Interplanetary travel question

[strider3]

Asking here because it really doesn't seem to fit in "gameplay". Using the Semi-major axis numbers for Kerbin, Dres and Jool, and doing a little subtraction, I find that the "straight line" distance between the orbits of Kerbin and Dres and between Dres and Jool, are almost identical. 27,239,508 km between Kerbin and Dres and 27,934,212 km between Dres and Jool. I'm obviously missing some very basic knowledge of interplanetary travel math because planning flights between these 3 planets results in a huge difference in travel times. Kerbin to Dres is around 1-2 years but Dres to Jool is 5 or more years. What basic knowledge of travel am I missing here?

Edited April 25, 2019 by strider3

[KerikBalm]

Travel time is not dependent upon the difference in SMA, its that simple

[Guest]

11 minutes ago, KerikBalm said:

Travel time is not dependent upon the difference in SMA, its that simple

OK, but can you tell me why the huge difference, in this case, in simple terms?

[KerikBalm]

The closer in you are, the faster your orbit.

To go from a lower orbit to a higher one with a standard hohman transfer requires half an orbital period (with an SMA equal to the average of your current one and the target one, assuming your current and target orbits are circular).

The farther out you go, the longer the orbital periods are

[Guest]

1 hour ago, KerikBalm said:

The closer in you are, the faster your orbit.

To go from a lower orbit to a higher one with a standard hohman transfer requires half an orbital period (with an SMA equal to the average of your current one and the target one, assuming your current and target orbits are circular).

The farther out you go, the longer the orbital periods are

Thanks, KB!

[KerikBalm]

Just to clarify on the half an orbital period: In a hohman transfer, your periapsis is at the lower orbit, and the apoapsis is at the higher orbit. Going from PE to AP, back to PE is an entire orbital period. Going from just PE to AP is half an orbital period. A perfect hohman transfer (circular non-inclined orbits, absolute minimum energy) always takes half an orbital period. The orbital period will obviously be somewhere between the orbital period of the smaller and larger orbits.

So going from 0.5 AU to 1.0 AU gives you a transfer with an SMA of 0.75 AU... whereas going from 1.0 AU to 1.5 AU gives you a transfer with an SMA of 1.25 AU.

Obviously the 0.75 AU SMA has a much shorter orbital period than the 1.25 AU SMA, and thus the transfers are much faster, even though both change the SMA by 0.5 AU.

 

The same applies for going to Mun vs Minmus... Minmus takes much longer to get to than the distance from kerbin would suggest, if you're doing an efficient hohman transfer (many people crank up the dV budget to get a faster transfer there)

Edited April 26, 2019 by KerikBalm

[Vexillar]

Hi @Victor3

As previous answers have said, the travel time is determined by the transfer orbit's period.  This is, (as represented by Kepler's third law) proportional to the square root of the cube of the semi-major axis, a.

[How much of the orbit's period is used in the transfer might vary according to timing, alignment or desired efficiency, but it remains the fundamental factor on which the other factors bear.]

Hence, the further from the Sun, the longer the period, independently of the difference in the start and end body's SMA.

So - using @KerikBalm's example above, if we take the the orbital period for an orbit with a of 1 AU to be 1 year, the period for an orbit of 0.75 AU would be about 0.65 years.  The period for an orbit with a equal to  a 1.25 AU orbit would be about 1.40 years. So the further out one is from the Sun (or the planet in whose moon system one is transferring), the longer the period.  Period.

 

 

[Snark]

On 4/25/2019 at 12:17 PM, Victor3 said:

OK, but can you tell me why the huge difference, in this case, in simple terms?

t = K * R^1.5

That is, the time taken for an orbit is proportional to the radius of the orbit raised to the 1.5 power.  If you have two orbits, A and B, and B is 4x the radius of A, it will have 8x the time to complete one orbit.

Orbits farther out from the primary get slower, by a lot.

[Vanamonde]

Although about in-game planets, this discussion will involve real orbital mechanics. Therefore, the thread has been moved to Science & Spaceflight. 

[wumpus]

On 4/29/2019 at 2:49 PM, Snark said:

t = K * R^1.5

That is, the time taken for an orbit is proportional to the radius of the orbit raised to the 1.5 power.  If you have two orbits, A and B, and B is 4x the radius of A, it will have 8x the time to complete one orbit.

Orbits farther out from the primary get slower, by a lot.

Try here: https://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion

for a more graphically based concept (Kepler had to work with geometry).  Note that for circular orbits you can derive Newton's law of gravitation with algebra, dealing with elliptical orbits appears to want a pair of partial differential equations (which meant that inventing calculus gave Newton a huge leg up on discovering the law of gravitation).