PDA

View Full Version : Travel duration to each of the planets



Preceden
2010-Feb-23, 11:21 PM
Hey all,

I have a website, Preceden, which let's you make online timelines and I need some help with a space travel timeline I want to build.

I just implemented a feature that let's you base the end date for an event on the present date, and I think an interesting way to showcase this feature would be to build a timeline showing the travel time to each of the planets assuming you left today.

The travel time is where I run into problems. I assumed for the sake of simplicity that we were traveling aboard the Voyager I (17.3 km/s) and that we were traveling in and straight line from Earth to each of the planets without the use of any Hohmann orbit transfers. That method, however, resulted some wildly inaccurate figures.

For example, Mars is about 78,000,000 km from Earth and traveling at 17.3 km/second, it would take about 52 days to get there. However, based on an article on this site, the actual time is about 214 days.

At first I was just going to go with the simple calculations, but as long as I'm doing this, I want to do it right. How can I calculate a good travel tme to each of the other planets so that I can make the timeline is as accurate as possible?

One last thing to note: This timeline can't take into account the position of the planets based on the time of year, so I need to come up with some kind of average regardless of when the hypothetical launch takes place. Or maybe we do it as a best case scenario for each of the planets. Any suggestions?

George
2010-Feb-24, 12:19 AM
I just implemented a feature that let's you base the end date for an event on the present date, and I think an interesting way to showcase this feature would be to build a timeline showing the travel time to each of the planets assuming you left today.That's cool.


For example, Mars is about 78,000,000 km from Earth and traveling at 17.3 km/second, it would take about 52 days to get there. However, based on an article on this site, the actual time is about 214 days. If you could stop the planets from revolving around the Sun and place them closest to Earth in their orbit, your math would be fine.

Imagine if Mars were frozen and only Earth revovled. Then you could fly off Earth as it approached Mars. If you were pointed straight at it and going 17.3 kps, then you would take longer to get there since you have the combination of the orbital distance difference, 78 million km, and the tangential distance to account for. The net outward speed (radial vector) would be 17.3 kps x sine of the angle to Mars.

But then Mars is not sitting still, so its motion must be taken into account.

I suspect someone, if not several, will be able to give you some nice references for these equations, but I thought I'd help explain why you see a difference in travel times.

tony873004
2010-Feb-24, 02:08 AM
You can't do it the way you suggest. George pointed out a few reasons, but there are more. Your speed wouldn't remain a consistant 17.3 km/s. As you climbed away from the Sun, your speed would decrease. As you dropped towards the Sun, you would pick up speed. You can't leave anytime you want, unless you have unlimited fuel. Some interplanetary trips use Hohmann transfers, or nearly-Hohmann Type I and Type II trajectories. Others are first sent to planets other than their target for gravity assists because the available fuel is not enough for a Hohmann transfer. This takes a lot longer. So the bottom line is that there is no simple answer to your question. Welcome to BAUT!

excaza
2010-Feb-24, 02:28 AM
You can get a ballpark answer using a hohmann transfer and some simple orbital mechanics.

I just pulled this out of one of the excel spreadsheets I made for one of my spacecraft design class a while ago.

Time of Flight = PI()*(((Rplanet1+Rplanet2)^3)/(8*muSUN))^0.5

Where:
PI() = pi, yum! (3.14159...)
Rplanet1 = Distance of planet 1 from the Sun, in km
Rplanet2 = Distance of planet 2 from the Sun, in km
muSun = Sun's gravitational parameter, approximately 1.33 x 10^11 km^2/s^2

EDIT: forgot to mention that the number given by the above equation is in seconds, so be sure to convert accordingly

This same equation also given here under calculations as 't_h on the Hohmann transfer wiki page: http://en.wikipedia.org/wiki/Hohmann_transfer_orbit

This is simply half the period of an elliptical orbit with one planet at the periapsis and another planet at the apoapsis.

This equation neglects a lot of things, but it will give you a good estimate. This will give about 259 days travel from Earth to Mars.

Hope this helps.

JustAFriend
2010-Feb-24, 02:38 AM
Google is your friend.... (http://www.google.com/search?hl=en&q=Mars+travel+times&aq=f&aqi=g-c1&aql=&oq=)

Jeff Root
2010-Feb-24, 02:51 AM
I suggest you assume that all the planets are in circular orbits; that Earth
and the destination planet are in exactly the right locations for a minimum
trip time; and that you don't worry too much about your speed at arrival.
Choose an appropriate launch speed from Earth for each destination for a
direct flight, not using gravitational slingshots, but accounting for the Sun's
gravity decreasing the speed to the outer planets and increasing the speed
to the inner planets. And let Tony or somebody calculate the numbers!

-- Jeff, in Minneapolis