- 22 Jul '09 22:34This new plasma rocket called Vasimr may allow trips to mars in a little over a month if they can power it up to 200 Mw levels, now ticking in at 50 Kw. I did the math, S=(AT^2)/s which solving for A (acceleration) is A=2S/T^2, setting the distance at 160 million Km, a mid level distance, from min of about 56 million Km to a max of roughly 300 million klicks, I did the calcs where you accelerate for half the distance, then turn around and decel for half which gets you to Mars with zero velocity relative to Mars.

According to my math, that clocks out to an outrageously low 5 milliG.

(.005G) Anyone want to confirm that? Seems like a very low G level to get to mars in 40 days. So it accelerates full bore for 80 million klicks and then the rocket motor faces Mars and decel's for 80 million klicks, each half taking about 20 days. Sound right? 5 MILLI G's? - 23 Jul '09 12:58

I have no idea what all those numbers really mean, but are you saying a plasma rocket's been*Originally posted by sonhouse***This new plasma rocket called Vasimr may allow trips to mars in a little over a month if they can power it up to 200 Mw levels, now ticking in at 50 Kw. I did the math, S=(AT^2)/s which solving for A (acceleration) is A=2S/T^2, setting the distance at 160 million Km, a mid level distance, from min of about 56 million Km to a max of roughly 300 million klick ...[text shortened]... s and decel's for 80 million klicks, each half taking about 20 days. Sound right? 5 MILLI G's?**

invented that assuming they can get 200 Mw will take people to Mars and back in little over a month?

The moon should be hours away then? - 23 Jul '09 17:38

Surprising indeed! I got a very similar number.*Originally posted by sonhouse***This new plasma rocket called Vasimr may allow trips to mars in a little over a month if they can power it up to 200 Mw levels, now ticking in at 50 Kw. I did the math, S=(AT^2)/s which solving for A (acceleration) is A=2S/T^2, setting the distance at 160 million Km, a mid level distance, from min of about 56 million Km to a max of roughly 300 million klick ...[text shortened]... s and decel's for 80 million klicks, each half taking about 20 days. Sound right? 5 MILLI G's?**

Assuming the acceleration is roughly symmetrical in both the time and distance domains, I started from the following:

a = k (0 < t < h)

a = -k (h < t)

Where "a" is the acceleration of the rocket in m/s2, "t" is the time in seconds, and "h" is the halfway mark in seconds (i.e. the point in time where the engines are turned around to slow down the rocket). From this, we can derive the other equations of motion:

v = int(0,t) a dt = (int(0,h) k dt) + (int(h,t) -k dt) = k(h-0) - k(t-h) = 2kh - kt

d = int(0,t) v dt = int(0,t) (2kh - kt) = 2kht - (1/2)kt^2

The distance given was 160 million km, so converting that to metres and subbing in we get:

1.60 x 10^11 m = 2kht - (1/2)kt^2

Simplifying, we have:

t^2 - 4ht + (1.60 x 10^11)/k = 0

Using the quadratic formula we have:

t = (4h +/- (16h^2 - (1.280 x 10^12)/k)^(1/2)) / 2

Since t = 2h, after simplifying we are left with:

16h^2 - (1.280 x 10^12)/k = 0

Solving for "k", we find:

k = (1.280 x 10^12) / 16h^2 = (8.0 x 10^10) / h^2

Since h = 20 days = 1728000 s, we find that:

k = (8.0 x 10^10) / (1728000^2) = 0.0268 = 0.03 m/s2

Converting this to g's, we get:

k = 0.0268 / 9.81 = 0.00273 = 0.003 g's

Surprisingly small! The maximum velocity attained during the trip will occur at the halfway point:

v(max) = kh = (0.0268 m/s2)*(1728000 s) = 46296 m/s = 1.67 x 10^5 km/hr

Surprisingly fast for such a small acceleration! The average commuting distance to work in the US is about 16 miles or 26 km, so travelling at this maximum speed it would take you about 0.56 ms to get to work, about the same amount of time it takes for a housefly to flap its wings 6 times. - 24 Jul '09 02:42 / 1 edit

Amazing, eh! I don't think you could even FEEL a few milliG's! Of course your trip to work would take a very brief time like you say but it would still take over a week to get up to that speed at 3 or 4 MilliG's!*Originally posted by PBE6***Surprising indeed! I got a very similar number.**

Assuming the acceleration is roughly symmetrical in both the time and distance domains, I started from the following:

a = k (0 < t < h)

a = -k (h < t)

Where "a" is the acceleration of the rocket in m/s2, "t" is the time in seconds, and "h" is the halfway mark in seconds (i.e. the point in time where the work, about the same amount of time it takes for a housefly to flap its wings 6 times.

Of course what saves your asss in space is no atmosphere to bog you down so ANY accel will get you going pretty fast. How about .003 G's going to PLUTO? How long would THAT take? (Of course, accel halfway and decel the other half)

The reason for such velocities is of course full time accel. A chemical rocket cranks out megawatts of power for say a half hour and then you have no other choice but to coast the rest of the way and then you have to get back down to relative zero speed at the other end so that represents even that much MORE fuel to haul around with you. So that gentle accel is continuous, that's the big difference. - 25 Jul '09 16:58

I think you would SEE the difference at that low an accel, if you were floating at the front of a spacecraft, you would slowly drift to the back of the ship. Here is a math problem for you: You are strapped, velcroed, to the front of a craft and there is a 10 meter space between you and the back of the craft, the back is where the rocket is thrusting at 0.003 G's. You unstrap from the velcro and you start drifting to the back end. How long does it take to go that 10 meters? Time starting the second you are free from the velcro tie downs.*Originally posted by twhitehead***Its actually not that surprising that the Gs experienced is very small. Just 1G as experienced on earth is enough to accelerate you to nearly 10m/s in just 1 second. (thats 36km/h I think)**