# Nerd question

Submitted by Brian W. Doss on Fri, 2004-08-13 01:52

If you had a spaceship capable of maintaining 1g of acceleration indefinitely, how long would it take to get from Earth to Mars, either at opposition or conjunction?

UPDATE:

To clarify: the spaceship would start in Low Earth Orbit and end up in Low Mars Orbit.

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## Do you want it to be capable

Do you want it to be capable of making orbit, or is it okay if it crashes into Mars at a few hundred thousand miles a second?

## Oh, and is this a trick

Oh, and is this a trick question, where you point out that a 1.0g-capable spacecraft stays hovering inches over the launchpad indefinitely, or can we assume it starts in earth orbit?

## yeah, its gotta make orbit.

yeah, its gotta make orbit. Tho I suppose that would create a big boom otherwise...

## and yeah, I was thinking

and yeah, I was thinking orbit-to-orbit transfer.

## If you had a spaceship

If you had a spaceship capable of maintaining 1g of acceleration indefinitely, how long would it take to get from Earth to Mars, either at opposition or conjunction?UPDATE

To clarify: the spaceship would start in Low Earth Orbit and end up in Low Mars Orbit.

An observation, but not an answer, is that a spaceship (or a paint flake) in Low Earth Orbit DOES indefinitely experience an acceleration of only somewhat less than 1g,

not making any progress towards Mars, but not needing any fuel either.

Regards, Don

## First question: why do you

First question: why do you care?

My CRC lists the semimajor axes of revolution for Earth as 145.97 million km and Mars as 227.84 million km. Ignoring the gravity of the two bodies and the Sun and their relative velocities, a mission at opposition would start at ~8000 m/sec and end at what, ~3500 m/sec? (Too tired to calculate circular-orbit velocity for Mars right now.) The distance to be covered is 81.87 million km, or 8.187e10 meters.

Assuming zero velocity at both ends gives you a trip time of 25 hours, 22 and a half minutes or so. I'm too tired to figure the decrease from assuming 8000 m/sec (in a favorable direction?) at the Earth end and ~3500 m/sec residual velocity at Mars' end; maybe someone else can do it. From the raw acceleration numbers it looks like it would cut at least 30 minutes off the trip time just from the head start.

For the bodies at conjunction, do you allow boring a hole through the Sun?

## No, no sun-boring holes.

No, no sun-boring holes. Just a regular earth-to-mars conjunction trip.

I want to know just because I'm curious. Right now it takes about 180 days to fall to Mars. Assuming your answer is roughly correct, a ship that could maintain earth gravity on the way would get there in a couple of days.

Which is kind of like the reduction in travel time in the past 300 years for someone traveling across the atlantic by boat, no?

## Robert Heinlein answered

Robert Heinlein answered this question, and a bunch of variants (0.1g instead of 1.0g, Jupiter instead of Mars, etc), in one of the articles collected in _Expanded Universe_. I don't have my copy handy, or I'd look up his answers.

## Cool, thanks :beatnik:

Cool, thanks :beatnik:

## No, the opposition-class

No, the opposition-class trip to Mars at 1 G would take a bit over a day. If you assume that the trip starts and ends from a stationary position at the semimajor axis distances of Earth and Mars (invalid, but probably close enough) and that Solar gravity is negligible (very close, it's only about 6 mm/sec^2 at Earth's distance) you get two half-trips of t = sqrt( d / 2a ) where a = 9.81 m/sec^2 and d = 4.094e10 meters. That gives t = sqrt( 2.087e9 sec^2) or t = 45680 seconds for the half-trip, or 91360 seconds for the full trip (at the center you turn over and decelerate the rest of the way). That is 25 hours and a bit.

For the conjunction-class mission the distance is more like 373.81 million km (3.7381e11 meters) and the time would be 195205 seconds (2 days 6 hours 13 and a fraction minutes); a bit over four times as far in a bit over two times as long, just what you'd expect from Newtonian physics.

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