I would not spend money at this point. As others say, something like this is every couple years on average, I think the last good one was 2020. HOWEVER: It's already fading, and the nearly full moon is already swamping it out. By this weekend when you'll get maybe 2–3 hrs after sunset, it is projected to no longer be naked-eye visible. Wait 'til the next one and plan ahead based on projections for that next one.

As a very gratuitous side-note, I will add it's always nice to see myself cited in Wikipedia. That 1.3M crater number comes from my lunar crater database.

To clarify, and I've made this comment on the OP, Pluto and Charon have an expected amount of cratering for bodies at the edge of the Kuiper Belt that are several billion years old. Impact speeds are much slower there, so craters are smaller from the same-sized impactor in the inner solar system, and their gravitational cross-section is much smaller than other, larger bodies on which you'd see more impact craters.

Pluto and Charon have an expected amount of cratering for bodies at the edge of the Kuiper Belt that are several billion years old. Impact speeds are much slower there, so craters are smaller from the same-sized impactor in the inner solar system, and their gravitational cross-section is much smaller than other, larger bodies on which you'd see more impact craters.

Just because ChatGPT said something does not mean it's true. I was at a concert last night where the opening act said they asked ChatGPT how many times they had performed there and it said once. The venue had records of 3 times. I suggest finding any sort of research paper that suggests a Mars magma ocean and go from that as your starting point.

I should've written 1/r^3. It means that the amount of tidal force you get goes down as the cube of the distance from the body doing the tidal pulling. Gravity is a 1/r^2 force, by comparison. That means that as you go farther away from the sun, the tidal force goes down really fast. All things being otherwise equal, if Earth were 2 AU from the sun instead of 1 AU, the tidal force experienced by it from the sun would be 1/8 as much (as it is now, the sun's tides on Earth are about 45% as strong as the moon's).

Tidal forces are proportional to r^3, meaning that at the distance of Mars, changing its already large eccentricity would not have the same effect as the same eccentricity change it would if Mars were at Earth's orbital distance. Mars is also smaller than Earth, and tidal force is the difference in gravity at one side of an object versus another, so again, conceptually, you'd have to increase eccentricity a lot more than you would for a larger planet.