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Comment by throwup238 | Hacker News Reader

It looks quite a bit bigger than Earth.

I don't think anyone will be leaving that atmosphere with chemical propulsion any time soon...

I learned something today :)

Gravity on surface of this planet will be 3(times heavier than earth)/(1.55(radius)^2), so 1.25g, however escape velocity also depends on planet radius so we get sqrt(1.55 * 1.25) = 1.4 earths escape velocity (so 15.7km/s). Assuming Isp specific impulse of engine of around 300s, let's compute m0/mf (initial mass of rocket / payload) = exp(vEsc/(Isp*g0)) (g0=9.8m/s^2 (on both planets)):

For earth, it's 45.

For this planet, it's 208.

So, the rocket (using our current chemical engine technology) needs 4.6 times more fuel than on earth. Much more difficult, but maybe possible...

edit: formatting

We also have to consider the thickness of the atmosphere. I presume a planet with 1.5x the radius of Earth and 3x the gravity would have an atmosphere that is far more thick and any rocket would have to contend with a significant amount of drag.

Plus, if it would take something like a Saturn V just to put the mass of a potato into low orbit I could easily envision an extraterrestrial civilization thinking it wasn’t worth the cost to try.

> I presume a planet with 1.5x the radius of Earth and 3x the gravity would have an atmosphere that is far more thick and any rocket would have to contend with a significant amount of drag.

I'm sure that's true holding all else the same, but all else isn't the same.

Maybe it's airless due to stronger stellar winds.

Actually, the atmospheric drag is not a big challenge. Yes, it is a consideration of course, hence the payload fairing etc, but much larger problem going to orbit is getting speed.

That's why it doesn't really help much to launch rockets from a balloon -- getting there is not a problem.

Just because it's bigger doesn't mean that the gravity is higher.

”Its mass is 3.02 Earths.”

Right, but you also have to account for density when calculating surface gravity.

TOI-715 b has a log g value of 5.0 +/- 0.2. Earth is 2.992.

Good to know. Thanks!

Correct, but it’s mass (weight) is 3x higher than earth’s, and since gravity is proportional to mass, you’d likely be much heavier there. If the radius of the planet was the same as earth’s, the gravity would be 3x higher, but if it’s bigger than earth, it might not be as bad. The formula is g= m/r^2 , where m is the mass of the planet and r is the radius of the planet.

Damn I'm gonna have to get serious about that diet before visiting :'(

r^2 is around 2.3 for the alleged size so it won't be too big a difference.

Yeah, but look at its period:

>>>ts mass is 3.02 Earths, it takes 19.3 days to complete one orbit of its star, and is 0.083 AU from its star. Its discovery was announced in 2023

https://i.imgur.com/ghLnFrH.png

What would gravity be on that thing, it doesnt say what its circadian rhythm is.. so how long is a day, if a "year" is 19 earth days (or is the period in 'day' local-solar-system-days?)

Orbital period is in Earth days. The same as when we say the orbital period of Venus is 225 days even though a Venetian day is longer than that.

It'd be pretty messy to instead define things in the database in object relative terms, particularly when you usually know things like the orbital period with significantly more accuracy than rotational velocity.

Well yeah, Venus is in our solar system.

Thanks, though - I wasn’t sure if we always use a local reference but then au is as local a unit as possible at these scales.

It's probably tidally locked, so the day will be the same as the year.

The gravity on the surface should be 3,02÷(1,55^(2)) times the gravity on earth. It is equal to ~1.257 so it's about 25% higher.

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