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I was reading about this asteroid (apparently, it has a moon, isn't that awesome?) and I started thinking about if I was on this asteroid, and I jumped, would I fall off?

It's been a while since I did something like this, so I went to Wikipedia to get some formulas. There I found

$$v_e = \sqrt{\frac{2GM}{r}}$$

Where

  • $G$ is the gravitational constant ($6.673 × 10^{-11}\ \mathrm{m}^3\mathrm{kg}^{-1}\mathrm{s}^{-2}$)
  • $M$ is mass of the body (found to be $4.2 × 10^{16}\ \mathrm{kg}$)
  • $r$ distance from center of mass (found to be $53.6\ \mathrm{km}$)

$$v_e = \sqrt{\frac{2(6.673\times 10^{-11})(4.2\times 10^{16})}{53600}} = \sqrt{104} = 10.22\,\frac{\mathrm{m}}{\mathrm{s}}$$

I'm a fairly fit guy, about 73kg and .6m vertical leap here on earth.

I found a formula for the speed of a free falling object to be

$$v = \sqrt{2gd}$$

If I assume that my speed when I hit the ground is the same as when I jump up, that means my jump speed is about 3.4 m/s, which is significantly less than the required 10.4 m/s.

So it would seem I would not be able to jump off of this asteroid.

Are my formulas, assumptions, and calculations correct here?

Also, how would I use this information to determine how high I would jump (would it be enough to freak me out? Well I'd be jumping off an asteroid, so it would freak me out anyway... although I imagine I would be going slow enough for (my ship?) to pick me back up.)

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2 Answers 2

up vote 6 down vote accepted

Looks as if your arithmetic is correct, and you could not escape 243 Ida. Given your leap velocity capability of 3.4 m/s, you would be able to leap 593 m above 243 Ida's surface, since its surface gravitational acceleration is only about 9.75E-3 m/s^2.

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Wow, I'd jump half a km up before falling back down? That would make for a fun ride! :-) –  corsiKa Apr 13 '11 at 14:44
1  
If I'm not mistaken, the total time you'd be in flight would be almost 700 seconds. Lot's of time to enjoy it. Scarey. –  Michael Luciuk Apr 13 '11 at 17:24

"So it would seem I would not be able to jump off of this asteroid."

That APPEARS to be correct. There are two adjustments that need to be made to the formula you used, plus other corrections; but I don't think they would get you permanently off 243 Ida.

"Are my formulas, assumptions, and calculations correct here?"

Calculations appear to be good, but there are two 'mis-assumptions' which invalidate the formulas you used, plus one error for constants (radius), plus plausibly one error in initial velocity: 1.) The geometry of 243 Ida is not even close to spherical. 2.) The formula you chose for speed of a falling object assumes a constant gravitational field, which is far from the truth. I think the easiest to use formula would be:

(v^2)/2 = Integral of (g(h))dh, where 'h' is height above asteroid surface, and 'g(h)' is a function of h.

3.) the dimesions of 243 Ida are roughly 54km x 24km x 15km; so the largest value for 'r' would be 54km/2 = 27km. 'r' could be as small as 7.5km. 4.) Your jump velocity of 3.4m/s is in Earth's gravity. That means you are fighting your own weight. The excess force OVER and above your weight is what is used to generate your jump velocity through the mechanism of F=ma. What do you think your initial jump velocity would be if you were hooked up to a bungie cord which lifted you with a force of (your body weight minus 1 ounce)?

"Also, how would I use this information to determine how high I would jump (would it be enough to freak me out? Well I'd be jumping off an asteroid, so it would freak me out anyway... although I imagine I would be going slow enough for (my ship?) to pick me back up.)"

The near field gravity on the surface of 243 Ida varies greatly, depending on where you are. To understand this better, look at a pic of the rock. It is roughly barbell-shaped. Try to imagine if we took this to the extreme, and 243 Ida was shape like two identical spheres, with a very small connecting waist. What would be the gravity at that waist? Very close to zero, since the forces of gravity of the two spheres would cancel each other out.

I'm not going to try to work out the integral, here; but will make a few assumptions which will show how much the answer given by the standard formula could be off: The standard answer for surface gravity on Ida 243 is 1.1 cm/s^2 down to 0.3 cm/s^2. Assuming it is 0.3 cm/s^2 at the waist and remains constant (which is not true, but wait...) AND you jumped with an inital velocity of 5m/s (a conservative estimate), you would attain a height of 4167 meters. Now, assuming r = 7500m at the waist, you might be tempted to think that gravitational pull would fall off by a factor of (4167/7500)^2 = 0.31, and so your jump would actually be higher; BUT, because of the geometry of the problem (two point sources of gravity instead of one), I think gravity would actually increase with height, as the two lobes of the asteroid started appearing more as a point source.

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The potential from two point sources is actually pretty easy to do. Also the asteroid is likely spinning, and you would want to jump in the direction of the spin. One possibility for the reason there are a lot of binary asteroids, is that they can get spun up by assymetries in solar heating and radiative cooling forces until some of the stuff falls off and becomes a satellite. –  Omega Centauri Apr 13 '11 at 18:11
    
Wow, Jeff, this is a very thorough post! I was assuming the gravity would be lowest "at the end of the barbell" so to speak, which is why I used it as the radius, and was assuming I'd jump straight up (as in, away from the center of mass) but your point about the extreme scenario is an interesting one. Also your point about initial velocity means I very well may be able to accelerate myself to 10 m/s or more! I would have to hope to bounce into another asteroid to get back. –  corsiKa Apr 13 '11 at 18:52

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