mountain is up and the projectile falls down. Imagine the same animation upside-down, with the mountain down and the projectile going up attracted by the Earth. That would be revelating.

[Anilkumar]

Having been a rocket scientist for 10 years, I choose the force of gravity and tangential velocity answer although the total velocity must be considered.

 

 

IN terms of a mathematical solution, this "equilibrium" is a matter of the distance from earth converted through its 1/R² relationship to obtain a force resulting in an acceleration (a=F/m) that is double-integrated to produce a change in position combined with integrating the satellite's velocity ("total" not just "tangential") to produce a change in position. If the new position is still outside the earth's surface (and atmosphere), you're still orbiting. The total velocity is then updated with the incremental velocity obtained by integrating the acceleration (v=at). I've done this graphically on paper by hand.

 

I've never heard of a catch me if you can game, and I don't know what it means. (I've even seen the movie and read about Frank Abagnale Jr and his company, including the interview with Australia's Radio National.) The surface of the earth is a spherical ellipsoid, and the escape velocity vector is normal to the surface. It all seems fairly static and non-specific (ie, not relative to a specific situation).

 

The 'catch me if you can' model is nothing but the thought experiment hypothesized by Isaac Newton titled 'Newton's cannonball'. Details here & here.

[swansont]

The gravitational force itself is the centripetal force. The question of equating them does not arise. They are one and the same in our case. And tangential velocity has no role in determining either forces.

 

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.