Free FallThe Free Fall Research Page
Margo Schulter provides this response in regard to a request on the Questions page:|
Please let me try to address one facet relating to the topic of freefalls on the Moon, a topic on which you have had some questions: the need for some braking device if one wants to fall a substantial distance such as the 2,000 feet you use in one of your scenarios.
Actually, one of your tentative replies sums up the governing principles quite well -- and I should note, following your wise example, that I write as a rank amateur when it comes to physics! The basic idea is that since the acceleration of gravity on the Moon is not quite a sixth that of Earth, it is necessary to fall for a tad more than six times as long, and to cover a tad more than six times the same distance, in order to attain the same freefall velocity, "all things being equal."
Of course, all things are emphatically unequal because of the drag imposed by the Earth's very substantial atmosphere, in contrast to the negligible lunar atmosphere! But, putting that aside, it's easy to show that we'll want a braking mechanism of some kind for that 2,000-foot lunar freefall.
Let's use the metric system, and make things a bit simpler by rounding our freefall distance of 2,000 feet to 600 meters, about 1968.5 feet. Under the lunar acceleration of gravity, during each second of our fall our velocity increases by 1.622 meters per second, which we'll round to 1.6 meters per second.
To determine the duration of our 600 meter fall, and the impact velocity if there were no braking, we can use the same formula as on Earth:
d = .5at^2
Here _d_ is the distance covered, _a_ is the constant rate of acceleration (here due to gravity), and _t^2_ is the square of the time. We can state this formula in something closer to English as follows:
distance = (half of the acceleration) x (time squared)
Thus we have:
600 = .8t^2
That is, our distance covered of 600 meters is equal to half the acceleration (half of 1.6, or .8) multiplied by the square of the time to cover this distance. Dividing both sides of the equation by 0.8, or multiplying them both by 1.25, we get:
750 = t^2
So we know that the time for the fall in seconds equals the square root of 750 -- about 27.4 seconds.
Our velocity increases at the constant rate of 1.6 meters per second, likely a more serviceable approximation on the Moon than on Earth with its nonnegligible atmospheric drag! So in 27.4 seconds, we would reach an impact velocity of about 43.8 meters per second.
A velocity of 43.8 meters per second -- around 98 miles per hour or 144 feet per second -- isn't too far from a typical terminal velocity on Earth of around 55 meters or 180 feet per second or 120+ miles per hour, and intuitively I'd guess that we'd want some braking device, possibly a retrorocket pack or the like as a first naive impression of one candidate technology.
We can also confirm the conclusion of our equation in more commonsense terms. From a terrestrial perspective, lunar gravity may seem very small, as shown in those leisurely jumps or hops of the Apollo astronauts. But in freefall, even an increase in velocity of 1.6 meters per second quickly adds up: 1.6 meters per second is about 3.6 miles per hour.
If we get into a car and accelerate from zero at 3.6 miles per hour each second, in 20 seconds we'll be up to 72 miles per hour; in 30 seconds, to 108 miles per hour; and in 40 seconds, to 144 miles per hour, the kind of velocity a skydiver might have during a terrestrial freefall. Any of these speeds seems "too fast for comfort" when colliding with a nonyielding surface, terrestrial or lunar.
I can hardly add to your intuitive suggestion that maybe "moon dust and sloping craters" might play a mitigating role analogous to that of terrestrial snow, trees, and sloping terrain; or your caution that our concerns in a lunar setting might include not only avoiding undue direct injuries from a freefall but preserving the integrity of a space suit or other portable life support system.
Finally, there might be another way of asking this question: "How far could I freefall on the Moon and land safely without any braking system?" Let's assume that our spacesuits or other life-sustaining gear, as well as ourselves, could safely sustain an impact at 14 miles per hour, about 6.25 meters per second, a typical impact velocity for a terrestrial parachute jump.
Under lunar acceleration of gravity, our velocity increases at 1.6 meters per second, so the duration of our safe freefall with impact at 6.25 meters per second will be about 3.9 seconds. From what height can we start in order to have a safe freefall of this duration?
We can again use our formula:
distance = (.5 acceleration) x (time squared)
Here we know the time for our freefall, 3.9 seconds, and also the acceleration at 1.6, of which half is 0.8. So the square of the time is 15.2 seconds, and we have:
distance = .8 x 15.2
Multiplying 15.2 by .8, we find that our safe lunar freefall distance is about 12.2 meters or 40 feet -- under the assumption that at a given lunar site, an impact at 14 miles per hour is safe for us and our life support systems!
In conclusion, even in the Moon's reduced gravity, freefall acceleration still "adds up."