Monday, 11 May 2020

gravity - Escape Velocity of Asteriod 243 Ida


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.)



Answer



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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