Since the rocket isn’t traveling in a straight line, how much actually ends up showing on the odometer. Same question for the probes launched to Mars?
Well, Apollo 11, after launch, orbited earth about 1.5 times, before the headed off toward the moon. By the time they got to Lunar Orbit insertion they were 220,000 nautical miles (253,000 miles) from earth. Then they orbited the moon 30 times before the Eagle descended toward the surface.
Not sure what that translates to actual miles traveled, but it might be a start.
Thanks, but that’s a little more that I was looking for. In the case of the Moon, the point at which they stomp on the gas to break out of orbit to the point that they arrive in lunar orbit. That would appear, on the surface, at least, to be somewhat more than the average distance from the Earth to the Moon. (Hmmm. . . sounds like a good title for something. . . .) But maybe not.
For a meaningful answer here, it will be necessary to specify the frame of reference: Earth? Moon? Sun? Other?
From Earth to Mars will be trickier, because it depends on where Mars is at in relation to the Earth, and from which bodies (and how many) the ship will get gravity assists from. Going to another planet makes the moonshot look like a straight line.
The simplest, least-energy transfer from one body to another is an elliptical orbit that touches (tangent) to the two (roughly) circular orbits. This assumes that boost time (when the engines burn) is an insignificant portion of the whole trip, which is likely with chemical engines. Any higher than the outer orbit, and you burned more fuel than you had to; any lower, and you don’t reach where you’re going. So timing is critical; you fire so that you arrive at the high point just as your target arrives too.
From earth to moon, this is an ellipse that grazes the outside of lower earth orbit and the inside of moon’s orbit. (Technically, probably about a thousand and a bit miles less, since it grazes the point at which you go into moon orbit.
Similarly, earth to mars is a half-ellipse that grazes earth orbit and Mars orbit. We ignore the slight amount of orbiting the earth, and orbiting Mars, since we’re talking a few thousand miles in tens of millions. Plus, there will be some distortion of the orbit to compensate for, due to earth’s gravitational effects leaving and Mars’ effect arriving; but generally - don’t count that.
An elliptical orbit is an orbit with a high point and a low point; because of orbital physics, it is an ellipse. The source gravitational point - in once case earth, the other - sun, has to be at one focus of the ellipse. (Technically, the focus of the earth-moon system is at their common center of gravity, but that’s still a point below the surface of the earth.)
When you try to do this, the question will be - “yes, but how wide an ellipse?” That’s where the “center of gravity at one focus” comes in. Recall the artsy method of drawing an ellipse- two nails or pins and a loose string between them - stretch the string with a pencil, and you can draw an ellipse. The two nails represent the focii. Any point on the circumference of an ellipse - the sum of distances to the two focus points is constant. A circle is just the special case where the two focii are at the same point.
So now, some numbers:
see - Ellipse Circumference Calculator
lower earth orbit - pick a spot, let’s for argument say we are 4,000 miles (about the diameter of the earth) from the center of the earth-moon system when we start, and 250,000 when we finish. This will be a very squashed ellipse.
Earth to Mars:
Earth orbit is 149.6Mkm, roughly.
Mars orbit is 228Mkm, roughly.
A much less eccentric ellipse on the path between these.
If I have the time, I’ll dig up the formula later on calculating the minor axis from the location of the focal points.
I wouldn’t stress too much about it… they’re mostly way-high miles.
According to this summary of the Apollo 11 mission on Nasa’s site:
http://history.nasa.gov/SP-4029/Apollo_11a_Summary.htm
(toward the bottom in the “Recovery” section)
I guess the distance travelled while orbiting the Earth and the Moon are included, but it seems to me that would be a small part of the total. Given that, and assuming the distance is about the same both ways (something I don’t know for sure at all) then the one way distance would be about 400,000 nautical miles, or about 460,000 statute miles.
Since the distance from the Earth to the Moon is about 238,000 miles, a “straight line” round trip would be 476,000 miles, which means they went an extra 352,000 miles, give or take.
A rough guesstimate. Take the straight line distance. Multiply by pi for round trip distance. Divide by 2 for one way distance. My WAG is that will get you to within 25 percent of the right number (and its probably on the high side).
I’d be curious to as to how many miles from earth they would be able to let off the gas from earth. At what point roughly, does the vacuum of space able to continue to propel the rocket or spaceship?
Pretty much immediately. For the Apollo missions, once in orbit, about twelve or thirteen minutes into the mission, the vacuum is good enough to essentially ignore any friction from air. If I recall correctly, about 150 miles or so up is pretty much air friction free.
After a couple of orbits around the Earth, the Apollo missions made a single burn of the engines lasting just a few minutes for “Trans-Lunar-Injection”, i.e. break out of Earth orbit and head for the moon. This single thrust of the engines (lasting about 4 minutes if I’m reading the timeline correctly) was enough to accelerate the spacecraft to a speed that would take them (coasting) all the way to the moon.
After the TLI burn, Apollo gradually slowed down from earth’s gravity pulling on it (pretty substantially too). It wasn’t until they got pretty close to the moon, that lunar gravity finally overcame earth’s, and started to accelerate them toward the moon.
The Apollo spacecraft was initially placed into an orbit 118 miles high with a speed of about 17500 mph. After checking all systems and confirming that there weren’t any problems, the third stage fired again for a bit over five minutes, boosting the spacecraft to about 24000 mph on a long elliptical orbit that would take it to the Moon. That’s the point where it’s at its top speed.
From then on the spacecraft coasts “uphill”, continuing to slow until it reaches a point where it’s close enough to the Moon for its gravity to be a stronger influence than the Earth’s (that’s at about 200,000 miles from the Earth). That’s the point it’s going the slowest, about 2000 mph iirc. It then begins to accelerate “downhill” to the Moon, until they fire the service module engine to slow it a bit and place it in lunar orbit. Their speed in lunar orbit was around 3600 mph.
At no point is the vacuum of space propelling the space vehicle. Rockets are fired to change the vehicle’s orbit, but the rest of the time it’s coasting. The vehicle is in a circular orbit around the Earth until the engines fired to place it on a long eliptical orbit that will intersect the orbit of the Moon. If a space vehicle fires it’s engine enough to leave Earth’s orbit headed for Mars, it would then be in an orbit of the Sun that is eliptical enough so that it will intersect the orbit of Mars.
Absolutely true. I interpreted the question as the more correct “At what point does the vacuum of space reduce the friction to the point where it no longer hinders the vehicle?”.
Yeah, didn’t word it as good as I should, but thanks, this space travel is interesting.
I don’t know if it’s fair to use the orbital distance traveled as distance ‘traveled’ by the rocket, since it would be traveling that distance even if were nailed to the pad at the Cape. So even though the transfer orbit is an ellipse that intersects the orbit of Mars, i think the true distance it travels is really better represented by the absolute difference in distance between Earth and Mars when the rocket arrives. Or perhaps a better measure would be taken by calculating the delta-V over time with respect to the Earth, or something like that.