# How do you derive the tidal force equation?

• student654321
In summary, the tidal force equation is a derivative of Newton's Law of Gravitation, and it's used to calculate the difference in force between two points on Earth. The approximation used is that the tidal force is proportional to the distance between the points. To get the Roche limit, you need to approximate the tidal force with two spheres being pulled apart.
student654321
I've been trying to derive the Roche Limit and have been successful for the most part but out of curiosity i was wondering how the tidal force equation was derived to be: -2GMmr/d^3 ??

Also on a related note i always get the constant at the front of the Roche Limit equation to be the cube root of two when apparently i should be getting two times this = 2.52, How is this constant found?

(i gather Roche gets 2.44 due to spheroid deformation).

Can anyone help with either of these problems.

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The tidal force equation you give is just given by the derivative of Newton's Law of Gravitation:

$$F=\frac{GMm}{d^2}$$

$$\Delta F=-\frac{2GMm}{d^3}\Delta r$$

where $$\Delta F$$ is the tidal force and $$\Delta r$$ is the radius of the object. The reason we care about the differential force is that we want to know how the outer edge of the object is being pulled relative to its center of mass. If they were both being pulled by equal forces, there would be no deformation. The above equation is only an approximation, by the way. It won't be valid unless $$\Delta r << d$$.

As for the Roche limit, the derivations I've seen that get 2.52 usually make some sort of approximation like putting two equal-radius spheres next to one another and finding the distance at which they need to be from a massive object to have the tidal force pull them apart. My impression is that it's a fairly arbitrary approximation, but I guess it gives values closer to the rigorous treatment of spherical deformation.

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Ah yes thankyou, i knew it was the derivative but didn't think to put the delta-r on the RHS.. i should have spotted this. I probably didn't because i was looking at websites that were taking the force at the front (d-r) of the object minus the force at the rear (d+r) and using this to calculate the differential force. I don't know if this is correct because there appeared to be some mistakes in their algebra with minus signs being misplaced etc, but would this account for the other terms that are missed out in this approximation?

To get 2.52 i did think that i would have to assume taking the whole diameter of the object but there would be problems with that. What you say about assuming two spheres being pulled apart makes sense more when picturing what's actually happening.

Thanks muchly!

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In My Resnik-Krane book, i was doing a problem in which They told me to CALCULATE THE DIFFERENCE IN FORCE DUE TO MOON ON A POINT ON ONE EXTREME OF THE EARTH AND FORCE AT A WATER DROP AT THE CENTRE(BUT OUTSIDE OF EARTH) OF EARTH. I calculated the expression and they told me that this was called the tidal force and is reponsible for this phenomena,you also try it , it will give you an exact expression.

student654321 said:
I've been trying to derive the Roche Limit and have been successful for the most part but out of curiosity i was wondering how the tidal force equation was derived to be: -2GMmr/d^3 ??

The tidal force is the difference between the gravitational forces at 2
different distances:

$$F_{tidal} =\frac{GMm}{d^2}-\frac{GMm}{\left( {d+\Delta r} \right)^2}$$

Find a common denominator

$$F_{tidal} =\frac{GMm\left( {d+\Delta r} \right)^2-GMmd^2}{d^2\left( {d+\Delta r} \right)^2}$$

FOIL the $$(d+\Delta r)^2$$ 's

$$F_{tidal} =\frac{GMm\left( {d^2+2\Delta rd+\Delta r^2} \right)-GMmd^2}{d^2\left( {d^2+2\Delta rd+\Delta r^2} \right)}$$

Factor out $$GMm$$

$$F_{tidal} =GMm\frac{d^2+2\Delta rd+\Delta r^2-d^2}{d^4+2\Delta rd^3+d^2\Delta r^2}$$

$$d^2$$ and $$-d^2$$ cancel each other

$$F_{tidal} =GMm\frac{\rlap{--} {d}^{\rlap{--} {2}}+2\Delta rd+\Delta r^2-\rlap{--} {d}^{\rlap{--} {2}}}{d^4+2\Delta rd^3+d^2\Delta r^2}$$

$$F_{tidal} =GMm\frac{2\Delta rd+\Delta r^2}{d^4+2\Delta rd^3+d^2\Delta r^2}$$

Eliminate that which is not significant. Take advantage that d >> r

$$F_{tidal} =GMm\frac{2\Delta rd+\rlap{--} {\Delta }\rlap{--} {r}^{\rlap{--} {2}}}{d^4+\rlap{--} {2}\rlap{--} {\Delta }\rlap{--} {r}\rlap{--} {d}^{\rlap{--} {3}}+\rlap{--} {d}^{\rlap{--} {2}}\rlap{--} {\Delta }\rlap{--} {r}^{\rlap{--} {2}}}$$

$$F_{tidal} =GMm\frac{2\Delta rd}{d^4}$$

Cancel the d the numerator with one of the d's in the denominator

$$F_{tidal} =GMm\frac{2\Delta r\rlap{--} {d}}{d^{\rlap{--} {4}3}}$$

$$F_{tidal} =GMm\frac{2\Delta r}{d^3}$$

Rearrange it to look like the formula in your post.

$$F_{tidal} =\frac{2GMmr}{d^3}$$

I don't get the negative sign that you get.

tony873004 said:
Rearrange it to look like the formula in your post.

$$F_{tidal} =\frac{2GMmr}{d^3}$$

I don't get the negative sign that you get.

It's a matter of the coordinate system you work in. If $$\Delta r$$ is relative to the body imposing the force, then there's a negative. If it's relative to the forced body, it's positive. In general, it's easier to get these results by doing a first-order taylor expansion (effectively what I did) instead of deriving the full formula and taking the limit (effectively what you did). It's not wrong, just more tedious.

SpaceTiger said:
It's a matter of the coordinate system you work in. If $$\Delta r$$ is relative to the body imposing the force, then there's a negative. If it's relative to the forced body, it's positive. In general, it's easier to get these results by doing a first-order taylor expansion (effectively what I did) instead of deriving the full formula and taking the limit (effectively what you did). It's not wrong, just more tedious.
I wish you'd read the homework help forums too. This is almost identical to the question I had asked last week. It took me hours to fail to figure this out. I only got it after visiting the teacher during office hours. That's why I was anxious to answer this thread with my new-found knowledge I think Doc Al was pointing me in the same direction you just described, but I'm no good at Calculus. Thanks ST.

tony873004 said:
I wish you'd read the homework help forums too. This is almost identical to the question I had asked last week. It took me hours to fail to figure this out. I only got it after visiting the teacher during office hours. That's why I was anxious to answer this thread with my new-found knowledge I think Doc Al was pointing me in the same direction you just described, but I'm no good at Calculus.

Oops, sorry 'bout that.

I was preparing a talk last week and was barely able to post at all.

Hi
I'm doing an eksamen exercise for tomorrow about the science fiction book "The Inetgral Trees" by Larry Niven and tidal forces. One of the quiestions are if it is realistic with a tidal force at 1/5g (little italic g - g-force, not gram) at one of the trees (it is big trees looking like integral symbols from another solar system http://en.wikipedia.org/wiki/Integral_Trees) and I really don't understand how to do. So I hoped maybe some of you know it. It would be such a big help. And it's urgent

And sorry for my bad english. I'm danish

## 1. What is the tidal force equation?

The tidal force equation is a mathematical formula that describes the gravitational force exerted by one astronomical body, such as a planet or moon, on another body due to their gravitational attraction. It takes into account the masses and distances of the two bodies to calculate the magnitude of the force.

## 2. How do you derive the tidal force equation?

The tidal force equation can be derived using Newton's law of universal gravitation, which states that the force between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. By applying this law to two interacting bodies, such as a planet and its moon, the tidal force equation can be derived.

## 3. What factors are involved in the tidal force equation?

The tidal force equation takes into account the masses of the two bodies, the distance between them, and the gravitational constant. It also considers the orientation of the bodies and the direction of the force. Other factors, such as the rotation and shape of the bodies, may also be taken into account in more complex versions of the equation.

## 4. How is the tidal force equation used in science?

The tidal force equation is used in various fields of science, including astronomy, oceanography, and geology. It helps scientists understand and predict the gravitational effects of celestial bodies on Earth and other planets, as well as the tides in the Earth's oceans. It is also used to study the formation and evolution of planetary systems and the dynamics of celestial objects.

## 5. Are there any limitations to the tidal force equation?

Like any mathematical equation, the tidal force equation has its limitations. It assumes that the bodies involved are point masses and does not take into account the effects of other forces, such as atmospheric drag or electromagnetic forces. It also becomes less accurate for bodies with irregular shapes or when they are very close to each other. However, it remains a valuable tool for understanding the dynamics of celestial bodies and their interactions.

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