# Quick and dirty way to estimate flight times for constant acceleration rockets

Thought experiment: Suppose we build a spacecraft with a low thrust electric propulsion and have a flight plan that involves constant acceleration on the first half of the journey, breaking on the second. Sounds ridiculous? Let's plug some numbers into the formula given here:

acceleration $a = 0.001 m/s^2$
Distance Earth-Jupiter $x=9.3 * 10^6 km$ Travel time $t=705d$
Start weight would be 80% fuel This assumes the exhaust velocity in the wikipedia article on the Vasimir, $v_e = 40km/s$

I think two years and 80% fuel ist not too terrible.

I'm not 100% sure such a craft could manage orbital insertion anywhere, I think leaving earth moon should be doable. The only beauty I see about such a craft would be that it could be built ultra light (in orbit) as it would never have to endure the stress of earth's gravity or a liftoff. but this is outside of the scope of my question.

My estimate does not take into account that such a craft would spend quite some time and fuel not simply flying from a to b but working against gravity wells: Earth, possibly earth-moon, sun. The proper way would be to simulate a flight plan. What is a non proper way that gives a reasonable estimate with a handful of iterations?

The scenario for the answer should be: starting from LEO and arriving at Jupiter in a heliocentric orbit using a low, constant acceleration.

This question tackles a similar scenario but with a fictional high-g drive where the considerations I'm interested in are not relevant.

• Could you state a scenario more clearly? Do you mean for example staring from LEO and arriving at Jupiter in a heliocentric orbit using a low, constant acceleration? – uhoh Apr 26 '18 at 10:04
• @SteveLinton that part is easy. It turns out, though it sounds too simple to be true, according to this extremely handy answer by a known "rocket scientist" that that time is equal to however long it takes for $at=\Delta v$. So if you are in LEO with an orbital velocity of 7700 m/s and your propulsion gives you 0.1 m/s^2, it takes 77,000 seconds to escape Earth orbit and become heliocentric. The twist is that your acceleration forward actually slows you down. – uhoh Apr 26 '18 at 10:52
• You then apply the same procedure a second time, where this time $\Delta v$ is the difference between the orbital velocity of the Earth, and the orbital velocity of Jupiter. Of course there are small corrections for phasing etc. but in the limit of slow acceleration, one doesn't need to actually calculate the trajectory explicitly to get the time. – uhoh Apr 26 '18 at 10:56
• You added these comments while I was typing the answer below, but I think they are pretty close to agreeing. – Steve Linton Apr 26 '18 at 10:58
• Following discussion on another question, it seems that $0.001 m/s^2$ is actually wildly unreaslistic, no power system available in the outer solar system produces enough power per kilogram to achieve that, or even come close to it. – Steve Linton Apr 28 '18 at 10:40

I posted this as a comment, but now I've found soem formulae, so I'll make it an answer.

I think what you need is some simplifications of the dynamics of spiralling orbits. So you would start in LEO, perhaps and spend some time in a spiralling Earth orbit until you reached escape velocity (or maybe reached a Lagrange point). Then you transition to a spiralling solar orbit, until you get close enough to Jupiter. Then you are ina spiralling Jupiter orbit, which you shrink until you are where you want to be, There are some inaccuracies at the transitions, but if you can find an expression for the time to spiral from radius R1 to R2 at given low constant acceleration, you are home.

This paper seems to have what we need in equation (5) $$r\approx {r_0\over \left(1-a_\theta t/v_0\right)^2}$$ and equation (15) and near it, which give formulae for the time required to escape from a body. $$1-{a_\theta t_{esc}\over v_0} = (2\epsilon)^{1/4}$$ where $\epsilon$ is the ratio of initial thrust to gravitional acceleration at the original orbital radius.

Applying this to your example, we start in LEO with $\epsilon = 10^{-4}$ and $a_\theta = 10^{-3}$. $v_0$ is orbital velocity, about 7800 m/s, So our time to escape from Earth is something like $$t_{esc} = \left(1 - 0.12\right) \times 7800/10^{-3}$$ which is roughly 7 million seconds, or about 80 days.

After escaping we will be more or less in the same solar orbit as the Earth, so we can do the calculation to widen our orbit out to Jupiter orbit:

Here $r_0$ is $1.5 \times 10^11$, $v_0$ is about $3\times 10^4$ and we want $r = 7.8 \times 10^11$. So $$1.5/7.78 = 1 - {10^{-3} t \over 30000}$$ or t = {6.3\over 7.8}\times 3\times 10^7 s$which is roughly 240 days. Finally, we need to spiral in to whatever Jupiter orbit we like. Suppose we want to go to Europa. This is the escape maneuver run backwards, and we get$v_0 = 13750$, and Jupiter's gravitational acceleration at Europa's orbit is about$2.5 ms^{-2}$so we get$\epsilon = 0.0004$. Plugging it in, we get something in the ballpark of 10 million seconds (another 100 days or so) to spiral in. • You can check this against my comment(s). – uhoh Apr 26 '18 at 11:04 • I think your comments are the case when$\epsilon$tends to zero. – Steve Linton Apr 26 '18 at 11:24 • yes, thus the "in the limit of slow acceleration" in the 2nd of my two comments I've linked to, as well as being (somewhere) in the answer I've linked to. I think the OP's value of ~1E-04 g is low enough for the OP's request for "quick and dirty". – uhoh Apr 26 '18 at 11:39 • This calculation is seriously flawed because the formulae work on the assumption that the orbit remains nearly circular. No orbit that gets to Jupiter in 16 months can possibly do that. – Steve Linton Apr 28 '18 at 10:41 A quick&dirty estimate that will work between distant targets - no short hops like between Earth and Moon. Earth-Jupiter at the least, so that direct-line distance is our primary concern, and climbing the Sun gravity well can be dropped the "quick&dirty" error chute. We assume delta-v much higher than for delta-v optimal transfer. 1. Pick a delta-V map that includes your route. Something like this: 1. Add up the delta-V for your route. Exclude the interplanetary segment (between Earth escape and Jupiter insertion; this is the area where lion share of our extra delta-v is applied, so it needs to be taken separately). Pick half of that sum as initial velocity value$v_0$. 2. Take the basic accelerated motion equation$s = {a t^2 \over 2} + {v_0 t}$and solve it for$t$, given$s$equal half the average distance between the bodies,$v_0$as the average between escape and insertion velocity, and$a$as the average acceleration of the spacecraft. Double that for the first approximation. 3. Add escape and capture times:$v_e = at+v_0$where$v_e$will be escape/capture velocity, and$v_0\$ - low orbt orbital velocity.

There. Quick, dirty, gives only a rough ballpark, completely neglecting alignment of the bodies - the time and delta-V are likely to be higher for worse alignment. Also, assumes constant acceleration, not variable but uninterrupted as would be the case with a spacecraft that depletes fuel.