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Looking at the times that are quoted to go to Mars (months) makes me wonder what speed do we need to get there in a week?

Chemical rockets can get to ~4.5 Km/s, thermal nuclear rockets is about twice that (~8 Km/s). Electric can be between ~15 Km/s to 100 Km/s ...not sure if range is right.

I'm taking about transit time from Earth orbit to Mars orbit. I'm not looking to deploy a bunch of refueling tanks in the way before leaving Earth (even if that helps). Or make a chemical rocket huge either (again ...even if that helps). I'm talking something somehow efficient. I understand that if we go really fast we need to break so if we accelerate half the way or more and then turn the vehicle around and start breaking that's fine. BTW I'm still thinking to go there the way we go there now where there is a window when Earth and Mars are closer. No direct travel at any point in Earth and Mars orbit yet. We'll do that later ...lol.

It looks like 50 Km/s gets us there in 39 days (vasimr engine?

)

What speed does it take to get to Mars in 7 (Earth) days? Thanks.

https://en.wikipedia.org/wiki/Nuclear_thermal_rocket

https://www.sciencedirect.com/topics/earth-and-planetary-sciences/nuclear-electric-propulsion

https://www.lanl.gov/science/NSS/issue1_2011/story5full.shtml

Rodo
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    One way to approximate it: Take the average distance between earth and mars during the window of opportunity and divide by 604800 (i.e. 7243600). That should give a good estimate of the necessary speed. – Paul Jun 11 '21 at 18:07
  • V= 55000000km / 604800s= ~91km/s ... looks possible!? – Rodo Jun 11 '21 at 18:54
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    @Paul That only works if your burns are short. For Brachistochrone trajectories your peak velocity has to be a lot higher. – Loren Pechtel Jun 12 '21 at 04:44
  • Different but related: Brachistochrone variation for Earth-to-Mars Orbit – uhoh Jun 12 '21 at 10:37
  • @lorenpechtel: Agreed. The OP appears to be concerned with bang-bang control for which this approximation is derived. And of course, this is only an “average” speed needed and does not account for non-instantaneous velocity changes. A more realistic estimate would require solving a constrained optimization problem. – Paul Jun 12 '21 at 14:14

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You are travelling in days a distance that planets travel in months. From a trajectory perspective, that means that neither the velocities of Earth and Mars, nor the gravity of the Sun are particularly relevant any more. That is, "point and thrust" trajectories.

You will be limited by either delta-v or acceleration.

  1. Delta-v limited case.

To minimise the delta-v for such fast trajectories, you would want to provide instantaneous impulses right at the start and end of the transfer, to have maximum speed for the coasting phase.

This is about as simple as it gets to model:

$$\Delta v = 2\frac{d}{t}$$

For your distance d of planetary alignment of ~70,000,000km (this varies due to the Earth and Mars meeting up at different points of elliptical orbits), this means a minimal $\Delta v$ budget of 230km/s. This would require a high thrust engine more performant than "regular" solid core nuclear rocket engine, but not outside the realm of realistic science fiction.

(you can forget about chemical rocket engines here. The mass ratio would be about 1.5e22, or 40,000 atoms of cargo per kilogram of fuel)

  1. Acceleration limited case.

This would be the trajectory you describe where the engines are kept running up to the halfway point, and then turned around to slow you down before arrival. No coasting. This requires a higher delta-v budget.

For constant acceleration, this means:

$$a = 4\frac{d}{t^2}$$

Which for your usecase is a mere 0.76m/s. That would be outrageously high for any ion engine unless we can find a denser power source than nuclear reactors.

SE - stop firing the good guys
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    Where does the "2" in the Δv formula come from? Is it because of round-trip? I'm talking one way. Thanks. – Rodo Jun 11 '21 at 21:44
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    @Rodo the 2 is to speed up and then slow down on arrival at Mars – BrendanLuke15 Jun 11 '21 at 21:49
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    For the acceleration-limited case Google for Brachistochrone trajectories. Info here: http://www.projectrho.com/public_html/rocket/torchships.php#id--Brachistochrone_Equations – Loren Pechtel Jun 12 '21 at 04:45
  • @LorenPechtel luckily they put "Brachistochrone" in quotes on that page; it's an excellent search term but I don't think it's exactly correct. In the past I wrote There can't be a "Brachistochrone curve to Mars orbit”... The use of the term is relegated to fiction on Earth and science on Kerbin to an earlier question. – uhoh Jun 12 '21 at 10:39
  • @uhoh We don't have a drive which can do a Brachistochrone trajectory, that doesn't mean the math isn't valid. – Loren Pechtel Jun 12 '21 at 14:59
  • @LorenPechtel math is always valid, my nitpick is that it can't technically be called a Brachistochrone unless the meaning of the term has evolved over time, and I didn't think that it has.That's why I said that it was good that they put "Brachistochrone" in quotations in your link, cluing us in that the usage of the term is at least a bit stretched when applied to orbital mechanics. – uhoh Jun 12 '21 at 15:02
  • @LorenPechtel A Brachistochrone curve is defined by fixed start and end points, and a bead frictionlessly constrained to a string in a uniform gravitational field. For any other least-time mathematical problem we need to use "scare quotes". – uhoh Jun 12 '21 at 15:13