How does the rotation of Earth affect the height of a tidal bulge?

In summary, the conversation discusses the gravitational field strength due to the moon at different points on Earth and how to calculate it. Different approaches are considered, including using the field strength and potential, but it is ultimately determined that considering the dynamics of the system and calculating the gravitational potential energy at different points gives the correct answer.
  • #1
etotheipi
Homework Statement
Derive expressions for the gravitational field due to the moon at the near and far sides of the Earth, and use these to estimate the height ##h## of the tidal bulge of water. Assume the Earth is covered with a layer of water 1000m deep.
Relevant Equations
##g = \frac{GM}{r^{2}}##
I'm having some conceptual difficulty with this; here's what (little!) I've done so far.

Suppose the distance between the centres of the Earth and the moon is ##x## and that the radius of the Earth is ##r##, and let the gravitational field strength due to the moon at the near side, far side and centre of the Earth be ##g_{a}##, ##g_{b}## and ##g_{c}## respectively. Then

##g_{a} = \frac{GM_{m}}{(x-r)^{2}} \approx \frac{GM_{m}}{x^{2}}(1+\frac{2r}{x}) = g_{c} + \frac{2GM_{m}r}{x^{3}}##

and

##g_{b} = \frac{GM_{m}}{(x+r)^{2}} \approx \frac{GM_{m}}{x^{2}}(1-\frac{2r}{x}) = g_{c} - \frac{2GM_{m}r}{x^{3}}##

I'm at a loss for what to do next. I considered transforming into a rotating frame centred on the moon and imagined a mass element ##dm## at each bulge which would have a fictitous ##dm(x-r)\omega^{2}## acting (assuming ##h\ll r##) in addition to ##dmg_{a}## and the weight due to the Earth at a height ##h## above the surface, which I obtained to be ##dmg_{h} = dm(g_{E} -\frac{2GM_{E}h}{r^{3}})##. However, when I did the force balance and substituted in the known constants (plus taking ##\omega = \frac{2\pi}{(28)(24)(60)(60)}##), I got an answer for ##h## which was many orders of magnitude greater than common sense would suggest.

I then thought to see if I could work out the effective gravitational potential energy at the surface of the bulge and equate this to that at the poles (taking the surface of the water to be an equipotential). This threw out, when assuming the depth at the poles is 1000m,

$$h(g_{E} - \frac{GM_{m}}{x^{2}}(1+\frac{2r}{x})) = 1000g_{E} \implies h = 1000.0005m$$

which is only ##0.0005## metres higher than at the poles. So this has to be wrong as well.

I was wondering if someone point me in the right direction, thanks!
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
It is not apparent to me how to solve this using the field strength. More obvious is to use potentials.

If you take it as static then you get that for a given distance from Earth's centre the GPE is least on the side facing the moon and greatest away from the moon, so that is clearly not going to give the right answer.
Allowing for the dynamics puts the least potential of all at Earth's centre, so I considered two parcels of surface water at the middle position (viewed moon on the horizon), then migrated one to the point nearest the moon and one the opposite way, taking both to end up at height x above where they started. It gave the right answer.
 
  • Informative
Likes etotheipi

Related to How does the rotation of Earth affect the height of a tidal bulge?

1. How does the rotation of Earth affect the height of a tidal bulge?

The rotation of Earth affects the height of a tidal bulge by creating a centrifugal force that causes water to bulge away from the center of rotation. This results in two high tides and two low tides per day.

2. Why do high tides occur on opposite sides of the Earth?

The high tides occur on opposite sides of the Earth due to the gravitational pull of the Moon and the Sun. As the Earth rotates, different parts of the planet experience these gravitational forces, resulting in two tidal bulges on opposite sides.

3. Does the rotation of Earth affect the timing of high and low tides?

Yes, the rotation of Earth does affect the timing of high and low tides. As the Earth rotates, different locations on the planet are exposed to the Moon and Sun's gravitational pull, causing tides to occur at different times.

4. How does the rotation of Earth affect the strength of tides?

The rotation of Earth does not directly affect the strength of tides. However, the tilt of Earth's axis and the shape of the ocean basins can affect the strength of tides in different locations.

5. Can the rotation of Earth cause changes in tidal patterns?

Yes, changes in the rotation of Earth can cause changes in tidal patterns. For example, changes in the Earth's rotational speed or axis tilt can affect the strength and timing of tides, resulting in different patterns over time.

Similar threads

  • Introductory Physics Homework Help
2
Replies
63
Views
2K
  • Introductory Physics Homework Help
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
241
  • Introductory Physics Homework Help
10
Replies
335
Views
8K
  • Introductory Physics Homework Help
Replies
19
Views
854
  • Introductory Physics Homework Help
Replies
13
Views
517
  • Introductory Physics Homework Help
Replies
9
Views
928
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
1K
Replies
2
Views
3K
Back
Top