Hammer & Feather II

Time I posted some video of my favourite astronaut experiment. This is such a simple demonstration, and Dave Scott carries it off well. He actually carried two feathers with him as he wasn’t sure how the static building up as he moved around would allow him to drop the first one. He didn’t need it, of course, but no-one knows what happened to the second feather. Do they?


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8 Responses to “Hammer & Feather II”

  1. John Smith Says:

    h = 1.5 m, g_moon = 1.622 m/s^2, then the free-fall time interval should be t = sqrt(2h/g_moon) = 1.36 sec.
    However, the hammer is in free fall for at most 1.1 seconds.
    Is the moon gravity at that particular location stronger than usual or there is some other explanation?

  2. saturn5 Says:

    Well, I haven’t checked your maths, but 1.5m is almost head height – 1.1m is more like it, wouldn’t you say?

  3. John Smith Says:

    According to the NASA web page, the height was even 1.6 m possibly due to the lunar boots.

    “During the final minutes of the third extravehicular activity, a short demonstration experiment was conducted. A heavy object (a 1.32-kg aluminum geological hammer) and a light object (a 0.03-kg falcon feather) were released simultaneously from approximately the same height (approximately 1.6 m) and were allowed to fall to the surface.”

  4. saturn5 Says:

    Well, that’s a remarkable figure. It suggests to me that the astronauts were walking on soles that were 30 cm or more thick (I would guess they were two or 3 cm)! I can’t explain your question away. Possibilities may be: film speed (I get less than a second timing from the clip above); it really is only about a metre or so; Scott dropped them into a hole/he was standing on a mound; weird gravity (seems unlikely to be that different); the whole thing was a hoax.

  5. Doug Says:

    Does anyone have a reference to a site that will explain the math behind this assertion that the hammer and the feather fall at the same rate. I don’t buy it. They appear to fall at the same rate becuse the earth (or moon in this case) is the largset contributor to the gravitational attraction, but the reality is that when the feather and hammer are dropped, the moon is moving toward the hammer and feather at one rate (based on the combined weight of the hammer and feather) and the hammer and feather are moving at a different rate. I realize the difference is very, very small, but still I think the argument is and has always been inaccurate. btw; I did the whole calculation in physics calss where the masses cancel out, but this was ignoring the fact that G and g are not the same thing and also very conveniently used the formula Fa = Gx(m1xm2/r^2) where G is the gravitational constant for the earth (or moon)…but G for the feather and hammer are conveniently left out. The only thing this feather and hammer question answered is..if you can dictate an answer by how you ask the question.

  6. saturn5 Says:

    Hi Doug. I’m not sure I understand your problem here, but let me try. As you can see from the video (or by repeating the experiment in vacuum on Earth) the objects (whatever they are) do fall at the same rate. You go back to Galileo and Newton here. Galileo’s famous demonstration with cannonballs showed that, although there was a tiny difference in times of landing due to air resistance, there wasn’t the big difference that Aristotle et al had asserted.

    Newton explained that F=ma (2nd Law). Simple really – is this where ‘masses cancel out’? Two objects – say 1kg and 10kg – fall. The 10kg mass is pulled 10x harder than the 1kg mass. But the 10kg mass is 10x harder to accelerate. So the acceleration is the same for both as a result – they both fall at the same time. a=F/m.

    It’s safe to consider only one side of the system and ignore the Moon or Earth here as ‘each force has an equal and opposite force’ – 3rd law. The Moon/Earth will of course accelerate toward the dropped object, but due to the enormous mass difference the acceleration is very tiny (the force between them is the same). So the difference in acceleration is in fact massive, not, as you say very small.

    Don’t know why you’d need the equation you quote or need to know g (or G) for the hammer. Good old F=ma is all you need. If you want to know how big the acceleration due to gravity is, that’s g (Earth 9.8m/s², Moon 1.6m/s²). Doesn’t matter what you drop, that’s how fast it accelerates, if you can ignore air resistance (as you can, of course, on the Moon). Strictly, it is how fast the Moon and the hammer accelerate together, but, as I say, you can ignore how much the Moon accelerates as it is negligible.

    Does that answer your question, whatever it was? 😉

  7. saturn5 Says:

    …as for G, that is the universal gravitational constant, and is the same whatever body you consider, be it planet or feather. The equation you quote gives the force between two objects a certain distance apart, so was appropriate for working out the force between the hammer and the Moon, say. More usually it is used to calculate the force between planets and the Sun, or between the Earth and the Moon.

    The value of G is approx 6.67×10^-11 m^3kg^-1s^-2. Look it up at Wikipedia.

  8. Mark Toner Says:

    Love the site and this is my favourite experiment, too. Looks like a good opportunity to spread the apparently counter-intuitive result of Galileo’s work. The weird distorted view we get of the universe due to living in a sea of air…

    Anyway, I thought you might like to check out my song, also called “The Hammer And The Feather” over on our band’s website (www.arbelos.eu). I wrote it for the big Apollo anniversary this year. Hope you like it.

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