Satellites in orbit

Anilkumar · November 28, 2011

Orbital motion. Or orbit. Yes.

Ah.

It is that simple.

Someone has rightly said - "Keep it simple, stupid".

My prejudice that science can't be simple made me search for complicated things.

But why only orbital motion? It could be any circular motion. When a stone is tied to a thread and given a circular motion, only the Gravity is replaced by the Centripetal force caused by the tension of the thread. The same thing happens there also. Right?

**swansont** · November 28, 2011

Ah.

It is that simple.

Someone has rightly said - "Keep it simple, stupid".

My prejudice that science can't be simple made me search for complicated things.

But why only orbital motion? It could be any circular motion. When a stone is tied to a thread and given a circular motion, only the Gravity is replaced by the Centripetal force caused by the tension of the thread. The same thing happens there also. Right?

Right. In any example of uniform circular motion, the net force is a centripetal force.

Anilkumar · November 28, 2011

a = v²/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.

This is what I was arguing with Swansont in #25.

Anilkumar, on 25 November 2011 - 07:25 PM, said: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.

Swansont 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.

But now I understand why Swansont was saying that.

Right. In any example of uniform circular motion, the net force is a centripetal force.

How can Centripetal force be the net force?

The part of the Initial force that is giving the Initial Tangential velocity, counter balances the Centripetal foce & the residual part of that Initial force is the net force that is giving the Satellite the forward motion.

Don't we need some force to keep the satellite in motion?

The centripetal force only plays the part of continuously correcting & counter-balancing that component of the Initial force that is giving the Satellite the Tangential path. The centripetal force is completely spent at every instant in correcting the Tangential path into a Circular path.

So the net force has to be that component of the the Initial force that is giving the Satellite the Velocity with which it is moving forward.

If Centripetal/Gravitational force is the net force here, then the Satellite, instead of moving forward in the circular path, should trace a path along the direction of the Centripetal force towards the center of the Earth i.e. it should fall down.

Edited November 28, 2011 by Anilkumar

ewmon · November 28, 2011

How can Centripetal force be the net force?

The part of the Initial force that is giving the Initial Tangential velocity, counter balances the Centripetal foce & the residual part of that Initial force is the net force that is giving the Satellite the forward motion.

Don't we need some force to keep the satellite in motion?

The centripetal force only plays the part of continuously correcting & counter-balancing that component of the Initial force that is giving the Satellite the Tangential path. The centripetal force is completely spent at every instant in correcting the Tangential path into a Circular path.

So the net force has to be that component of the the Initial force that is giving the Satellite the Velocity with which it is moving forward.

If Centripetal/Gravitational force is the net force here, then the Satellite, instead of moving forward in the circular path, should trace a path along the direction of the Centripetal force towards the center of the Earth i.e. it should fall down.

The gravitational force = the centripetal force = the only force = the net force. And this holds true with any orbital motion ... circular, elliptical, linear (ie, straight into the central body), etc.
The initial force that provided the initial tangential velocity no longer exists, and if did exist to provide more tangential velocity (such as a rocket engine), it couldn't counterbalance the centripetal force because the two forces are orthogonal.
No force is needed to keep the object in motion.
Again, the initial force no longer exists. The centripetal force is continuously acting upon the object. Your idea of a force to maintain the tangential velocity is wrong.
Again, the initial force no longer exists. It has effectively been converted into the tangential velocity. The initial force no longer exists.
Yes, the centripetal force does accelerate the object toward the central body; however, the tangential velocity causes it to also move "to the side", which results in an elliptical or circular orbit.

If you would just stop thinking about an initial force still existing, all the physics would become very simple.

Additionally, the tangential velocity has absolutely nothing to do with the centripetal force because the centripetal force is the gravitational force, which is not a function of tangential velocity. However, in the extremely unlikely case of a perfectly circular orbit, you can back-calculate the centripetal acceleration using the a=v²/r equation.

Please, I'd like to alert everyone here that focusing on the special case of a circular orbit and the a=v²/r relationship that seems to apply (but does not) is causing undue confusion here. The physics involved here is being misrepresented, and I'm sure it's being done sincerely, but it's still being misrepresented.

Anilkumar · November 28, 2011

The gravitational force = the centripetal force = the only force = the net force. And this holds true with any orbital motion ... circular, elliptical, linear (ie, straight into the central body), etc.

The initial force that provided the initial tangential velocity no longer exists, and if did exist to provide more tangential velocity (such as a rocket engine), it couldn't counterbalance the centripetal force because the two forces are orthogonal.

No force is needed to keep the object in motion.

Again, the initial force no longer exists. The centripetal force is continuously acting upon the object. Your idea of a force to maintain the tangential velocity is wrong.

Again, the initial force no longer exists. It has effectively been converted into the tangential velocity. The initial force no longer exists.

Yes, the centripetal force does accelerate the object toward the central body; however, the tangential velocity causes it to also move "to the side", which results in an elliptical or circular orbit.

If you would just stop thinking about an initial force still existing, all the physics would become very simple.

Additionally, the tangential velocity has absolutely nothing to do with the centripetal force because the centripetal force is the gravitational force, which is not a function of tangential velocity. However, in the extremely unlikely case of a perfectly circular orbit, you can back-calculate the centripetal acceleration using the a=v²/r equation.

Please, I'd like to alert everyone here that focusing on the special case of a circular orbit and the a=v²/r relationship that seems to apply (but does not) is causing undue confusion here. The physics involved here is being misrepresented, and I'm sure it's being done sincerely, but it's still being misrepresented.

Thanks Ewmon for the painstaking explanation.

I do not understand how to clear my doubts without expressing them. If my arguments are percieved as misrepresentation of physics, then I am being unfairly misunderstood.

Having said that;

OK, we shall consider there is no atmospheric drag & the the Initial force need not exist continuously to maintain the Tangential velocity & ceases to exist after giving an initial thrust & the Satellite will continue to be in a state of motion with that intial thrust. We shall stop talking about the Initial force.

Now,

Let us make things about physics , clear first.

"Any motion occurs in the direction of the net force". [Am I representing Physics rightly here? Correct me if Iam wrong.]

Then, if Centripetal force is the Net force; the motion should take place in the direction of the net force, which is towards the center of the Earth.

The component of the force that is making the object move "to the side" at any instant; must be the Net force, because the only motion that is taking place is "to the side", at any instant.

Thank you.

**swansont** · November 28, 2011

How can Centripetal force be the net force?

Not to be intentionally snarky, but how can it not? If an object moves in a circle at constant speed , it must have a net force directed toward the center of its motion such that F=mv^2/r. That comes from the analysis of how the velocity changes over time — there must be an acceleration, and it must conform to that equation.

However, this is a condition, not a new type of force. The centripetal force will be the sum of whatever forces are acting on the body.

The part of the Initial force that is giving the Initial Tangential velocity, counter balances the Centripetal foce & the residual part of that Initial force is the net force that is giving the Satellite the forward motion.

That's not part of the problem before us. How circular motion was initially achieved is a separate problem. A satellite changing its tangential speed is not in uniform circular motion.

Don't we need some force to keep the satellite in motion?

A force acting through a displacement must do work. But the energy of a satellite in a circular orbit is constant: the speed is constant, so there is no change in KE, and r is constant, so there is no change in PE. The force is acting toward the center, and so is perpendicular to the instantaneous displacement, which means no work is done. A force is necessary for an orbit, but not to keep the satellite in motion.

The gravitational force = the centripetal force = the only force = the net force. And this holds true with any orbital motion ... circular, elliptical, linear (ie, straight into the central body), etc.

No. If the motion is something other than circular at constant speed, then the centripetal force is not equal to the net force. By definition.

Please, I'd like to alert everyone here that focusing on the special case of a circular orbit and the a=v²/r relationship that seems to apply (but does not) is causing undue confusion here. The physics involved here is being misrepresented, and I'm sure it's being done sincerely, but it's still being misrepresented.

It would be helpful if you were clear about what parts you think are being "misrepresented" because a=v²/r most certainly applies to a circular orbit. It does not, however, apply in general to orbits.

Let us make things about physics , clear first.

"Any motion occurs in the direction of the net force". [Am I representing Physics rightly here? Correct me if Iam wrong.]

Then, if Centripetal force is the Net force; the motion should take place in the direction of the net force, which is towards the center of the Earth.

The component of the force that is making the object move "to the side" at any instant; must be the Net force, because the only motion that is taking place is "to the side", at any instant.

Thank you.

No. Force does not tell you the direction of motion (i.e. the velocity), it tells you the direction of the acceleration. If you drive in a car and come to a stop, your direction of motion is forward, but the acceleration is backward. For a satellite in a circular orbit, the instantaneous direction of motion is tangent to a circle, but the force is perpendicular to it (which is what makes this a special case where an additional equation holds)

Anilkumar · November 28, 2011

Not to be intentionally snarky, but how can it not? If an object moves in a circle at constant speed , it must have a net force directed toward the center of its motion such that F=mv^2/r. That comes from the analysis of how the velocity changes over time — there must be an acceleration, and it must conform to that equation.

However, this is a condition, not a new type of force. The centripetal force will be the sum of whatever forces are acting on the body.

That's not part of the problem before us. How circular motion was initially achieved is a separate problem. A satellite changing its tangential speed is not in uniform circular motion.

A force acting through a displacement must do work. But the energy of a satellite in a circular orbit is constant: the speed is constant, so there is no change in KE, and r is constant, so there is no change in PE. The force is acting toward the center, and so is perpendicular to the instantaneous displacement, which means no work is done. A force is necessary for an orbit, but not to keep the satellite in motion.

No. If the motion is something other than circular at constant speed, then the centripetal force is not equal to the net force. By definition.

It would be helpful if you were clear about what parts you think are being "misrepresented" because a=v²/r most certainly applies to a circular orbit. It does not, however, apply in general to orbits.

No. Force does not tell you the direction of motion (i.e. the velocity), it tells you the direction of the acceleration. If you drive in a car and come to a stop, your direction of motion is forward, but the acceleration is backward. For a satellite in a circular orbit, the instantaneous direction of motion is tangent to a circle, but the force is perpendicular to it (which is what makes this a special case where an additional equation holds)

Let us keep everything aside;

let's first see how an object responds to forces;

if an object is hit by a force 'Fa' in one direction;

and hit by another force 'Fb' at the same instance in another direction;

there will be a resultant force say, 'Fr' in a third direction other than the directions of either 'Fa' or 'Fb';

now will the object not move in the direction of the net resultant force;

So if the net force in a circular/orbital motion is the Centripetal force, then will the object not move in the direction of the net centripetal force?

ewmon · November 28, 2011

OK, we shall consider there is no atmospheric drag & the the Initial force need not exist continuously to maintain the Tangential velocity & ceases to exist after giving an initial thrust & the Satellite will continue to be in a state of motion with that intial thrust. We shall stop talking about the Initial force.

"Any motion occurs in the direction of the net force". [Am I representing Physics rightly here? Correct me if Iam wrong.]

Then, if Centripetal force is the Net force; the motion should take place in the direction of the net force, which is towards the center of the Earth.

The component of the force that is making the object move "to the side" at any instant; must be the Net force, because the only motion that is taking place is "to the side", at any instant.

Ok, and in a more general sense, no force needs to exist to maintain any kind of velocity.
No, stop talking about a "net force" as if there is more than one force acting on the object. There is only the gravitational force, which produces an incremental velocity (F=ma, a=F/m, and v=∫a=∫F/m) that combines with the existing tangential velocity to produce a new velocity. Velocities are additive.
The centripetal force is the net force, and it produces an incremental velocity towards the central body (Earth) that combines with the existing velocity. Velocities are additive.
No. No sustaining force is required to maintain velocity. Think of a hockey puck, or an air hockey puck, which maintain their motion without the constant influence of a force.

If an object moves in a circle at constant speed , it must have a net force directed toward the center of its motion such that F=mv^2/r. That comes from the analysis of how the velocity changes over time — there must be an acceleration, and it must conform to that equation.

However, this is a condition, not a new type of force. The centripetal force will be the sum of whatever forces are acting on the body.

If the motion is something other than circular at constant speed, then the centripetal force is not equal to the net force. By definition.

It would be helpful if you were clear about what parts you think are being "misrepresented" because a=v²/r most certainly applies to a circular orbit. It does not, however, apply in general to orbits.

What "net" force as if there's two or more forces combining to produce a "net" force? The term "net force" is misleading. There is only one force: gravitational force. The amount of centripetal force produced by gravity can be back-calculated (F=mv²/r) from the characteristics of the circular orbit(that is, from v) and r) . The velocity does change over time, and there must be an acceleration, and it must conform to that equation. However, this acceleration is a product of gravity, but not from any kind of v²/r effect.
The back-calculated value F=mv²/r is not a new type of force. The condition of v²/r exists only because the circumstances (gravity and tangential velocity) produced the circular orbit in the first place. False, the centripetal force will only be the gravitational force, but not "the sum of forces are acting on the body". (Swansont, what "sum of forces"? No other forces exist in the example. Please describe any so-called "other forces".)
False. The centripetal force is the gravitational force totally independent from the characteristic of the motion/orbit (circular, elliptical, linear, etc). For example, the Sun's gravitational force alone produces the highly elliptical orbit of Halley's comet. "By definition"? What definition? Certainly not v²/r. The relationship of a=v²/r does not produce the circular orbit.
I think I've been clear in this post. The relationship of a=v²/r does not produce the circular orbit, it merely back-calculates the gravitational force that produced the special case of a circular orbit in the first place. And you are correct, a=v²/r does not "apply" (ie, back-calculate) to any other kind of orbit.

Let us keep everything aside;

let's first see how an object responds to forces;

if an object is hit by a force 'Fa' in one direction;

and hit by another force 'Fb' at the same instance in another direction;

there will be a resultant force say, 'Fr' in a third direction other than the directions of either 'Fa' or 'Fb';

now will the object not move in the direction of the net resultant force;

So if the net force in a circular/orbital motion is the Centripetal force, then will the object not move in the direction of the net centripetal force?

Yes, F_r = ΣF = F_a + F_b.

Actually, the object will accelerate in the direction of the resultant force, so there will be some movement (change in position) and there will be some change in velocity, both in the force's direction.

There are no multiple forces to produce a "net" force, there is only the gravitational (centripetal) force (which, okay, is the "net" force) that causes the object to accelerate toward the central body, and yet, the tangential velocity also causes it to move "sideways".

Edited November 28, 2011 by ewmon

**swansont** · November 28, 2011

Let us keep everything aside;

let's first see how an object responds to forces;

if an object is hit by a force 'Fa' in one direction;

and hit by another force 'Fb' at the same instance in another direction;

there will be a resultant force say, 'Fr' in a third direction other than the directions of either 'Fa' or 'Fb';

now will the object not move in the direction of the net resultant force;

So if the net force in a circular/orbital motion is the Centripetal force, then will the object not move in the direction of the net centripetal force?

If the object is already moving, the motion need not be in the direction of the net force. What you are thinking of is actually a special case, i.e. the only time that you must have the force and velocity be in the same direction. In the example we are discussing, we are ignoring how the satellite got its speed v. Once it has that speed and is in a circular orbit, you know that the force is perpendicular to the velocity.

[*]What "net" force as if there's two or more forces combining to produce a "net" force? The term "net force" is misleading. There is only one force: gravitational force. The amount of centripetal force produced by gravity can be back-calculated (F=mv²/r) from the characteristics of the circular orbit(that is, from v) and r) . The velocity does change over time, and there must be an acceleration, and it must conform to that equation. However, this acceleration is a product of gravity, but not from any kind of v²/r effect.

In this case, but not in general. What I wrote is true in general: in any case of uniform circular motion, the net force is given by mv^2/r. In this example, the net force is simply the gravitational force.

[*]The back-calculated value F=mv²/r is not a new type of force.

Which is precisely what I wrote.

[*]False. The centripetal force is the gravitational force totally independent from the characteristic of the motion/orbit (circular, elliptical, linear, etc). For example, the Sun's gravitational force alone produces the highly elliptical orbit of Halley's comet. "By definition"? What definition? Certainly not v²/r. The relationship of a=v²/r does not produce the circular orbit.

Sorry, no. If you do not have a circular orbit, the centripetal force equation is not equal to the gravitational force. If the orbit is elliptical, there is tangential acceleration as well. If the "orbital" motion is linear you can trivially see this is false: at the turning point, the speed is zero, yet there is still a force on the object. You appear to agree below, where you say that it does not apply to any other kind of orbit.

[*]I think I've been clear in this post. The relationship of a=v²/r does not produce the circular orbit, it merely back-calculates the gravitational force that produced the special case of a circular orbit in the first place. And you are correct, a=v²/r does not "apply" (ie, back-calculate) to any other kind of orbit.

I'm not seeing where anyone has claimed otherwise. Centripetal acceleration is a condition that is true when you have uniform circular motion. As I previously wrote.

Yes, F_r = ΣF = F_a + F_b.

Actually, the object will accelerate in the direction of the resultant force, so there will be some movement (change in position) and there will be some change in velocity, both in the force's direction.

In the limit of small time increments (i.e. if we use calculus), the movement is perpendicular to the force. If there was movement in the direction of the force, the force would be doing work.

ewmon · November 28, 2011

False. The centripetal force is the gravitational force totally independent from the characteristic of the motion/orbit (circular, elliptical, linear, etc). For example, the Sun's gravitational force alone produces the highly elliptical orbit of Halley's comet. "By definition"? What definition? Certainly not v²/r. The relationship of a=v²/r does not produce the circular orbit.

Sorry, no. If you do not have a circular orbit, the centripetal force equation is not equal to the gravitational force. If the orbit is elliptical, there is tangential acceleration as well. If the "orbital" motion is linear you can trivially see this is false: at the turning point, the speed is zero, yet there is still a force on the object.

I said that the centripetal force is the gravitational force. Okay, let me rephrase it: The gravitational force is centripetal in nature. But you're right, if the orbit is not circular, the centripetal force equation cannot back-calculate to obtain the value of the gravitational force. This is why the centripetal force equation is of such extremely limited use when discussing satellite orbits.

Again, the centripetal force equation (that you're using here) is not generally applied to ballistic (aka, "free fall") trajectories that satellites experience. It generally applies to objects restricted to move in a circular path (ie, tethered objects, objects restricted by a circular surface, parts of a spinning wheel, etc). For free-falling satellites in all sorts of simple orbits, the centripetal force is produced by gravity. In the real world, the equation used to compute the force attracting the satellite to the central body is the gravitational force equation.

Gravitational force is the only force in the simple orbits (circular, elliptical, linear ,etc) considered in this thread. By definition, and as I said, the gravitational force is the centripetal force (ie, it intersects the central body's center of mass — by definition), independent of whether the object's velocity is normal to it. I'm using the term "centripetal force" generically, and I think you're using it as only the force computed from the centripetal force equation.

When the orbit is not circular, then the velocity vector isn't always normal to the gravitational force, and as you say, this gravitational force (which is centripetal in nature) can be broken into a tangential component of force which accelerates and decelerates the object along its line of travel (ie, along its velocity vector) and also a normal component of force (ie, normal to the velocity vector) that does not intersect the center of mass of the central body.

Anilkumar · November 29, 2011

. . . in a more general sense, no force needs to exist to maintain any kind of velocity.

This holds good only in ideal conditions where there is no atmospheric drag or in the absence of any resistance.

. . . There is only the gravitational force, which produces an incremental velocity (F=ma, a=F/m, and v=∫a=∫F/m) that combines with the existing tangential velocity to produce a new velocity. Velocities are additive.

The centripetal force is the net force, and it produces an incremental velocity towards the central body (Earth) that combines with the existing velocity. Velocities are additive.

. . . gravitational force (which is centripetal in nature) can be broken into a tangential component of force which accelerates and decelerates the object along its line of travel (ie, along its velocity vector) and also a normal component of force (ie, normal to the velocity vector) that does not intersect the center of mass of the central body.

. . . .If an object moves in a circle at constant speed , it must have a net force directed toward the center of its motion such that F=mv^2/r. That comes from the analysis of how the velocity changes over time — there must be an acceleration, and it must conform to that equation.

However, this is a condition, not a new type of force.

So on the basis of these; I can say . . .

Let me see if I have got it all correct.

In case of an object moving in circular orbit arround Earth [or any body]:

If I say the object is moving due to its velocity 'v' arround the Earth - I am wrong.
If I say the Earth's Gravity is holding the object from escaping - I am wrong
If I say the object is moving in any particular direction - I am wrong.
If I say the object is moving at an uniform acceleration arround the Earth - I am right.

------------------------------------So

" A circular motion is a SYSTEM in itself where all the forces, any velocities or any other entities [whatever their names - just don't mention them] acting on an object, have together given the object an ACCELERATION due to which it is going arround in an orbit".

Or simply, we should say;

"It is a "System" where the satellite is ACCELERATING [can't say moving] arround the Earth." [Due to all the forces, velocities etc inherent to the System]

Am I correct?

Edited November 29, 2011 by Anilkumar

ewmon · November 29, 2011

In case of an object moving in circular orbit arround Earth [or any body]:

If I say the object is moving due to its velocity 'v' arround the Earth - I am wrong.

If I say the Earth's Gravity is holding the object from escaping - I am wrong

If I say the object is moving in any particular direction - I am wrong.

If I say the object is moving at an uniform acceleration arround the Earth - I am right.

------------------------------------So

" A circular motion is a SYSTEM in itself where all the forces, any velocities or any other entities [whatever their names - just don't mention them] acting on an object, have together given the object an ACCELERATION due to which it is going arround in an orbit".

Or simply, we should say;

"It is a "System" where the satellite is ACCELERATING [can't say moving] arround the Earth." [Due to all the forces, velocities etc inherent to the System]

Am I correct?

An object moving only due to its tangential velocity will travel in a straight line, and thus, cannot orbit the Earth.
The Earth's gravity can prevent an object from escaping by keeping it in an orbit or by causing the object to collide with the Earth.
The object can move, and it can move in a particular direction.
Yes, the object is moving at an uniform acceleration arround the Earth.

Well, the satellite is both moving and accelerating. It can't accelerate without moving because δv=∫a (the change in velocity v is the integration of the acceleration a).

Anilkumar · November 29, 2011

. . . . Yes, the object is moving at an uniform acceleration arround the Earth.

So when the object is in a net acceleration, there is a net force. And that net force is the Centripetal force (that was what, Swansont was repeatedly pointing out, shown below).

Right. In any example of uniform circular motion, the net force is a centripetal force.

Now that things have settled to some extent; how about testing them a little bit, if I am permitted.

I have two Cannonballs A & B of mass 1kg & 2kg respectively.

I want to put both of them in the Earth's orbit.

If I take them up in a rocket along with the Cannon to a height 'h' above the Earth [Newton took them atop a tall mountain. But I am in a Rocket age now.];

I give them each an initial thrust Ta & Tb respectively with the help of my Cannon in a direction tangential to the orbit of radius 'h'.

Their tangential velocities would be Va & Vb respectively.

They start moving with accelerations Aa & Ab respectively in an orbit arround the Earth.

What should/would be the minimum values of the following quantities?

h, Ta, Tb, Va, Vb, Aa, & Ab.

Edited November 29, 2011 by Anilkumar

**swansont** · November 29, 2011

Now that things have settled to some extent; how about testing them a little bit, if I am permitted.

I have two Cannonballs A & B of mass 1kg & 2kg respectively.

I want to put both of them in the Earth's orbit.

If I take them up in a rocket along with the Cannon to a height 'h' above the Earth [Newton took them atop a tall mountain. But I am in a Rocket age now.];

I give them each an initial thrust Ta & Tb respectively with the help of my Cannon in a direction tangential to the orbit of radius 'h'.

Their tangential velocities would be Va & Vb respectively.

They start moving with accelerations Aa & Ab respectively in an orbit arround the Earth.

What should/would be the minimum values of the following quantities?

h, Ta, Tb, Va, Vb, Aa, & Ab.

We can see that to be in a circular orbit, F = mv^2/r must hold. And that this force is from gravity, F = GMm/r^2

All we have to do is set them equal to each other. The mass of the cannonball cancels, and we are left with the mass of the earth, M.

v^2 = GM/r where r is measured from the center of the earth. Whatever speed the cannonball needs, it depends only on r (G and M being constants)

Anilkumar · November 29, 2011

We can see that to be in a circular orbit, F = mv^2/r must hold. And that this force is from gravity, F = GMm/r^2

All we have to do is set them equal to each other. The mass of the cannonball cancels, and we are left with the mass of the earth, M.

v^2 = GM/r where r is measured from the center of the earth. Whatever speed the cannonball needs, it depends only on r (G and M being constants)

At an height of 'h', I suppose the equation becomes;

v²=G_hM/(r+h), [Where G_h=Gravity at height 'h']

So from this we can infer that;

The Tangential velocity that the Cannonball needs to remain in orbit is;

independent of its own mass;

inversely proportional to the radius of the orbit.

Now does this also mean that;

At any height, small or big, if sufficient Tangential velocity, calculated according to the above formula, is given to any object of any mass; it will remain in orbit arround the Earth.

For a given height, all objects of all masses will need the same Tangential velocity to remain in orbit.

Edited November 29, 2011 by Anilkumar

**swansont** · November 29, 2011

Now does this also mean that;

At any height, small or big, if sufficient Tangential velocity, calculated according to the above formula, is given to any object of any mass; it will remain in orbit arround the Earth.

For a given height, all objects of all masses will need the same Tangential velocity to remain in orbit.

Yes. It's why a space capsule/shuttle can dock with the ISS, even though that have different masses. They can be put into essentially identical orbits.

Edited November 29, 2011 by swansont
typo

ewmon · November 29, 2011

You (Anilkumar) might want to hear from me, and so, I agree with what has been discussed about these cannonballs/docking scenarios.

Anilkumar · November 29, 2011

Yes. It's why a space capsule/shuttle an dock with the ISS, even though that have different masses. They can be put into essentially identical orbits.

You (Anilkumar) might want to hear from me, and so, I agree with what has been discussed about these cannonballs/docking scenarios.

Fantastic;

& that last piece of information, nice one, 'the docking process';

Swansont, Ewmon;

you're great;

Thank you both & everybody, for being with me.

Edited November 29, 2011 by Anilkumar

michel123456 · November 29, 2011

This is the most wonderful thread I have seen since my arrival here. I am touched. (no sarcasm)

For the first time, I have spend my available amount of + reputation for the day, I didn't even know there existed a limit.

Anilkumar · November 30, 2011

This is the most wonderful thread I have seen since my arrival here. I am touched. (no sarcasm)

For the first time, I have spend my available amount of + reputation for the day, I didn't even know there existed a limit.

Sorry Michel123456,

I forgot to thank you personally,

You gave me the first break by,

pushing me sideways,

while I was walking on the street in a straight line;

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?

michel123456 · November 30, 2011

Sorry Michel123456,

I forgot to thank you personally,

You gave me the first break by,

pushing me sideways,

while I was walking on the street in a straight line;

Thanks.

Now that you think you have understood how things are going, just figure:

I am not pushing you any more: I am attracting you (1) from a distance (2).

(1) how is that possible? I am not a magnet and you are not made of steel.

(2) how is that possible? How can a force act "magically" through distance ?

That's the beauty of gravity.

Anilkumar · November 30, 2011

Thanks.

Now that you think you have understood how things are going, just figure:

I am not pushing you any more: I am attracting you (1) from a distance (2).

(1) how is that possible? I am not a magnet and you are not made of steel.

(2) how is that possible? How can a force act "magically" through distance ?

That's the beauty of gravity.

There is a force of attraction between any two objects with mass according to Newton's 'Law of universal Gravitation';

Given by :-

----------------F=GMm/r²

Edits: Grammar

Edited November 30, 2011 by Anilkumar

michel123456 · November 30, 2011

There is a force of attraction between any two objects with mass according to Newton's 'Law of universal Gravitation';

Given by :-

----------------F=GMm/r²

Edits: Grammar

Yes.

Isn't that amazing?

Think about it twice.

Anilkumar · December 1, 2011

Yes.

Isn't that amazing?

Think about it twice.

Yes. It is the most mysterious thing in the universe.

And, is there any empirical evidence for the existance of Gravitational waves?

Have they been detected?

If so, how do they behave?

The relativists say they exist.

Janus · December 1, 2011

Yes. It is the most mysterious thing in the universe.

And, is there any empirical evidence for the existance of Gravitational waves?

There is indirect evidence in the form of a pair of orbiting neutrons stars. Theory says that they should radiate energy away in the form of gravity waves, and this should cause them to spiral closer to each other. Careful measurements have shown that they are approaching each other at the same rate as predicted by gravity wave emission.

Have they been detected?

Not yet. but this is not surprising as they would be very weak and it takes extremely sensitive equipment to detect.

If so, how do they behave?

They travel at c and are essentially "ripples in space".

The relativists say they exist.

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