# Calculate Height of Ocean Tides if Earth Had a Larger/Smaller Radius

For context, I'm coming at this question from the Worldbuilding stackexchange, where I've posted a related question. I'm not a physicist or scientist by training, so if I make any glaring errors, please do let me know...

# I want to understand how to calculate the theoretical height of the ocean tides on Earth if you vary the radius of the Earth.

For illustrative purposes let's use 4,000km (Earth A) and 8,000km (Earth B) (Earth's real radius is 6,371km). To keep things simple we can use the tides as caused by the sun and discount moon tides. The same principles would apply to the moon anyway.

I want to take into account that ocean tides experience friction, Earth has continents, etc. In other words, that Earth is a real planet, and not a perfect heavenly sphere. I don't mean we need to put all this into the equation, but what I do want to figure out a back-of-the-napkin calculation that is accurate enough to predict average tide height to within a reasonable range of error.

Edit: I know there are complex harmonics at play, so if the best that can be done without complex modeling are establish likely ranges like "10-25% less" or "more than a 10% decrease but less than a 50% decrease" then that would be reasonable I think!

Lastly...

I've found reference to the solar tide height of Earth in multiple locations. They seem to contradict themselves to some extent. The two most common numbers I'm seeing are:

• Principal solar semidiurnal = 179.05mm (0.179m)
• Theoretical solar tidal amplitude = 250mm (0.25m)

I don't know if these are accurate, but I've listed them in case they prove useful. Perhaps someone can even explain why—if they both are accurate—they differ.

## How I'm Approaching This

Basic Calculation

I believe the basic calculation of RELATIVE tidal force on Earth A and Earth B is easy to compute. It would be:

T = M/(d^3)

Where M = mass of sun & d = distance.

Considering the mass of sun and earth's orbital distance don't vary in this scenario, the tidal force would be the same for Earth A and Earth B. However that's not the full story. This is the tidal force as measured from the center of Earth. But based on my understanding of tides, it's the differential in tidal force between sides of the planet that causes the tides, not just the absolute tidal force. So if you vary the distance between sides of Earth (through changing the radius) the size of the tides must necessarily change.

By by how much? What is the principle at play and how do you calculate the new tide heights?

There's a second wrench I can throw into the works here...

Does the change in tidal force = the change in tide height?

That is, as experienced in a real planet like Earth. My intuition is that factors like friction and even the time between tides would effect the actual realized tidal height. I don't have any hard numbers or models to back that intuition up though.

Thanks!

• Read physics.stackexchange.com/a/121858/520, first. Then realize that because there is a resonant system involved there are no easy answers, any more than whatever nominal global value you find can be easily used to predict the local tides at any given place. – dmckee Mar 4 '18 at 20:17
• see The Oceanography of Tides The depths of high and low tides are given by equations I.44, 45. – sammy gerbil Mar 4 '18 at 21:36
• Those equation (which desperately need some typesetting help!) are under highly unrealistic assumptions. They are exactly the sort of thing that doesn't reliably predict real tidal variation, and the authors say as much just a few pages later on. – dmckee Mar 4 '18 at 21:53
• @dmckee Do you think there is a "back of the napkin" method to determine a reasonable range for the increase/decrease in average tidal height? If you shrink Earth's radius to 4k km the corresponding tide height decrease is not going to be 99% nor 1%. Those numbers are obviously unreasonable. So they would be in between. It just seems to me there should be some relatively straightforward method to establish a reasonable upper-bound and lower-bound to the range. Maybe I'm mistaken? – n_bandit Mar 4 '18 at 22:04
• The system is resonant AKA non-linear. That means that you can ball-park it several different ways, but there are no guarantees. Look at the top map in the link I gave you: on Earth today you can find places where the ocean tidal amplitudes are less than $10\,\mathrm{cm}$ and other places where they are upwards of $130\,\mathrm{cm}$. That's more than a full order of magnitude. Your small globe has a weaker forcing function (which argues for smaller tides), but also shorter runs between extremes of the forcing function (which argues for larger). – dmckee Mar 4 '18 at 22:33