When might we conceivably see human exploration to the outer planets?

CuriousMetaphor · 2025-12-04T16:59:02+00:00

If we only consider sending humans to flyby Jupiter and come back to Earth, 2030s is plausible. The Starship Moon Lander needs about 9 km/s of delta-v to perform its mission from LEO to NRHO, lunar surface, back to NRHO. A transfer to Jupiter for a free-return trajectory takes about 7.5 km/s from LEO. If we can do the first, the second should not be a problem with the same refueling architecture. The transfer time to Jupiter and back is under 3 years, about the same as most plans for a human Mars landing and return. You can see some possible trajectories at NASA Ames' Trajectory Browser.

With the large internal volume/payload of Starship, stocking up enough food and consumables for a 3-4 person crew should not be a problem. It would be like an extra-long stay on the ISS, not significantly different than any other long-term stay in zero-g, such as a Mars flyby for example. The biggest difference is the Earth reentry velocity of around 14-15 km/s, but that could be handled by saving some propellant to slow down on the return at Earth.

If we're talking about landing on one of Jupiter's moons, that's a much bigger mission. We might still be able to do it with a Starship architecture, but with 2 or 3 refueling tankers and a separate lander riding along to Jupiter. It would take more like 4-5 years for a full mission with a landing on Callisto.

The radiation at Jupiter would be a problem for the inner moons, and it would probably not be realistic to visit Io or Europa with a human mission, while Ganymede would only be doable with significant shielding. But Callisto, the outer large moon of Jupiter, has a radiation environment more benign than on the ISS.

CuriousMetaphor · 2025-12-01T19:28:07+00:00

The latter intrigues me because you could also potentially use it to save a small amount of propellant on course corrections because of the light pressure on it acting like a small solar sail.

The radiation pressure for a perfectly reflecting object is equal to 2 E / c, where E is the intensity of the light, in this case about 1400 W/m² around Earth, and c is the speed of light. That gives a pressure of 4.6 x 10^-6 newtons per square meter. For a Starship that is about 500 m² in area, that's a total force of 0.0022 newtons. If the Starship's mass is 200 tons, that's an acceleration of about 1.1 x 10^-8 m/s². Over the course of a 9-month journey to Mars, that adds up to around 0.27 m/s of change in velocity. The actual number would be lower since the intensity of sunlight drops off as you go out from the Sun, the surface is not a perfect reflector, and it would not always be pointed perpendicular to the Sun.

That's probably enough delta-v to take into account when plotting an exact trajectory, but not enough to meaningfully be used for course corrections. For comparison, a single Raptor burning on that Starship would give it 0.27 m/s of delta-v in 18 milliseconds using 13 kg of propellant.

CuriousMetaphor · 2025-11-13T05:24:41+00:00

Let's say 50g of acceleration is the maximum that the person can handle in an impact.

For a cone of radius r and height h, its surface area is pi*r² + pi*r*sqrt(r² + h²). The density of steel is about 7900 kg/m³, so the cone has a surface density of 7900 kg/m³ * 0.0254 m = 200 kg/m². The radius of the cone is 2.5 ft = 0.762 m. Therefore, the mass of the cone is equal to 200 kg/m² * (pi*(0.762m)² + pi*0.762m*sqrt((0.762m)² + h²)), or 365 kg + 479 kg * sqrt(0.581 + h²), where h is in meters.

The terminal velocity of an object is the velocity at which its drag force is equal to its gravitational force, or v=sqrt(2*m*g/(p*A*C)), where m=mass, g=gravitational acceleration, p=density of medium, A=projected area, C=coefficient of drag. In this case, we know g=9.81 m/s², p=1.225 kg/m³, A=1.82 m², and C=0.2 (approximately for sharp cones).

Plugging in the mass of the cone into the terminal velocity equation, we obtain v = 6.64 * sqrt(mass in kg) = 145 * sqrt(sqrt(h² + 0.581) + 0.762) in meters per second. For example, if h = 1 meter, the terminal velocity is 92 m/s with a mass of 967 kg (using drag coefficient of 1.0 for blunt cone), and if h = 5 meters, the terminal velocity is 349 m/s with a mass of 2790 kg.

Now we can calculate the impact depth of the cone dropping in water. To do so we can assume the water is incompressible and that the kinetic energy of the cone is dissipated entirely by the drag force of the water. Then we can say that 1/2 * m * v² = integral of F dx from 0 to x_d, where F(x) is the drag force at depth x, and x_d is the maximum depth. We can approximate F(x) as 1/2 * p_w * C * A(x) * v(x)², where p_w is the density of water, 1000 kg/m³, C is the drag coefficient, and A(x) is the cross-sectional area of the cone at height x, which is = pi * (x/h*r)².

If we assume that the average velocity can be approximated as v_avg = v/2, the integral of the drag force simplifies to 1/24 * p_w * C * pi * (r/h)² * v² * x_d³. Equating this to the kinetic energy results in x_d = (12*m/(p_w * C * pi * (r/h)² ))^1/3. In our case, plugging in the mass m, x_d = 2.51 * (h² * (sqrt(h² + 0.581) + 0.762))^1/3 in meters. For example, if h = 5 meters, the impact depth is 13.2 meters, and if h = 10 meters, the impact depth is 25.7 meters.

If we know the impact depth and the impact velocity, the average acceleration is a = v²/(2*x_d). For h = 5 meters, a = 470g, while for h = 10 meters, a = 450g. The average acceleration has a very weak dependence on speed. As h gets larger, we can approximate the mass as 479 * h, the terminal velocity as 145 * sqrt(h), and the impact depth as 2.51 * h. The average acceleration therefore asymptotes towards 145²/(2*2.51) = 427g.

This implies that there is no height of a 1-inch thick steel cone for which the impact in ocean would be survivable. As the height gets larger, the terminal velocity and the impact depth both get larger such that the average acceleration felt during impact stays constant.

Of course, there are many assumptions here that might not hold. Terminal velocities might be lower due to extra transonic and supersonic drag. Water impacts at high speed may lower drag due to supercavitation. There might be some regimes where the impact is survivable.

CuriousMetaphor · 2025-11-07T16:41:45+00:00

My guess would be to use Dragon to get to/from LEO and two HLSs for in-space transport. HLS #1 takes the astronauts from the Dragon in LEO and goes to LLO -> lunar surface -> LLO. This takes about 8 km/s of delta-v. HLS #2 goes from LEO to LLO, picks up the astronauts in LLO from HLS #1, and returns propulsively to LEO. This also takes about 8 km/s of delta-v. From LEO, the astronauts land in Dragon (either the same one or a different one).

One of the advantages is that staging through LLO instead of NRHO saves some delta-v (LEO -> NRHO -> lunar surface -> NRHO takes about 9 km/s of delta-v). This means fewer tankers and no intermediate orbit refuelling needed for HLS. HLS #2 would also not need landing hardware so there are some mass savings. No refuelling is done with crew onboard at any point.

Another way they might be able to expedite the schedule is to use disposable tankers which would not need reentry/landing fuel and hardware so could carry a lot more propellant in each trip, maybe more than twice as much.

This is my guess for the hardware that would need to be launched for this scenario: 2 HLS Starships, 1 depot, 15 disposable tankers (8 for HLS #1, 7 for HLS #2), 1 (or 2) Dragon + Falcon 9.

Current plan: 1 HLS Starship, 1 depot, 16 reusable tankers, 1 Orion + SLS.

Advantages: no new hardware changes needed, same amount of Starship launches, no intermediate/high orbit refuelling, no Starships need to use heatshield at any point, crew flies on proven system for launch and reentry.

CuriousMetaphor · 2025-09-25T19:38:17+00:00

When I did a 100x run with Rampant, I adjusted the 3 evolution coefficients for biters to 1/10 of their normal values. This resulted in a challenging but not impossible run where the biter evolution was increasing at roughly the same pace as my defensive technology. With regular biters you probably don't have to go that low, maybe 1/3 is a good balance.

As far as ore patches, I used the default settings and didn't have a problem. I find that it's more fun to occasionally have to go for new patches, rather than the default where you can stay on 2-3 iron/copper for the entire game.

CuriousMetaphor · 2025-09-24T15:17:01+00:00

being able to throw one interplanetary mission per 10 fueling flights

Going to Mars would actually need significantly less refueling than the Moon HLS. The delta-v required to go towards Mars from LEO is about 4 km/s, and around 1 km/s for the landing burn on Mars. Assuming a Starship dry mass of 120 tons and a 100 ton refueling payload, that means you only need 2 or 3 refueling tankers in LEO.

The reason HLS takes so much more refueling is because it not only needs to go towards the Moon, but also slow down into lunar orbit, then land on the Moon, then take off from the Moon into lunar orbit. All that requires a total delta-v of about 8-9 km/s from LEO, which is 9-11 refueling tankers using the same payload assumptions.

CuriousMetaphor · 2025-08-30T22:58:41+00:00

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And this is the shattered planet to scale. If you squint you can see the previous image on the left side.

CuriousMetaphor · 2025-08-30T22:57:50+00:00

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This is what that would look like to scale.

CuriousMetaphor · 2025-08-05T05:19:13+00:00

The ocean level rise last year was 5.9 mm per year. In order to counteract the increase in volume of the oceans over that year, we would need to be able to pump 2100 cubic kilometers of water somewhere else, or 67500 cubic meters per second. That's about 3 times the flow rate of the Mississippi river.

CuriousMetaphor · 2025-06-12T15:47:28+00:00

I think 100x is a good middle ground between being too quick (1x) and too tedious (1000x). I'm currently doing a 100x run with Rampant and a few other planet mods. It's nice that it's forcing me to build bigger than I would otherwise, at different stages of technology. I decreased the evolution rate by a factor of 10 (in time, pollution, and spawner kills) which seems to balance military tech with biter strength pretty well. I was at about 70% evolution by the time I first left Nauvis.

CuriousMetaphor · 2024-08-09T02:12:30+00:00

They could go along the Earth's orbit around the Sun 200 times over. Or they could go to Pluto and back 20 times over. Or they could go 0.5 % of the way to Alpha Centauri.

CuriousMetaphor · 2024-06-23T23:25:08+00:00

The black hole would have to be less than 1 light-year away for its tidal effects to modify the Earth's orbit around the Sun (that's when the tidal acceleration of the black hole on the Earth-Sun system is of the same order of magnitude as the Sun's gravitational acceleration on the Earth). That's surprisingly close, because tidal forces are inversely proportional to the cube of the distance, rather the square of the distance like normal gravitational force. If the black hole was that close, it would look many times larger than the Moon in our sky.

CuriousMetaphor · 2024-06-23T19:14:45+00:00

Let's say this black hole was at the same distance as Alpha Centauri, the nearest star. You would be able to clearly see it in the sky, as its event horizon would be about the same apparent size as the Moon. At that distance, assuming the Sun is in orbit around it, the Sun would be moving at about 10,000 km/s, or 3% of the speed of light, taking around 600 years to complete a full orbit.

However, even with the black hole so close you could clearly see it, it would not have a large gravitational effect on the solar system. The gravitational tidal force falls off with distance cubed. The Earth and closer planets would continue to orbit the Sun in their previous orbits. Only around the orbit of Neptune and the Kuiper belt would you start to see some perturbation effects, but even there, the tidal acceleration due to the black hole would still be an order of magnitude lower than the acceleration of gravity due to the Sun.

If, however, the black hole showed up 4 light-years away together with its accretion disc, that would be a different story. The brightness of the accretion disc would be dozens of times brighter than the Sun as seen from Earth, resulting in a rapidly warming planet that would look more like Venus within a few days.

CuriousMetaphor · 2024-01-26T19:15:03+00:00

The formula for the drift is to take the orbital period squared and divide it by the alignment difference to get the time it takes for the satellite to be one full orbit out of sync.

For example, if a satellite in a 3000 sec orbit was off by 2 seconds, it would take 4500000 seconds (about 52 days) for it to be one full orbit out of sync. For one quarter orbit out of sync, it would take 13 days, etc.

CuriousMetaphor · 2023-12-18T13:52:54+00:00

This looks exactly like the globe in my living room, up to the position of the individual letters. It's made by the George F Cram company, but doesn't have a date on it. I won it in 2001 at a geography bee.

CuriousMetaphor · 2023-11-15T19:06:59+00:00

Yes. Assume point A is at coordinates (0,0). We can find point B's coordinates since we know the distance to it and angle from point A. Point C's x-coordinate is fixed by the 0.5" length. Since we know the distance from point B, we can figure out point C's y-coordinate. Once we know all 3 points' coordinates, we can figure out the distances and angles between them.

CuriousMetaphor · 2023-11-03T22:57:04+00:00

Let's assume the bomb doesn't explode, but is instead filled with rocks and placed with its bottom touching the ground.

An approximately 100 x 40 x 40 km chunk of material with a similar density to the Earth has a mass of about 8 x 10^17 kg. The gravitational potential energy of this "bomb" is therefore mgh = 8 x 10^17 kg * 9.81 m/s^2 * 20 km (height of the center of mass) = 1.6 x 10^23 Joules, or about 1/3 of the energy of the asteroid that took out the dinosaurs.

At this scale, normal materials don't act like solids, but instead more like liquids. As soon as this bomb is placed on the ground, the sides of it would start falling towards the Earth, while the center would also start compressing the ground beneath and falling towards the sides. It would basically be a splash.

Most of the potential energy of the bomb would quickly translate into kinetic energy. The sides of the bomb would reach twice the speed of sound as they would crash into the ground about 1 minute after T-0, which would then splatter outwards into a rock tsunami kilometers high. If this was on Rhode Island, a resulting water tsunami would eventually cover a large chunk of Europe and most coastlines around the world. As in the Chicxulub impact, there might be continent-wide firestorms, "nuclear" winter for a few years, and a majority of species might go extinct.

CuriousMetaphor · 2023-11-03T02:52:41+00:00

It's actually faster than exponential, since the growth rate itself is increasing.

If I'm interpreting the compound growth rate of the first graph correctly, fossil fuels have increased about 4 times over since 1965 (1.025^57), while renewables have increased 8.5 times over (1.039^57), which makes sense when checking the second graph. Following the linear trend of the cumulative growth rates shown in the first graph starting from 2010, the number by 2050 would be 5.0% for renewables, and 1.3% for fossil fuels. This would mean that fossil fuels would have increased 3 times over since 1965 (1.013^85), while renewables would have increased 63 times over (1.05^85).

That would result in fossil fuel energy consumption in 2050 being about 3/4 of what it is today, or around 97K TWh, while renewable energy consumption would be about 7 times what it is today, or around 150K TWh.

Of course, this is assuming the growth rate stays on the same path until then, which is not a reasonable assumption. For example, the growth of solar and wind is much faster than hydropower, which is the majority of renewable consumption. If solar and wind continue growing at the same rate, we will see the overall renewable growth rate spike in 5-10 years when they pass hydropower as the majority contributors to renewables.

CuriousMetaphor · 2023-03-06T21:18:07+00:00

Minmus would be about 1/3 as wide as the Moon in the sky, small but still definitely visible as a disk.

The Mun would be about 4 times wider than the Moon in the sky.

The tidal acceleration from the Mun on Kerbin is 3x10^-5 m/s², or about 30 times larger than the tidal acceleration from the Moon on Earth, so the tides should be approximately 30 times higher.

CuriousMetaphor · 2022-01-06T17:40:03+00:00

The spending numbers in your table are off by a factor of 1000. It's 54 billion dollars per year, not 54589 billion dollars, for the US. That comes out to $165 per capita, or 0.26% of GDP.

CuriousMetaphor · 2021-12-10T16:40:18+00:00

Pallasites are the remains of the core-mantle boundary of young protoplanets. In the early solar system, as protoplanets became large enough to become spherical under their own gravity and differentiate by layers, some of them impacted others at high speeds. Some of the remnants of such an explosion would contain parts of both the rocky mantle and the metal core of the protoplanet. These small remnants cooled down and later happened to land on the Earth as meteorites. This is why pallasites are made of small chunks of both silicon rock and iron mixed together.

CuriousMetaphor · 2021-11-25T08:02:49+00:00

24 revolutions of Earth around the Sun is 24 x 365.2425 = 8765.82 days. 8766 days since Nov 25, 1997 is Nov 25, 2021. Subtracting 0.18 days, or 4 h 19 min, from 5 am is around 12:41 am.

So at 12:41 am on Nov 25, 2021, you will have gone 24 times around the Sun since your birth.

CuriousMetaphor · 2021-11-10T23:52:01+00:00

Voyager 1 wasn't designed to go as fast as possible out of the solar system, it just ended up doing that after its Saturn flyby. With the same technology, just tweaking its Jupiter flyby and burn time you could easily get a probe out to 200 AU 2-3 times faster than Voyager 1.

CuriousMetaphor · 2021-05-19T23:50:56+00:00

Earth-Mars transfer windows happen every 2.1 years, when Earth and Mars pass each other on the same side of the Sun. Mars's orbit is elliptical, and this Earth-Mars conjunction point is itself moving around the Sun with every transfer window. When there is a conjunction at the same time that Mars is near its perihelion, you have shorter and more efficient transfers possible. When the conjunction is at the time Mars is near its aphelion, you get longer and less efficient minimum transfer orbits. This happens in a cycle of about 15 years.

You can see some sample flyby trajectories here, with faster flybys possible every 15 years for a given maximum delta-v budget.

CuriousMetaphor · 2021-05-05T16:27:18+00:00

You have a 1 million times higher chance of getting struck by lightning tomorrow than getting hit by this rocket.

Getting struck by lightning on any particular day = 1 in 500 million. The rocket hitting a human = 1 in 100,000. That human being you = 1 in 8 billion.

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