A spacecraft is in LEO and I would like to reach a certain point in space; let's say Mars.

If I look at the diagram in this answer I see that the delta-v from low Earth orbit to low Mars orbit is roughly 2.44 + 0.68 + 0.39 + 0.67 + 0.34 + 0.40 + 0.70 = 5.6 km/sec.

If I wanted to do that using solar-electric propulsion how could one estimate the total energy needed in kilowatt-hours (kW-h) that would be required to achieve that delta-v?

You can assume my satellite is similar in size to other deep space spacecraft that use solar-electric propulsion.

  • $\begingroup$ There are a few problems with your question, both with English and some physics. I've made an edit to improve on both. Have a look and see if you'd like to make any further changes. $\endgroup$
    – uhoh
    Commented Feb 9, 2019 at 7:55
  • $\begingroup$ solar-electric propulsion, you think about an ion engine using electric power from solar cells? What about mass of spacecraft, fuel mass and the speed of the ions? These will be needed for calculation too. $\endgroup$
    – Uwe
    Commented Feb 9, 2019 at 8:37
  • $\begingroup$ After you get mass of spacecraft as @Uwe suggested, this could be a crude way for rough estimation - Depending on the thrust levels and Isp you could get the time it takes to achieve that $\delta V$ and then looking at the power specifications of the engine, you could estimate the energy by multiplying this power with the time it is fired for. You will have to consider thrust efficiency and PPU efficiency. $\endgroup$
    – MyTwoCents
    Commented Feb 9, 2019 at 9:59
  • $\begingroup$ Errata $\delta V$ -> $\Delta V$ $\endgroup$
    – MyTwoCents
    Commented Feb 9, 2019 at 10:10
  • $\begingroup$ @Uwe please re-read the last sentence! $\endgroup$
    – uhoh
    Commented Feb 9, 2019 at 11:21

1 Answer 1


Ion thrusters are usually specified with two numbers: thrust and electrical power. As an example, let's take the NSTAR engine of DS1. According to Wikipedia it produced 92 mN thrust at a power of 2.3 kW. Now we can apply Newtons well-known formula to determine the acceleration $$F = m \cdot a \quad \rightarrow a = \frac{F}{m}$$ as well as the velocity change if the engine runs for a time T: $$\Delta v = a \cdot T = T \cdot \frac{F}{m}$$ Let's call the 'efficiency' of the engine $\epsilon = \frac{F}{P}$, i.e. the force generated from a given amount of power. Please note that this is not a fixed number, but varies - doubling the power usually does not double the force. $$\Delta v = \epsilon \cdot P \cdot T \cdot \frac{1}{m}$$ or, in terms of energy expended: $$E = P \cdot T = \frac{m}{\epsilon}\cdot \Delta v$$

As an example, to accelerate DS 1 (m = 500 kg) by $\Delta v$ = 100 m/s, we need: $$E = \frac{m}{\epsilon}\Delta v = m \cdot \frac{P}{F} \cdot \Delta v$$ $$\quad = 500 \rm{kg} \cdot \frac{2.3 \rm{kW}}{92\rm{mN}}\cdot 100\frac{\rm m}{\rm s} = 1250 MWs = 347 kWh $$

As you can see, we are assuming constant vehicle mass and are not employing the rocket equation. When using ion thrusters, the fuel consumption is quite low for smaller adjustments - in our example about 0.3% of the total vehicle mass. For larger $\Delta v$ we have to come back to the conventional rocket equation - or do a break down manually by dividing the change into several smaller changes.

The diagram you cite is unfortunately not applicable for low-thrust ion propulsion. It is only valid for instantaneous propulsive manoveurs, but not for extended burn times as needed here. In general, the required $\Delta v$ is larger the lower the thrust is, but the precise numbers have to be calculated using all the details of the planned journey and can't be predicted from a simple chart like yours.

  • $\begingroup$ You are assuming constant mass of spacecraft and no fuel consumption of ion thruster? $\endgroup$
    – Uwe
    Commented Feb 9, 2019 at 16:13
  • $\begingroup$ @Uwe Should have mentioned that. But that's justified by the fact that only 0.3% of the vehicle mass needs to be accelerated for the calculated maneuver. $\endgroup$
    – asdfex
    Commented Feb 9, 2019 at 17:34
  • $\begingroup$ This is a really great answer, and the last paragraph about delta-v for Hohmann-like transfers not being appropriate for low-thrust spirals is important. This may have been explored in a Q&A here somewhere, I'll start looking around. $\endgroup$
    – uhoh
    Commented Feb 10, 2019 at 4:10
  • $\begingroup$ I did a little bit here but it is not conclusive, so I've just asked Ratio of low-thrust slow spiral to Hohmann transfer delta-v? $\endgroup$
    – uhoh
    Commented Feb 10, 2019 at 4:34
  • $\begingroup$ There's a cool answer, you may want to mention it in your answer here? space.stackexchange.com/a/34115/12102 $\endgroup$
    – uhoh
    Commented Feb 10, 2019 at 11:02

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