[spaceman spiff]

No REer has suggested describing rotational motion as simple linear motion (I think that what you meant by "linearly".).

Your whole schpiel has been that the ISS has linear acceleration relative to the earth. Do you mean to say that its motion relative to the earth isn't rotational?

Think hard about that. Your previous response indicates their is a severe gap in your critical thinking.

No. The ISS rotates about the RE. Why do you ask? Do you think an object can't rotate about the RE and have linear acceleration?

ur dense

Someone correct me if I'm wrong, but as far as I know linear acceleration is just the force divided by the mass, and it's the second derivative of the position vector w.r.t. time. Since there's only one force acting on the ISS and its velocity vector changes with time, it follows that it does have linear acceleration. The term linear has nothing to do with the movement, it's just to differentiate it from angular acceleration, which can be defined as the torque divided by the moment of inertia, or the derivative of the angular velocity w.r.t. time. Any object in uniform circular motion has a non-zero linear acceleration and zero angular acceleration.

[spoon]

No REer has suggested describing rotational motion as simple linear motion (I think that what you meant by "linearly".).

Your whole schpiel has been that the ISS has linear acceleration relative to the earth. Do you mean to say that its motion relative to the earth isn't rotational?

Think hard about that. Your previous response indicates their is a severe gap in your critical thinking.

No. The ISS rotates about the RE. Why do you ask? Do you think an object can't rotate about the RE and have linear acceleration?

ur dense

Someone correct me if I'm wrong, but as far as I know linear acceleration is just the force divided by the mass, and it's the second derivative of the position vector w.r.t. time. Since there's only one force acting on the ISS and its velocity vector changes with time, it follows that it does have linear acceleration. The term linear has nothing to do with the movement, it's just to differentiate it from angular acceleration, which can be defined as the torque divided by the moment of inertia, or the derivative of the angular velocity w.r.t. time. Any object in uniform circular motion has a non-zero linear acceleration and zero angular acceleration.

Mostly irrelevant. I'd say about 85%.

However, the bit in bold is important because that approximates the motion of the ISS.

The term linear has nothing to do with the movement

This part is just wrong.

[markjo]

No REer has suggested describing rotational motion as simple linear motion (I think that what you meant by "linearly".).

Your whole schpiel has been that the ISS has linear acceleration relative to the earth. Do you mean to say that its motion relative to the earth isn't rotational?

Think hard about that. Your previous response indicates their is a severe gap in your critical thinking.

Yes, the circular orbit of the ISS has a (more or less) constant angular velocity, but it also experiences centripetal acceleration.  Why is that so hard to understand? 

[Gulliver]

No REer has suggested describing rotational motion as simple linear motion (I think that what you meant by "linearly".).

Your whole schpiel has been that the ISS has linear acceleration relative to the earth. Do you mean to say that its motion relative to the earth isn't rotational?

Think hard about that. Your previous response indicates their is a severe gap in your critical thinking.

No. The ISS rotates about the RE. Why do you ask? Do you think an object can't rotate about the RE and have linear acceleration?

ur dense

Someone correct me if I'm wrong, but as far as I know linear acceleration is just the force divided by the mass, and it's the second derivative of the position vector w.r.t. time. Since there's only one force acting on the ISS and its velocity vector changes with time, it follows that it does have linear acceleration. The term linear has nothing to do with the movement, it's just to differentiate it from angular acceleration, which can be defined as the torque divided by the moment of inertia, or the derivative of the angular velocity w.r.t. time. Any object in uniform circular motion has a non-zero linear acceleration and zero angular acceleration.

Did you maybe forget that the object in a perfectly circular orbit is still accelerating, at a right angle to its instantaneous velocity, toward the center of the orbit? If it weren't, it would fly off (standard pedantic disclaimer on FoRs) and not be in orbit any longer.

[spaceman spiff]

The term linear has nothing to do with the movement

This part is just wrong.

This whole discussion is basically a discussion on semantics but you're wrong. In usual terminology, linear acceleration is any change in velocity, which is a vectorial quantity. Therefore, any rotational movement will have linear acceleration.

Did you maybe forget that the object in a perfectly circular orbit is still accelerating, at a right angle to its instantaneous velocity, toward the center of the orbit? If it weren't, it would fly off (standard pedantic disclaimer on FoRs) and not be in orbit any longer.

No, and that's why I said that any object in uniform circular motion will have a non zero linear acceleration.

[Gulliver]

The term linear has nothing to do with the movement

This part is just wrong.

This whole discussion is basically a discussion on semantics but you're wrong. In usual terminology, linear acceleration is any change in velocity, which is a vectorial quantity. Therefore, any rotational movement will have linear acceleration.

Did you maybe forget that the object in a perfectly circular orbit is still accelerating, at a right angle to its instantaneous velocity, toward the center of the orbit? If it weren't, it would fly off (standard pedantic disclaimer on FoRs) and not be in orbit any longer.

No, and that's why I said that any object in uniform circular motion will have a non zero linear acceleration.

Sorry, I erred. Please ignore that post.