So i'm building this planet and i want to know how the tides work in there!

Moon absolute info (comparing to our moon)
Mass 8.663x10²²kg (1.18)
Radius 1794.12km (1.03)
Density 3583kg/m³ (1.07)
Distance from planet 224734.26km (0.59)
Planet absolute info (comparing to Earth)
Mass 8960.4x10²¹kg (1.5)
Radius 7294.5km (1.15)
Density 5514kg/m³ (1)
Distance from sun 60,000,000km (0.4)
Star absolute info (comparing to Sun)
Mass 1x10^30kg (0.7)
Radius 0.6x10^9 (0.86)
Density 1,11x10³kg/m³kg/m³ (0.79)

I know the tides are influenced by the gravitational forces of the moon and star, but i simple don't know how to measure the height! I suppose they would be stronger than in Earth, but it's a different planet and star to compare.
How do i measure the tides in this planet?

  • $\begingroup$ It looks like it would be the same as earth, just larger and more freqent $\endgroup$
    – Madman
    Mar 30 at 17:51
  • $\begingroup$ As in the variables in play are exactly the same ones present on earth, they're just stronger $\endgroup$
    – Madman
    Mar 30 at 17:52
  • $\begingroup$ Yes, i thought the tides would be stronger, but i want to know how much stronger it is... i was searching about tides in Earth and the highest is near 12m in Bay of Fundy. But here en.wikipedia.org/wiki/Tidal_range it says the tidal range can be 0 to 16+m. Can i calculate an average tide for the planet? $\endgroup$ Mar 30 at 18:10
  • $\begingroup$ This might help manoa.hawaii.edu/exploringourfluidearth/physical/tides/… $\endgroup$
    – Madman
    Mar 30 at 18:21
  • $\begingroup$ The actual tides on Earth are only about 1.2m and that is from lowest to highest.. The effect is hugely amplified when deep water meets shallow coastal regions, especially if the coastline forms a funnel. $\endgroup$
    – PcMan
    Mar 30 at 18:33

The mass and radius of the proposed planet are such that the gravitational acceleration on the surface is quite similar to Earth's -- about 1.5 / 1.15 = 1.3 times as large. The larger the gravitational acceleration, the more force is needed to lift water; heavily handwaving because fluid mechanics, we can say that the same tidal force would produce a tidal range only 2/3 of the tidal range on Earth.

For the tidal force, only the mass of the star + distance to the star and the mass of the moon + distance to the moon count. Tidal force is proportional to the mass of the body producing it and inversely proportional with the cube of the distance from it.

Taking the Sun-Earth-Moon system as baseline:

  • In the Sun-Earth-Moon system tidal range is about 0.6 m (about 2 feet) in the ocean far from the shore; of this, about 1/3 (0.2 m or 2/3 foot) is due to the tidal force of the Sun, and about 2/3 (0.4 m or 1 1/3 foot) is due to the tidal force of the Moon.

  • In the proposed system, the star is only 70% of the mass of the Sun, but it is 2.5 times as close. The tidal force due to the star is then about 2.5^3 * 0.7 = 11 times as strong; which means that the tides due to the star will be about 11 * 2/3 = 8 times as large, that is 1.6 m (5 feet) instead of 0.2 m.

  • In the proposed system, the satellite is both more massive and closer than our Moon; so that the tidal force due to the satellite is about 1.2 / 0.6^3 = 5.6 times as strong as the tidal force due to Earth's Moon. This means that the tidal range due to the satellite will be about 5.6 * 0.4 * 2/3 = 1.5 m (5 feet) instead of 0.4 m.

Overall, in the middle of the ocean, far from the shore, the average tidal range will be about 3 m (10 feet) instead of about 0.6 m (2 feet), that is, about 5 times as large.

However, because the tidal components due to the star and to the satellite are about equal, the tide cycle will be much stronger than on Earth. On Earth, the difference between the smallest tides (neap tides) and the largerst tides (spring tides) is about 1 to 3. (Again, in the middle of the ocean). In the proposed system, when the satellite and the star are in quadrature their tidal forces annihilate each other almost completely, with the effect that neap tides will be barely perceptible, whereas the spring tides will be a whopping 6 meters (18 feet). In the middle of the ocean.

Tides on ocean shores are very much higher than in the middle of the ocean. On Earth, that's about let's say 2 m (7 feet) on the average, about 3 times the tidal range in the middle of the ocean; with a maximum tidal range of about 15 m (50 feet) in selected places, and close to zero in other places. (Tidal range on the shore depends very very very much on the particular geography of the particular point of interest.)

In the proposed system, tidal range on the ocean shore will be expected to be about 10 m (33 feet) on the average, in a cycle with about zero neap tides and about 20 m (66 feet) spring tides, with selected points going to a breathtaking 70 m (230 feet) spring tides! (And other points having no tides to speak of...)

This will have dramatic effects on the geography of ocean shores. Expect to have larger, more numerous and more extended tidal flood plains (similar to our Wadden Sea off the Dutch coast). Shores which are not solid rock will be smoother. Places with good natural sea ports will be fewer.

  • $\begingroup$ @PcMan: I already did -- second and third bullet points. It won't do to do it twice. $\endgroup$
    – AlexP
    Mar 30 at 18:38
  • $\begingroup$ @PcMan: And I already did it. Tidal force due to the star = 11x (second bullet point). Tidal range due to the star = 11x * 2/3 = 8x (same second bullet point). $\endgroup$
    – AlexP
    Mar 30 at 18:41
  • $\begingroup$ So the sailing in this world is ruined LOL At least it will be great to evolve water life to land... Thank you very much! $\endgroup$ Mar 30 at 19:37

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