I am curious about the smallest object in space that would have a functional gravitational force. My operational definition of “functional gravitational force” is that if I was on the object, I couldn’t achieve escape velocity by jumping vertically with my own power. FWIW I’m 160 pounds (I’m assuming that doesn’t matter, but I don’t know what I don’t know here), and say on earth I can get a vertical jump of 30 cm. (Yes, pathetic, I know)
Bonus questions - if I did jump with that force, how far would I go from the object before being returned to its surface? And if I did jump from the surface of the body, would I would visibly displace it from the equal and opposite reaction?
I did some googling on this, but found the equations for calculating escape velocity beyond me.
No, this isn’t homework. It’s something I’ve wondered about ever since seeing The Little Prince as a kid, with that scene on his home planet where the little prince is taller than the diameter of the planet (if memory serves).
A Neutron star’d hold you down, and some of them are only 7.5 miles in diameter.
For something made of ordinary matter:
Ceres (D= 487 km) would hold you down with 0.028g, and an escape velocity of 0.51 km/s (60 mph = 0.026 km/s).
Phobos, with a diameter of 14 miles, and an escape velocity of 25 mph is about at the limit of jump-offable size. Surface gravity there is about 0.008g, so moving around would probably involve a lot of inadvertent orbiting.
Phobo orbits at only about 9500km from Mars. Taking that into account, the escape velocity would be lower when jumping off the Mars-facing surface, and higher when jumping away from Mars. I’m not sure if the difference would be significant.
Also, Phobos is an irregular shape, so you could maximise your chance of being able to jump off it by finding the point furthest from it’s centre of mass.
I get all confused when I dabble in understanding actual physics stuff, but by your definition, it seems to me even a pencil eraser would eventually pull you back down (assuming the eraser and you were the only two objects in the universe involved in the gravitational equation at hand). By “pull you back down” I mean reunite with you; obviously what is going where is a frame-of-reference definitional issue.
As you make your feeble leap off the eraser, its (putative) gravitons start exchanging, and as massless particles they have no physical distance limit. Your initial escape speed starts slowing down, and although you and the eraser (or anything any amount smaller that still has mass) may get pretty far apart, eventually you are going to stop, accelerate toward one another down the gravitational field, and come together again. (Assuming we can, for the sake of looking at the concept, subtract all the other gravitational fields acting on you and the eraser).
Your dilemma, I think, is that by your definition you are not adding any energy beyond the initial jump…it’s not clear to me you’d be able to jump your butt permanently off even a speck of bee poop.
Yes, but it’s in orbit, which means that the gravitational force from Mars is, on average, exactly balanced by the centrifugal force. The balance doesn’t occur equally over the entire body, though, leaving you with tidal forces that would make it easier to jump off from either the point facing Mars, or the point directly opposite that, than from points in between.
To the OP, two points of clarification: First, what do you mean by “smallest”? Do you want lowest mass, or smallest radius? Second, do you want objects that are specifically known to exist (like Phobos), objects that aren’t specifically known but which presumably do exist somewhere in space (like “A spherical iron asteroid with a radius of X”), or objects which aren’t even known to exist at all but which might (like “A primordial black hole with mass of X”)?
No, this isn’t true. By reaching escape velocity, you would never be pulled back to the parent body, even if the two bodies were the only two in the universe. People get confused over this, because they think gravitational attraction will always bring two objects together, given an infinite amount of time. However, gravitational attraction is inversely proportional to the square of the distance between the bodies. Gravity will forever be reducing the speed of the escaping object, but the rate at which it does so reduces as it recedes, so it’s velocity never reaches zero.
No, while he and the pencil eraser would slow down relative to each other, they would asymptotically approach some constant speed, and continually always get further and further apart. Escape speed (often mistakenly called escape velocity) is the speed of initial jump such that that asymptotic speed is zero: Any initial speed less than that, and the two objects will eventually re-collide, but at escape speed or greater, they never will. The escape speed of a pencil eraser is tiny, though, so D18 isn’t going to have any difficulty at all jumping off of it, and the asymptotic speed will be very, very close to the initial speed of the jump.
I think my limit for functional gravity would have to much larger than the minimum to prevent you from jumping into orbit. Imagine how frustrating it would be to cross a room if every step had you floating ten feet off the ground and slowly descending. On Phobos, I’m pretty sure you’d navigate just like folks on the space shuttle - you would just push off of walls or pull along ropes. A little vertical push every so often would keep you from settling on the ground. After all, .008 g is only 8 mm/s^2. That’s 3 seconds just to “fall” the first inch of height. If you were the proverbial kid jumping off his roof, you could practically read a book on the way down.
So… if I were the OP, I think I’d be looking for places with at least 0.1 g or so.
Phobos is so small you could easily throw a baseball at greater than escape velocity. But you probably couldn’t jump off unaided, it might be just within the capabilities of some people, the problem is that you couldn’t get a running start because running would be impossible. And you’re right that Phobos would seem almost like a zero-g environment, except you’d eventually drift towards the surface really slowly.
Oh, I thought of a way to launch yourself via muscle power alone. You set up a track of blocks. You push off from the first block, not upwards, but parallel to the ground. As you reach the second block, push off from that, then the next and the next and the next, until you reach arbitrary velocity. It would be tricky because if you give yourself any upwards velocity your legs won’t be able to reach the later blocks, and if you have too much downward velocity you’ll plow into the surface at a pretty good clip, like crashing a bike. On the upside you’d probably bounce off the surface on Phobos, rather than absorbing all that kinetic energy like on Earth.
Assuming you are the only objects in the universe, there isn’t a distance past which you would never return. The distance at which you go will grow exponentially as you get closer to escape velocity, but barring some weird quantum gravity thing, as long as no objects interfere you can get arbitrarily far away from the other object and still come back, provided you don’t accidentally give yourself escape velocity by mistake (since to achieve long distances you will be pushing off extremely close to EV)
My physics is a little rusty, so any physicists out there should feel free to correct me.
Its probably easier to talk about escape energy rather than escape velocity. Lets say on earth you have mass M and can jump 1 meter into the air, then the potential energy you can gain by jumping is M*9.8(m/s)^2.
If you are on an planet with radius R and average density D, then escape energy is M4/3piGD*R^2
Solving for all the numbers we get that for a jump to equal escape velocity
R in meters =187,286/Sqrt(D) and that
mass is kg is 2.7512^16/Sqrt(D)
where D is in units of kg/m^3
So the denser the better both in terms of minimum volume and minimum mass.
Black hole material would be the most dense, having infinite escape velocity, and 0 volume for any finite mass, but it would either suck you into a singularity or evaporate via Hawking radiation.
Next best would be Neutronium with a density of 4x10^17 kg/m^3. This would yield a radius of 0.3mm and a mass of 43 million kg. But this too is probably unstable (physicists help me here).
For stable elements the densest would be Osmium at 22,610 kg/m^3. This yields a radius of 1.24km with a mass of 1.83*10^14kg
For a natural planet type material, using the density of the earth 5.515kg/m^3 as a starting point leads to a radius of 2.5km and a mass of 3.7*10^14kg
Your first bonus question: There is no answer. If your leap were at the exact escape velocity, your speed would decrease asymptotically to zero as you receded infinity far from the asteroid. You could increase the mass any infinitesimal amount you want to get any return time you want.
Thank you all for the discussion so far! I’ve enjoyed reading it.
Chronos in answer to your request for clarification, I was thinking along the line of “a spherical iron asteroid with a radius of X” as that’s what I had in mind when I conceived the question
Buck So adjusting for my jump of 30 cm, I take it that the object of earth-like substance would become about 2 km radius. Wow. I had no idea it would be that small
Dracoi I see what you mean about my idea of functional, but what I had in mind was an object that basically I wouldn’t float away from if I was traveling through the solar system on it and which I could move around on if I did so gingerly. That’s why I tried to frame it as being able to avoid escape velocity if I purposefully didn’t jump
Now, I see what’s being said about bonus question 1, so let’s reframe it this way: if my maximum jump on earth is 30 cm, and that would create the exact escape speed (Chronos, I was paying attention), then let’s say I cut that back to the equivalent of a 25 cm jump on earth. How far from the surface of this spherical earth-material-like object would I go before starting my descent? I really like the idea of this huge, lazy jump which takes me way out into space but then allows me to gently fall back to the object!
This could come in handy one day when we are mining the asteroids.
Or in a hard SF story about mining the asteroids.
My guess is that the latter will precede the former.
Actually it would be about 1.3 km radius (0.54*2.5). Incidentally the rule is an increase/decrease in jump height by a factor of x, will increase/decrease the radius by a factor of Sqrt(X) and the mass by a factor of X^3/2.
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Let E the earth height of your escape velocity jump, and j be the earth height of your current jump, R is the radius of the planet, then your jump height will be
j*R/(E-j)
So for a 1.3km planet, a reduction of your earth jump height from 30 to 25 would give you a jump altitude of 6.5km, assuming you didn’t accidentally start orbiting the planet.
If you lived on a 2km planet (escape jump height = 71cm) your jump height of 30cm on earth would yield a jump of merely 1.46 km straight up.
Yes, it would be unstable. Neutron stars are only kept in the state they’re in by their extreme gravity, so there’s a minimum stable size for a neutron star. The precise value of this minimum size is unknown (we don’t know all the details of the equation of state of neutronium), but it’s somewhere in the vicinity of 1 solar mass and 10 km radius.
By the way, D18 you didn’t ask, but if my calculations are correct (a somewhat dubious assumption) the 6.5 km jump should take approximately 6 hours to complete
Hey Buck, you’re a step ahead of me! I had read your answer earlier, but didn’t get to respond, meanwhile I got to thinking about how long the round trip would take and had just sat down to ask that very question.
