Calculating Fuel-Payload Ratio for Martian Rocket Lift-off

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Discussion Overview

The discussion revolves around calculating the fuel-payload ratio for a rocket lifting off from Mars, utilizing Tsiolkovsky's rocket equation. Participants explore the implications of using different gravitational constants for Mars and Earth, the specific impulse of rocket engines, and the necessary delta-v for achieving Mars escape velocity. The conversation touches on theoretical calculations and practical considerations for space travel between Mars and Earth.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about using Mars' gravity in Tsiolkovsky's equation, noting that a lower gravity should not lead to a lower delta-v.
  • Another participant clarifies that the "g" in the equation is necessary for converting specific impulse into exhaust velocity, as it incorporates Earth gravity into the unit.
  • A participant proposes calculating Mars-standard specific impulse by adjusting for the ratio of Earth to Mars gravity, while questioning the final velocity required for a rocket to leave Mars.
  • Another participant argues that specific impulse remains constant regardless of the launch planet and suggests calculating the required delta-v for lifting a lander into orbit around Mars.
  • Discussion includes calculations for delta-v needed for various maneuvers, including leaving low Earth orbit (LEO) and entering Mars orbit (LMO), with emphasis on the similarities in total delta-v required for both outbound and return journeys.
  • One participant expresses difficulty in grasping the complexity of the calculations involved.

Areas of Agreement / Disagreement

Participants exhibit a mix of agreement and disagreement regarding the application of gravitational constants and specific impulse in calculations. There is no consensus on the final approach to determining the fuel-payload ratio, as various methods and assumptions are proposed.

Contextual Notes

Participants highlight the need for clarity on the definitions of specific impulse and delta-v, as well as the assumptions made in calculations. The discussion remains open-ended, with unresolved mathematical steps and varying interpretations of the equations involved.

Who May Find This Useful

This discussion may be of interest to those studying rocketry, space travel, or gravitational physics, as well as individuals curious about the complexities of interplanetary missions.

  • #31
Flisp said:
Oh, that was a surprise! Does that also mean that if we assume atmospheric breaking on both ends of the travel, it will be the same fuel-payload ratios both ways? And that that is the very simple answer to my question?
No. It is asymmetric. You gain a big advantage with atmospheric braking on Earth (high escape velocity/orbital velocity and good available atmosphere). Not so much on Mars (much lower escape velocity/orbital velocity and thinner atmosphere) and none at all on the moon (even lower escape velocity/orbital velocity and no atmosphere).

So there is some advantage to be had on the outbound trip but a much greater advantage to be had on the return.
 
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  • #32
But if we would go from orbit to orbit (outside atmosphere) it would be symmetrical?
 
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  • #33
Flisp said:
But if we would go from orbit to orbit (outside atmosphere) it would be symmetrical?
Yes, indeed.
 
  • #34
jbriggs444 said:
Yes, indeed.
Great, thank you. And if I want a more accurate comparisson I have to piece the trips together the way you discribed it before. Earth to orbit, orbit to orbit, orbit to Mars surface and then back.
 
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  • #35
Flisp said:
Great, thank you. And if I want a more accurate comparisson I have to piece the trips together the way you discribed it before. Earth to orbit, orbit to orbit, orbit to Mars surface and then back.
Credit where credit is due, I think that was @Janus, not me.
 
  • #36
Well, thanks to you all. Great forum! If I publish something on my blog I will mention you guys just the way I did with ny article about interstellar flight. Owe you a lot.
 
  • #37
Flisp said:
What? How can that be? The fuel-payload ratio was certainly not the same both ways when we flew to the moon!

In terms of leaving Earth orbit, to arriving at Moon orbit, compared to leaving Moon orbit and returning to Earth orbit, it was. The difference between going to and returning from the Moon is that with going to the Moon, part of the mass of what would be considered "payload" was due to the fuel needed for the return trip.
From LEO to the Moon or back the other way requires a delta V of ~4 km/s. The Service module had a ISP of 314s, or an exhaust velocity of ~3077 m/s. This would require a fuel to payload ratio of ~2.7 ( 2.7 kg of fuel for every kg of payload. So this mean's that for every kg you wanted to return from the Moon, you had sent an additional 2.7 kg in fuel to the moon. So leaving LEO, you had to start with ~10 kg of fuel for every kg you would want to return from the Moon. ( For the actual Moon missions, we left quite a bit of the mass in the form of the LEM back at the Moon, so this reduced the return trip fuel cost some, allowing us to trim some of the total fuel cost. )

Apollo was of course able to save on return fuel by using aero-braking while returning from LEO. If they had been forced to use a powered landing like the takeoff was, this would have required a much different approach. Possibly by docking the Command module to a fueled lander module parked in LEO (no sense in carrying fuel that you don't need to all the way to the Moon.
 
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  • #38
Janus said:
From LEO to the Moon or back the other way requires a delta V of ~4 km/s. The Service module had a ISP of 314s, or an exhaust velocity of ~3077 m/s...
Thank you, Janus, great to have some real numbers to check against.
 

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