# Revolution under the influence of a central force

1. Nov 17, 2018

### src2206

Hello friends and fellow Physics enthusiasts

At the beginning let me confess that I am more of a sleeping member though I do follow this forum quite closely. Today, I am posting to seek some help from fellow Physics educators. I hope this is the right section of the forum to post, if not, I would like to request the moderator to be kind enough to move it to an appropriate section.

A few days back, I had an argument with a person on a social media platform regarding the effect of the absence of diurnal motion (rotation of Earth on its own axis). According to him, there would be a change in "Day-Night" even if only the Earth's revolution around the Sun is present. In support, he also showed a diagram which I am enclosing with this post. I disagreed, arguing that it is not possible that a point on the periphery of Earth will change its position relative to the Sun if there is no rotation of Earth. He then argued that if any other external star system/star is taken as a reference frame then it is possible.

I would like to request you all to help me out in the following two points:

1. Is he right? Am I making any mistake in understanding? Is the diagram attached right or wrong?

2. I am unable to formulate a pointed and logical reply in an organized manner to counter his argument. Provided he is wrong, I would like to request you to help me on this.

Under the influence of a central force and in absence of any torque, how can the position of a point on the periphery of the object change with respect to the central point or any other reference frame?

I hope I have clarified my dilemma and question clearly to the members. If you require any other input, please feel free to ask. I am unable to post the conversation, as it was in my mother tongue "Bengali" and I am sure most of the members here would have no use for that.

Thank you all, waiting eagerly for your reply.

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2. Nov 17, 2018

### phinds

yes
you have to be since you're wrong.

Think about this. You stand at a point on the Earth facing the sun. As the Earth rotates around the sun you continue to point towards exactly the same distant star. When The Earth gets to the other side of the sun 6 months later, you are facing directly away from the sun.

3. Nov 17, 2018

### src2206

Let's put it in another perspective. If I tie a pebble with a string and rotate it around my finger, will the position of the knot change?
Please understand, I am not trying to argue, but I am trying to clear the misunderstanding.

Last edited: Nov 17, 2018
4. Nov 18, 2018

### Tom.G

Try putting your hands in front of you and point both index fingers up, finger nails facing away from each other.
Consider one fingertip to be the Sun and the other as the Earth.
You are the finger nail on the Earth.
Now without rotating either hand, cross your arms at the wrists.
Are you now facing the Sun?

5. Nov 18, 2018

### src2206

Thanks so much, Tom. I totally agree. But why the example of rotating something by a string is not conforming with this idea?

6. Nov 18, 2018

### Tom.G

Probably because in keeping the string taut, you are rotating the pebble thereby keep the knot always pointing to your finger.

7. Nov 18, 2018

### src2206

Nope, there is no rotation of the pebble about its own axis

8. Nov 18, 2018

### Tom.G

If there was no rotation it would behave as your finger nail did. Can you draw a sketch or something to better describe the situation with the string and pebble?

9. Nov 18, 2018

### PeroK

The attached diagram shows what would happen if the Earth were not rotating on its axis.

If point P were attached to the Sun, then the Earth would have to rotate once per year. For example, if this were an experiment with the central force being provided by a rope between a central point and an object, then the object would have to rotate once per orbital revolution.

If you analyse this carefully, then you'll see that in this case you do not have a perfect central force. The force is applied to a single point on the surface, not to the centre of mass. The rope, therefore, also provides a torque in order to rotate the object.

10. Nov 18, 2018

### Staff: Mentor

Your friend is correct.

The question can be addressed using a laser ring gyroscope attached to the earth to measure the rotational velocity of the earth. If the rotational velocity were exactly $2\pi/86400 \: rad/s = 7.2722 \: 10^{-5}\: s^{-1}$ then if it were changed to 0 there would be no day night cycle. The actual measured value is $7.2921150 \: s^{-1}$. Not coincidentally these two values differ by about one part in 365.

Last edited: Nov 18, 2018
11. Nov 18, 2018

### A.T.

The force of the string is usually not acting on the center of mass of the object, so it can create torques, which force the object to spin at the same same rate it revolves around the center.

12. Nov 18, 2018

### A.T.

The red part is your misunderstanding. It should be "relative to inertial coordinate axes". In the image the body axes of the planet stay fixed relative to the axes of the inertial image axes, so the planet is not rotating.

13. Nov 18, 2018

### sophiecentaur

I'm not sure that I understand what the OP is trying to say (or his on-line friend). Rather than trying to unpick all that, I think it's best to state the accepted situation and then see if that first post fits - and it doesn't seem to.
Despite the fashion to discuss these things in terms of various different frames of reference, I think the frame of the Fixed Stars is by far the easiest one to discuss and most familiar with most people. In a period of a year, the Sun doesn't move significantly so let's call the Sun stationary and non rotating. To a first approximation, the Sun has no influence on our rotation. (Tidal locking with the Moon etc. is not relevant to this first stab at things)
If the Earth were in a circular orbit round the Sun but not rotating relative to the fixed stars, then one point on the Earth would be facing in the direction of one particular star all the time. We would go round the Sun once a year and our 'day' would last exactly one year too.
In fact, we do rotate - daily. We have 365 1/4 Noons (Sun overhead) each year and the fixed star we were using will be overhead one more time than the Sun, during our solar year and the sidereal day (time between our star being overhead) is about 4 minutes shorter. There's enough difference in the time scales of the Solar System and the Galaxy to allow this approximation to work well enough for starters.
There could be endless Q and A about this topic but, unless the OP actually asks the 'right' questions, he will not get the 'right' answers. I suggest the OP Googles the term Sidereal Day and gets familiar with what it all implies. This topic is not altogether intuitive and there are a number of YouTube videos that could help - take your pick.
PS The title of the thread seems to involve two concepts that are not related.

14. Nov 21, 2018

### src2206

Thank you all for your replies and insights. I have been extremely busy during the last few days, so could not reply in a timely manner, my apologies for that. I would surely go through the points raised here in detail and come back. Though one thing I must disagree, that my title consists of two different concepts.

15. Nov 22, 2018

### sophiecentaur

Translation and rotation are two different concepts - as are Momentum and Angular Momentum. I would say that common scientific usage of the word "rotation" (and especially in the context of astronomical motion) implies angular momentum around the axis of the object itself. If the term is not used exclusively for that, it would need to be qualified every time, to distinguish it from "orbit" or "trajectory" under an external force.
Can you find a significant number of examples where an orbit is described as rotation? (You may well find one or two but that goes for many examples of inappropriate use of words.)

16. Nov 22, 2018

### Mister T

The pebble will rotate once per revolution. If that pebble were a planet the length of the day and the length of the year would be the same.

There's an old riddle that goes something like this. You have two coins, one has a diameter of one inch and the other has a diameter of three inches. Both lie flat on a table top, touching. The smaller coin is rotated in such a way that it doesn't slip against the larger coin as it rolls around it. When the coin first returns to its original location how many times will it have rotated?

17. Nov 22, 2018

### A.T.

Yes
No, there would be no day-night-cycle in that situation (if the finger is the sun).

18. Nov 22, 2018

### Staff: Mentor

It would have no solar day, but it would have a sidereal day equal to its year.

19. Nov 22, 2018

### A.T.

Right, and in the diagram in the OP it would be the other way around.

Last edited: Nov 22, 2018
20. Dec 16, 2018

### src2206

My dear friends and fellow Physics enthusiasts

At the very beginning, I offer my unconditional apologies for such a delayed response. Unfortunately, I was too tied up to come back here and post a coherent reply to this discussion.

What I am looking for is a mathematical way to represent this mathematically in the equation of motion, the way we can mathematically show the Coriolis force while analyzing rotational motion. I understand that there is going to be a change in the location of a point on the periphery of Earth relative to the Sun. I am trying to figure out a way to mathematically show/express the "origin" of this "pseudo-rotation".

I hope I have made my point clear. Thank you again for all your support.

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