So, we have an imaginary super-earth planet twice the mass of earth, but with a density and volume meaning experiences only 1.05 g surface gravity. This planet has a single Luna-mass moon that orbits it at a distance of 395,000 kilometres every 35 days. The planet rotates every 23 hours, 48 minutes. Year length is 244 days.
What would the tides on this planet be like? That is, how long would its high/low tides be, and would they be, on average, stronger or weaker than those on earth?
First, calculate how much bigger the radius of this Super-Earth would be. The tidal force scales linearly with that radius. On top of that, the tidal bulge height is proportional to tidal force and Earth radius. The stronger the Earth gravity though, the smaller the bulge.
In total, $tideAmplitude = \frac{constant*radius^2}{surfaceGravity}$.
I think the result is that the amplitude is about two times bigger.
You can find references to the calculations here:
https://physics.stackexchange.com/questions/269038/height-of-tidal-bulge-reference-needed
The timing would be similar because the rotation speed is similar.
The equation for acceleration due to gravity is:
$g = G M_p / d^2$
$g$ = acceleration due to gravity
$G$ = Newton's Gravitational Constant
$M_p$ = Mass of the planet
$d$ = distance between the centers of the objects
This can be rearranged and solved to give your planet a radius of 8800km. This is approximately $\sqrt[]2$ times the Earth's radius, which makes sense because if the surface gravity was exactly 1g, then the radius would be exactly $\sqrt[]2$ times the Earth radius. Now that we have that, we can calculate the tidal forces.
Tidal forces have an x-component and a y-component, both of which depend on location. Since you are interested in the tides, only the x-component matters. According to this, we can approximate that as:
$\Delta F_x = (2 G M_p m_m R/d^3)cos\theta$
$\Delta F_x$ = x-component of the tidal force
$M_m$ = Mass of the moon
$R$ = Radius of the planet
If we are on the equator, then we can drop the $cos\theta$. Shove this equation and the values you gave into Wolfram Alpha and we get 1.67 * 10^19 Newtons. For comparison, the same calculation for Earth give 7.02 * 10^18 Newtons. Now, I want to say that because the former number is over 2 times the size as the latter number, the tides on your planet are twice the size as the tides on Earth. Unfortunately, NOAA gives evidence to the contrary:
The tidal range of a particular location is dependent less on its position north/south of the equator than on other physical factors in the area; topography, water depth, shoreline configuration, size of the ocean basin, and others.
Remember that $cos\theta$ in the second equation that we ignored. Well, that is the part of the equation that deals with how north/south of the equator you are. By saying how north/south a location is has less impact on tidal range than other factors, the NOAA is saying that the strength of the tidal force is less important than other factors.
In other words, even though tides are caused by tidal forces, the strength of those tidal forces are not the most important factor in determining tidal range. A huge piece of evidence for this that many places with the largest tidal ranges are far from the Equator, with the Bay of Fundy, Canada have the Earth's largest.
