Calculating Tidal Range (Gravitation)

In summary: What you are asking for is the difference in tide height at the two points, not the difference in tidal height between the point nearest to the moon and the point farthest from the moon.
  • #1
phantomvommand
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Homework Statement
I am trying to recall as much of this question as I can. If some of the information I state below is physically false/impossible, do change the question as you see fit.
Relevant Equations
Newton's Laws of Gravitation
Effective Potential
Consider only the Earth-Moon system, where both the Earth and Moon are spheres. A horizontal line joins the centres of the Earth and Moon. Consider a point P that lies on the surface of the Earth. The line joining P and the centre of the Earth meets the horizontal line joining the centres of the Earth and Moon at an angle Θ.

(i) Assume the Earth and Moon orbit about the common centre of mass in circular motion. Find the angular frequency w.
I solved this by finding centre of mass, and equating centripetal force required with gravitational force of attraction.

(ii) Find the effective Potential at point P.
I googled and found that effective potential = L^2/2mr^2, but I am not sure how to substitute m and r in the case of 2 planets. m is apparently the reduced mass. Is r the radius of earth, or distance from Earth to centre of mass? I think the latter, but I'm not sure. Furthermore, in circular motion about the centre of mass, there is no radial component of motion. Would the effective potential then just simply be the total energy of the Earth?

(iii) Calculate the difference in tidal height between the point nearest to the moon, and the point farthest from the moon.
I am not sure how to do this at all. However, the answer to part (ii) is relevant.

You are welcome to introduce symbols for radius of earth, distance between Earth and moon, radius of moon, etc.

All help is appreciated. Thank you everyone!
 
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  • #2
I think you'll find this video interesting. Tides are more complicated than most people think.
 
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  • #3
phantomvommand said:
googled and found that effective potential = L^2/2mr^2, but I am not sure how to substitute m and r in the case of 2 planets. m is apparently the reduced mass. Is r the radius of earth, or distance from Earth to centre of mass?
I assume you want the effective potential in the rotating frame.
If you are not sure how to use the formula, it is not hard to work it out from first principles. Just write down the potential wrt each body and add them and the centrifugal potential.
https://web.njit.edu/~gary/321/Lecture17.html

For (iii), what do you think the relationship is between the potential function and the tidal height?
 
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  • #4
haruspex said:
I assume you want the effective potential in the rotating frame.
If you are not sure how to use the formula, it is not hard to work it out from first principles. Just write down the potential wrt each body and add them and the centrifugal potential.
https://web.njit.edu/~gary/321/Lecture17.html

For (iii), what do you think the relationship is between the potential function and the tidal height?
Thanks for the helpful reply! Is the relationship simply: Difference in effective potential energy = mgh?
 
  • #5
phantomvommand said:
Difference in effective potential energy = mgh?
The effective potential energy includes GPE already, so no need to bring it in again as mgh.
Are you familiar with the term equipotential surface?
 
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  • #6
haruspex said:
The effective potential energy includes GPE already, so no need to bring it in again as mgh.
Are you familiar with the term equipotential surface?
I don’t see how equipotential surfaces are used here.
 
  • #7
phantomvommand said:
I don’t see how equipotential surfaces are used here.
If you pour water into a bowl, what determines the shape of the surface?
 
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  • #8
haruspex said:
If you pour water into a bowl, what determines the shape of the surface?
I suppose the effective potential at both the point nearest and point farthest from the moon is the same. I think this solves the question.

however, in the derivation of the effective potential, I notice that the rotation of the Earth is ignored. To account for it, (in the case where the point in question is located on Earth) would there now be 2 “centrifugal potential energies”?
 
  • #9
phantomvommand said:
I suppose the effective potential at both the point nearest and point farthest from the moon is the same.
Why do you suppose that? What do the equations say?
phantomvommand said:
the rotation of the Earth is ignored.
The rotation of the Earth on its own axis will have the consequence for the effective potential at all points at the same latitude, so you can ignore it here.
 
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  • #10
haruspex said:
Why do you suppose that? What do the equations say?
If they were not equal, water would move towards the point of lower potential? So the point of maximum tide height must have the same potential as the point where tide height is minimum, otherwise water would move until such a situation is achieved.
 
  • #11
phantomvommand said:
If they were not equal, water would move towards the point of lower potential? So the point of maximum tide height must have the same potential as the point where tide height is minimum, otherwise water would move until such a situation is achieved.
Yes, the sea surface forms an equipotential surface.
I may have misunderstood your remark in post #8.
You are asked to "Calculate the difference in tidal height between the point nearest to the moon, and the point farthest from the moon."
I.e., the difference in distance to Earth's centre between a point of the equipotential nearest the moon and the point of the equipotential furthest from the moon.
I read your post #8 as saying there would be no difference.
 
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  • #12
haruspex said:
Yes, the sea surface forms an equipotential surface.
I may have misunderstood your remark in post #8.
You are asked to "Calculate the difference in tidal height between the point nearest to the moon, and the point farthest from the moon."
I.e., the difference in distance to Earth's centre between a point of the equipotential nearest the moon and the point of the equipotential furthest from the moon.
I read your post #8 as saying there would be no difference.
Thanks for the reply. So by equating the 2 potentials, i suppose a relationship between the radius of the 2 points can be found. I realized my question was erroneous :/ The point on Earth with the lowest tide, and that with the highest tide, is not the nearest and farthest from the moon. The 2 points in question are at 90 and 0 degrees respectively. Nonetheless, I suppose the same method should work.
 
  • #13
phantomvommand said:
Thanks for the reply. So by equating the 2 potentials, i suppose a relationship between the radius of the 2 points can be found. I realized my question was erroneous :/ The point on Earth with the lowest tide, and that with the highest tide, is not the nearest and farthest from the moon. The 2 points in question are at 90 and 0 degrees respectively. Nonetheless, I suppose the same method should work.
Yes.
 

Related to Calculating Tidal Range (Gravitation)

1. What is tidal range?

Tidal range refers to the vertical difference between the high tide and low tide levels in a body of water.

2. How is tidal range calculated?

Tidal range is calculated by taking the difference between the highest and lowest water levels recorded at a specific location over a certain period of time, typically a 24-hour period.

3. What factors affect tidal range?

The main factor that affects tidal range is the gravitational pull of the moon and the sun on the Earth's oceans. Other factors such as the shape and depth of the ocean basin, the Earth's rotation, and weather patterns can also have an impact on tidal range.

4. Why is calculating tidal range important?

Calculating tidal range is important for various reasons. It helps us understand the patterns and variations in tides, which is crucial for navigation and marine activities. It also plays a role in coastal engineering and predicting potential flooding events.

5. Can tidal range be predicted accurately?

While tidal range can be predicted to a certain extent, it is not an exact science. Factors such as weather patterns and local conditions can affect the accuracy of tidal range predictions. However, advancements in technology and data collection have improved the accuracy of tidal range predictions in recent years.

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