To my semi-educated brain, tidal forces seem to appear when there's a large difference in the gravitational acceleration of one end of an object compared to another or, as you use smaller objects, the differential of gravitational acceleration. Wikipedia says that the magnitude of gravtitional acceleration is [math]a_g=GMr^{-2} \therefore \frac{da}{dr}=\frac{-2GM}{r^3}[/math]

However, I've ran into a couple of problems with this. Mostly, as far as I can tell, it's measured in nonsensical units. (Annoyingly, I haven't been taught how to do units. I'm just trying to generalize from [imath]\frac{d}{dt}(1m)=1ms^{-1}[/imath]) Since a is measured in ms^-2, differentiating by r involves dividing the units by metres, so [imath]\frac{da}{dr}[/imath] is measured in [imath]s^{-2}[/imath].

The other problem is that Wikipedia's discussion of this produces an [imath]\Delta r[/imath] term, and I'm not following where it comes from. It seems to be important, since setting it to zero in their example gives an answer of zero.

Any help with this?

## Misunderstanding tidal forces

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### Misunderstanding tidal forces

Last edited by Robert'); DROP TABLE *; on Thu Dec 09, 2010 10:31 pm UTC, edited 1 time in total.

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### Re: Misunderstanding tidal forces

You are correct in how differentiation affects units. Remember that G is in units of m^3*kg^-1*s^-2. This will make the units work out.

The delta-R is the difference in distance between two points on the object. If delta-R=0, then the two points are actually the same point, and there is obviously no difference in acceleration.

The delta-R is the difference in distance between two points on the object. If delta-R=0, then the two points are actually the same point, and there is obviously no difference in acceleration.

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Last updated 6/29/108

Last updated 6/29/108

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### Re: Misunderstanding tidal forces

jmorgan3 wrote:You are correct in how differentiation affects units. Remember that G is in units of m^3*kg^-1*s^-2. This will make the units work out.

[math]m^3*kg^{-1}*s^{-2}*\frac{kg}{m^3}[/math]

[math]=\frac{m^3}{kg}*\frac{kg}{m^3}*\frac{1}{s^2}[/math]

[math]=\frac{1}{s^2}[/math]

I don't think inverse square-seconds corresponds to any physical quantity. (But thanks for the explanation of delta-R)

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### Re: Misunderstanding tidal forces

Hertz per second?

she/they

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### Re: Misunderstanding tidal forces

Robert'); DROP TABLE *; wrote:jmorgan3 wrote:You are correct in how differentiation affects units. Remember that G is in units of m^3*kg^-1*s^-2. This will make the units work out.

[math]m^3*kg^{-1}*s^{-2}*\frac{kg}{m^3}[/math]

[math]=\frac{m^3}{kg}*\frac{kg}{m^3}*\frac{1}{s^2}[/math]

[math]=\frac{1}{s^2}[/math]

I don't think inverse square-seconds corresponds to any physical quantity. (But thanks for the explanation of delta-R)

Yeah, that's correct units. And note that in the discussion, the [imath]\Delta r[/imath] multiplier to that will work it out to [imath]\frac{m}{s^2}[/imath] which is precisely what's needed to make it an acceleration.

The [imath]\Delta r[/imath] is because the distance between the two objects matters, more generally it's actually the distance parallel to the direction of gravitational force. THis makes sense when you think about it, a long rod with big [imath]\Delta r[/imath] is going to have strong tidal forces, the gravitational field is going to be very different at the two ends. Whereas a very short rod is going to have both ends be almost equally pulled by gravity, and this is reflected in the small [imath]\Delta r[/imath].

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### Re: Misunderstanding tidal forces

Just because the units have a certain form doesn’t necessarily tell you what the object does. Consider torque (N•m = kg(m^2/(s^2)) and work (J = m•N = kg(m^2)/(s^2), which have the same units but represent very different physical concepts.

In the present case, the units are exactly what you would expect them to be: “(change in) acceleration per meter”. Yes, it happens to come out as 1/(s^2) in SI base units, but don’t make the mistake of thinking there’s anything fundamental about our base units, nor anything insightful that comes from writing things in terms of them. If your base units included acceleration and distance, for example, then the tidal forces would be in perfectly logical units of acceleration per distance. But if you tried to calculate how quickly the rate of oscillation of a pendulum was changing, you’d get the same units, which wouldn’t mean much. In SI, of course, that would just be Hertz per second. You get the idea.

In the present case, the units are exactly what you would expect them to be: “(change in) acceleration per meter”. Yes, it happens to come out as 1/(s^2) in SI base units, but don’t make the mistake of thinking there’s anything fundamental about our base units, nor anything insightful that comes from writing things in terms of them. If your base units included acceleration and distance, for example, then the tidal forces would be in perfectly logical units of acceleration per distance. But if you tried to calculate how quickly the rate of oscillation of a pendulum was changing, you’d get the same units, which wouldn’t mean much. In SI, of course, that would just be Hertz per second. You get the idea.

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