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Tidal Acceleration and planetary rotation

by AotrsCommander
Tags: acceleration, planetary, rotation, tidal
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AotrsCommander
#1
Jun7-14, 08:09 AM
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I'm struggling to get my head around tidal acceleration. I get what is IS, I know how to calculate it, but what I'm not sure is how to apply it in a way that's meaningful to me.

So. If we took a stationary Earth (i.e. non rotating, tide-locked) and have the moon magically appear in orbit and applying the moon's ≈1.1e-06N tidal acceleraton, how would I work out what effect that has on the rotation speed (and thus period) of the Earth?

For, example, how would I calculate how long it would take for the Earth to become tide-locked to the moon (because that'd be the stable end-point, yes?) Or how long a day would be some time period (say, 4 billion years) after the moon appeared?
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paisiello2
#2
Jun7-14, 09:35 AM
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Can you show how you derived that number for tidal acceleration?
AotrsCommander
#3
Jun7-14, 10:08 AM
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##a=GM\frac{d}{R^3}##

where a = tidal acceleration, G is the gravitational constant (6.67384e-11), d is the diameter of the distorted body (in this case Earth, 2*6371000m), M is the mass of the distorting body (in this case the moon, 7.347E+22kg) and R is the distance between the two bodies (384399000m).

(It is only a first-order equation, as I understand, but sufficent to my purposes.)

I calculated the numbers via spreadsheet, which gave me 1.09996E-06N

I also found this page, which gave consistent results (and in fact made me realise my spreadsheet have missed G being to the E-11...!)

(I could use the more complex equation on the linked page for greater accuracy - and may do later, but for the moment, first-order is enough for me to get a handle on the numbers.)

DrStupid
#4
Jun7-14, 10:36 AM
P: 476
Tidal Acceleration and planetary rotation

Quote Quote by AotrsCommander View Post
For, example, how would I calculate how long it would take for the Earth to become tide-locked to the moon (because that'd be the stable end-point, yes?)
That's impossible without additional information. You need the rate of energy dissipation due to the cyclic deformation of Earth. That mainly depends on offshore water and the geological structure of Earth. A rigid, perfectly elastic or superfluid Earth would never be tidal locked.
AotrsCommander
#5
Jun7-14, 10:46 AM
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Quote Quote by DrStupid View Post
That's impossible without additional information. You need the rate of energy dissipation due to the cyclic deformation of Earth. That mainly depends on offshore water and the geological structure of Earth. A rigid, perfectly elastic or superfluid Earth would never be tidal locked.
Ah. Right.

I'm guessing there's no easy way to make an estimate of that without some significant mathamatical (or computer) modelling, is there?

I'd ask if we know what the rate is for Earth, but I assume it would be different for a non-rotating Earth, yes? (During my reading around astrophysics these last few weeks, I saw something on tidal locking that said Earth would eventually become tide-locked to the moon, albeit on a timescale past the end of the life of the sun, which I why I thought it might be possible to estimate.)


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