# Is there a simple relation between delta-v and travel time?

For example, if it takes a delta-v of 4 km/s to fly from LEO to Mars in 8 months, according to one mission plan, would it take half the time, 4 months, if the delta-v was doubled to 8 km/s? Is there any rule of thumb relationship, or must a completely new trajectory be calculated for any change of delta-v?

## 3 Answers

You seem to be thinking in terms of speed, a scalar quantity. Of course if you double speed, you cover a given distance in half the time.

But velocity is a vector quantity. It has direction as well as magnitude. Here is a diagram of adding vector quantities: In the above illustration, the red vector would be the delta V (change in velocity) in going from the blue vector to the black vector.

So changing speed isn't the only delta V. Changing direction also boosts delta V.

For example imagine two cars with different speeds 30 mph and 40 mph. If they're going the same direction in parallel lanes, the difference is 10 mph. If they T-Bone at an intersection, the difference is 50 mph. If they meet head on, the velocity difference has a 70 mph magnitude. Direction can make a huge difference.

Here's an illustration of an earth to Mars Hohmann: Look at the transfer orbit's velocity vectors at earth and at Mars. Notice the velocity vectors are pointing the same direction, so only a change in speed is needed.

Now here's an illustration of a non Hohmann earth to Mars transfer: Notice that the mars velocity vector is nearly the same size as the transfer orbit's velocity vector when it crosses Mars' orbit. We don't need to speed up or slow down much. But changing direction requires a lot of delta V.

Deviating from Hohmann for a 4 month trip would boost your delta V by a lot more than double. This is due to direction change non Hohmann transfers require.

Here are three more examples, transfer orbits depicted by Rikki-Tikki-Tavi in another answer to this question. In Rikki's illustration 3 Vinfinity quantities are given, those with regard to earth at departure. His illustration lacks the Vinfinity numbers with regard to Mars.

My illustration shows Vinfinity vectors as red arrows. It has Vinfinity quantities for both earth and Mars.

Outside of a planet's sphere of influence, the transfer orbits can be modeled as ellipses or a parabola about the sun. But within a planet's sphere of influence, the path is better modeled as an ellipse about the planet. Vinfinity is a hyperbola's velocity at an infinite distance from the gravitating body. A hyperbola's speed can be found by $\sqrt{V_{esc}^2+V_{inf}^2}$. Escape velocity is $\sqrt{2GM/r}$. As you can see escape velocity grows larger as the planet gets closer. At the edge of a planet's sphere of influence, escape velocity is close to zero and the hyperbolic velocity is very close to Vinfinity.

Escape velocity is around 5 km/s near Mars' surface. So a hyperbolic orbit grazing Mars' atmosphere would have a speed of $\sqrt{5^2+V_{inf}^2}$ km/s.

For Rikki's 3 examples this becomes:

$\sqrt{5^2+2.65^2}$ km/s which = ~5.7 km/s (the Hohmann orbit)

$\sqrt{5^2+6.23^2}$ km/s which = ~8 km/s

$\sqrt{5^2+20.31^2}$ km/s which = ~21 km/s (the parabolic transfer orbit)

The 5.7 km/s is what needs to be shed for Mars landing coming from a Hohmann orbit. Aerobraking can accomplish this for smaller pay loads but for more massive payloads this is hard. Only .7 km/s is needed to brake into a capture orbit about Mars.

Given the 8 km/s hyperbola periapsis velocity, you will need to shed twice as much kinetic for a soft landing. 3 km/s delta V would be needed to brake into a capture orbit about Mars. Rikki seems to believe this is trivial, but it's not.

For a 21 km/s hyperbola periapsis velocity I don't think it's practical to use aerobraking for a soft landing. 16 km/s would need to be shed to brake into a capture orbit. 16 km/s is about what it takes to get from earth's surface to the moon's surface.

• I disagree with you about the delta V for mars orbit injection being the major factor. You can use aerobrakeing for most of that $\Delta v$, as described in my post. Nice illustration though. Did you draw it yourself? – Rikki-Tikki-Tavi Aug 26 '14 at 23:49
• @Rikki-Tikki-Tavi the quantities in red are V infinities. Near Mars surface, hyperbolic velocity would be sort(vinf^2 + vesc^2). Near Mars surface Vesc is about 5 km/s. So speed would be sort(5^2 + 2.6^2) for or about 5.5 km/s Hohmann. Martian aerobraking can shed that for small payloads. – HopDavid Aug 27 '14 at 0:25
• For the other transfer orbit I depicted, hyperbolic speed at Mars surface would be sort(11^2+5^2) or about 12 km/s. It would be tough for Martian aerobraking to shed that much speed. – HopDavid Aug 27 '14 at 0:28
• Yes, as I described in my post, there are limits to aerobraking. But for reasonable trajectories, it works well. – Rikki-Tikki-Tavi Aug 27 '14 at 0:30
• A transfer orbit from earth to Mars would be hyperbolic with regard to Mars. So there is a vinf wrt Mars. That's not shown in your diagram. Do you know what the Vinfs wrt to Mars would be? – HopDavid Aug 27 '14 at 1:03

A particular delta-v (relative to a given celestial body, such as the Moon) actually implies a particular transfer orbit toward the target body.

Also, delta-v is the difference in velocity (produced by a particular maneuver), it is not an absolute velocity.

So let's say you start out at 11 km/s relative to the Earth (in an approximate Earth-Moon system transfer orbit) and then near the Moon apply a further delta-v of 4 km/s along your direction of travel. Ignoring orbital effects, that will push your orbital speed to 15 km/s. If you instead double the delta-v (which means you need to bring along a lot more fuel, which increases the fuel requirement for achieving the initial E-M transfer orbit, which means you need even more fuel just to get off the ground, and round and round we go in the rocket equation...) to 8 km/s, not only do you end up in a different orbit (so have to redo all the orbital calculations) but also your final velocity is only 19 km/s at most. You gain about 26% in terms of absolute velocity by doubling the delta-v.

An astronaut on a spacewalk in Earth orbit might start out with an orbital velocity of on the order of 7-8 km/s (the orbital velocity of their spacecraft) and apply a delta-v of a fraction of a m/s in order to get to where they want to go. And so on.

For a perhaps somewhat extreme but very real example, consider that Apollo could abort the 11 km/s transfer orbit to the moon by applying an approximate 1.8 km/s delta-v about a third to half of the way from the Earth to the Moon. (True "abort" would require a delta-v of twice the current velocity, to reverse the direction of travel while in the end resulting in the same orbital velocity in a reversed orbit. Even ignoring whether it's practical to "reverse" the orbit like that in the first place, applying an unplanned delta-v of well over 20 km/s to something as massive as the Apollo CSM, let alone the CSM/LM pair, is not an easy feat.) That abort wasn't an option in the case of Apollo 13, but it (and the illustrative graph shown on the Wikipedia page linked) goes to show some of how it's not quite as simple as "double the delta-v, halve the time" or even "double the velocity, halve the time".

• This is incorrect. Yes, trans lunar insertion starts at ~11 km/s when departing from LEO. But as you near the moon you are no longer traveling 11 km/s. Near apogee you are traveling around .2 km/s. Applying a 4 km/s burn near the moon would push you to 4.2 km/s orbit wrt earth (neglecting effects of moon's gravity). If you applied the same 4 km/s while traveling 11 km/s just above earth's surface, delta V would be larger (due to Oberth effect) – HopDavid Aug 26 '14 at 17:35

It's not necessary to take an entirely different route every time. There is the minimum energy transfer, that you mentioned, but you can use more fuel(or better yet, use a more efficient engine) to take a similar, but different route that takes you there faster. The raise in energy requirement is moderate at first, but becomes uncomfortable if you want to cut your travel time by more than a month or so. I think there is a table in a book I have at home ("Astronautics" by Ulrich Walter). I will let you know.

Also, there are entirely different courses you can take, but I don't think any of them are faster. For example, in some constellations a Venus flyby gives you the chance to return after only a short stay at Mars, rather than waiting out a full cycle.

Edit:

I misremembered. In Astronautics I found the following two figures (which are even better; read below):

The image above makes three examples of orbits we could fly. You could fly even more extreme courses than the 70 day one, but you can see how the required earth escape speed $$v_\infty$$ becomes uncomfortable.

This graph gives us the the same in exact numerical terms. You can see that shortening the time a bit is quite cheap, but then it quickly becomes expensive. $$t_x$$ is the transfer time. $$t_H$$ is the time for the Hohmann-transfer. I won't explain all variables involved, I don't think anyone has the patience to read that.

It is true that you also need an burn to turn into Mars' orbit, but I disagree with HopDavid about this being the major factor:

The $$\Delta v$$ for the for the injection is $$v_\infty\sqrt{\frac{1-e}{2}}$$ where $$e$$ is the eccentricity of the orbit around mars and $$v_\infty$$ is the excess velocity with regard to mars now. This can be very high, because you can slowly circularize the orbit for free by aerobraking. For $$e=0.95$$ the Mars injection takes only 16% of $$v_\infty$$. Of course, there is a limit to how eccentric your orbit can be, because you may escape Mars' sphere of influence if you come in too fast.

• But one cannot take the same route faster, the (hyperbolic) trajectory will have another shape if one has a higher speed, and the distance traveled will be different because Mars is a moving target, and the Mars insertion orbit will be different, right? – LocalFluff Aug 26 '14 at 11:30
• Yes. Also, the starting time frame may be different. You can't realistically go to Mars without using the Earth's orbital speed, so you are dependent on the two bodies being in specific places. These aren't Hohmann-Transfers anymore, so there is quite a bit of optimization involved. – Rikki-Tikki-Tavi Aug 26 '14 at 12:43
• Escape velocity near Mars surface is about 5 km/s. You can get a Mars capture orbit's eccentricity as close as you like to 1, Let's call it .996. You're not going to get a periapsis velocity higher than 5.01 km/s. – HopDavid Aug 27 '14 at 0:46
• At periapsis of the hyperbola velocity is sort(vesc^2 + vinf^2). See en.wikipedia.org/wiki/Hyperbolic_trajectory#Velocity Even in the case of Hohmann this would be about 5.6 km/s. So for injection into a Mars capture orbit, you'd need .6 km/s. But 2.6 km/s * sqrt((1-.996)/2) is about .12 km/s. I believe your equation for injection to Mars capture orbit is incorrect. – HopDavid Aug 27 '14 at 0:53
• Your diagram has V infinities wrt to earth. But not the Vinfs wrt to Mars. – HopDavid Aug 27 '14 at 0:57