I’ve been trying to calculate how much propellant a spaceship from Earth /LEO/ to Mars /LMO/ at constant acceleration of 1g would need. Here’s the data given: dry mass mf=100 t; g=9,81 m/s^2; Isp=10000 s /ion thruster/; trip duration Earth to Mars at constant acceleration of 1g, decelerating halfway 1d 21h 13m1s = 162781 s, according to this source; Delta V from LEO to LMO = 6,6 km/s, according to this source.

Initial total mass with propellant mo=? Mpropellant=?

From the Isp=10000 s I get the exhaust velocity: Isp=ve/go; ve=Isp x go=10000s x 9,81m/s^2=98100 m/s.

Then I continue with the Tsiolkovsky rocket equation and solve for the initial total mass with propellant:

$$\Delta v = v_e \cdot \ln\frac{m_o}{m_f}$$

$$m_o = m_f \cdot e^{\frac{\Delta v}{v_e}} = 100,000kg \cdot e^{\frac{6,600m/s}{98,100m/s}} = 100,000kg \cdot e^{0.067278287} = 100,000kg \cdot 1.06959 = 106,959kg$$

That gives me the mass of the propellant – 6959 kg. With the total duration given at 162781s, I calculate the mass flow rate:

$$\dot{m = \frac{\delta m_{propellant}}{\delta t}} = \frac{6,959kg}{162,781s} = 0.042751kg/s$$

The mass of the propellant /6,9 tons/ and the mass flow rate seem to be too low – are the calculations correct or am I missing something? Also, the only connection to the constant acceleration of 1g was the duration of the trip – should the acceleration be considered in a different equation?

Thank you in advance.

  • 5
    $\begingroup$ Please don't use images for equations, you can type them here directly. $\endgroup$
    – asdfex
    Commented Mar 20, 2022 at 15:41
  • $\begingroup$ The trip would need constant acceleration (+1 g) as well as constant decceleration (-1 g). $\endgroup$
    – Uwe
    Commented Mar 20, 2022 at 18:35
  • 3
    $\begingroup$ A thruster able to achieve 1 acceleration is almost certainly not going to be describable as an ion thruster. Continuous 1g acceleration trajectories are the territory of thermonuclear or antimatter thermal rockets, which are, for now, totally science fiction (sometimes referred to as "torchships") $\endgroup$
    – ikrase
    Commented Mar 25, 2022 at 4:25
  • 1
    $\begingroup$ Also, a rule of thumb: Delta-V of around your exhaust velocity or less is easily doable, delta-V of more is rapidly ballooning your rocket and you can very easily get into "bigger than the universe" territory. $\endgroup$
    – ikrase
    Commented Mar 25, 2022 at 11:35
  • $\begingroup$ A simple sanity check on your data: Note the units of ISP: seconds. At 1g no rocket has a burn time more than a few times it's ISP. Your burn is more than 16x your ISP, your rocket needs a moonlet of fuel. $\endgroup$ Commented Oct 2, 2022 at 5:06

1 Answer 1


TL;DR $\Delta v$ is only defined for one specific trajectory, but will be vastly different for different transfers.

$\Delta v = 6600 \rm{m/s}$ is only valid for a typical transfer like we do it today. You calculated the amount of fuel needed for this type of transfer using the usual month-long transfer.

To accelerate for two days at 1 g, you need $\Delta v = 160 \rm{ks} \cdot 9.8 \rm{m/s^2} = 1.57 \cdot 10^6 \rm{m/s}$.

Tsiolkovsky leaves you with a fuel mass fraction of a whopping $e^{\frac{1.57 \cdot 10^6}{10^5}} = 8.9 \cdot 10^{6}$ which translates to 100 million tons of fuel.

And don't forget, this amount of fuel needs to be accelerated to $v_e$ within two days, so you need a power plant that is able to deliver $P = \frac{1}{2} \frac{m}{t} v_e^2 = 3.1 \rm TW$, about the global electricity production.

  • 1
    $\begingroup$ Hi asdfex, thank you - I think I get it now. Shouldn't it be 10^5 and not 10^4 in the denominator (approximation of 100000 m/s^2)? $\endgroup$
    – sea_for
    Commented Mar 20, 2022 at 16:15
  • 1
    $\begingroup$ @sea_for. Yes, fixed this 1 minute earlier. $\endgroup$
    – asdfex
    Commented Mar 20, 2022 at 16:16
  • 2
    $\begingroup$ OK, all clear now - I first calculate the Delta V by multiplying the whole duration with the g's needed, and then use the Tsiolkovsky equation to find my initial wet mass. Thank you for the correction & explanation! PS noted about the images :) $\endgroup$
    – sea_for
    Commented Mar 20, 2022 at 16:25
  • $\begingroup$ Correction of 1st eqn, ve is a velocity not an acceleration; i.e., m/s $\endgroup$
    – tckosvic
    Commented Mar 21, 2022 at 16:14
  • $\begingroup$ @tckosvic Where do you see the problem? All three parts are m/s. $\endgroup$
    – asdfex
    Commented Mar 21, 2022 at 17:06

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