How big would a planetoid have to be to keep me from jumping off and not coming down?

Gravity, as we know, is an incredibly weak force. So weak that it takes a whole lot of stuff in one place for its effects to become noticeable.

So, how much, really?

Let’s postulate that our planetoid has roughly the same density as the Earth, and is perfectly spherical; no high places to jump from. Let’s also postulate that I am a 165-lb. (~75kg) Olympic athlete who can run and jump with the best of them. And finally, let’s postulate that my life-support equipment doesn’t hinder my motion in any significant way. (After all, if this planetoid’s gravity is just barely strong enough to keep me down, it’s not going to have an atmosphere to speak of. If necessary for the problem, assume no atmosphere at all.)

So, how big does the planetoid have to be, to make sure that if I go up, I come back down?

Along the same lines, suppose I can drive a golf ball as well as anyone the PGA Tour has ever seen. How big does the planetoid have to be so that my golf ball must resurface, so to speak?

I think you also have to postulate and otherwise empty universe; otherwise, you might push off and be prevented from falling back toward the planetoid because of another gravity well (whereas otherwise, you might have fallen back, but only after a very very very long time).

I think what you want to calculate is escape velocity.

OK, easy part first: how much velocity can your legs produce. Let’s SWAG that you can leap 1 meter vertically on the Earth. Velocity = sqrt(2as) = sqrt(29.811) = 4.4m/s.

The density of the Earth, on average, should be = [3/(4[symbol]p[/symbol])]*(M/r[sup]3[/sup]) = 5500 kg/m[sup]3[/sup], which means that, to maintain the same density, the quantity (M/r[sup]3[/sup]) for your planetoid should equal 23,000.

Using the escape velocity formula V = sqrt (2GM/r) and substituting the value of G and the 23000, we get V = sqrt (2*6.67X10[sup]-11[/sup]*r[sup]2[/sup]*23000).

Since V = 4.4 m/s, the only unknown is r, and I get r = 2511 meters. So roughly 5km diameter, assuming my math is correct.

I’d have to ponder this and it is a wonderful question, but I don’t believe the same muscular effort that produces an upward velocity of 4.4 m/s on earth would produce the same upward velocity on your planetoid. I would think the initial upward velocity on the planetoid would be higher.

Hmmm. Good point, actually, but I tentatively think that the 4.4 m/s is still pretty reasonable, for two reasons: First, the speed with which you can flex your leg muscles is a limiting factor (although maybe not the limiting factor, if you see what I mean).

Second, and more important, the other limiting factor is the amount of energy your legs can produce. If you think about the mechanics of a vertical jump for a moment (squatting and then propelling yourself upwards), you’ll see that at the point when your legs leave the floor, the energy is in two forms: potential energy (the height you’ve lifted your center of mass out of the squat) and the kinetic energy (the speed with which you’ve accelerated your mass). Since you’re eventually jumping a meter off the ground, I’d argue that the kinetic energy is greater by a factor of around three.

On the planetoid, the kinetic energy required to accelerate to 4.4 m/s is still the same, because your mass is still the same. However, the potential energy required to rise from a squat would be less, by a lot, and thus this portion of energy would potentially (ha!) be available for conversion into additional KE.

However, since there’s only an extra 1/3 proportion of energy, and velocity is the square root of same, that would at best increase the velocity to about 5 m/s, and that’s ignoring my first point above. Granted, I’m making some assumptions about physiology here (is muscle speed a function of load? I dunno.), but I’d say that any error here is on the order of some other assumptions (like the 1 meter vertical distance), and using 4.4 m/s is still pretty reasonable.

You’re right, but it doesn’t make all that much difference.

Let’s play a bit with zut’s numbers.

Firstly, we’ll assume that your body’s center-of-mass travels upwards one metre before your feet leave the ground, and we’ll accept zut’s assumption that it travels another metre into the air after that.

You accelerate upward for 1 m at 1 g to reach 4.4 ms[sup]-1[/sup], then accelerate downward for 1 m at 1 g to reach 0 ms[sup]-1[/sup].

When you’re accelerating upward at 1 g, you’re overcoming the earth’s gravity as well, and so your leg muscles are exerting a force equivalent to 2 g. You’re able to exert that force regardless of the prevailing gravity.

Now let’s try it with lower gravity, say 0.1 g. In the first metre when you’re accelerating, you get

v[sup]2[/sup] = 0[sup]2[/sup] + 2 x (2 - 0.1) x 9.8 x 1

v = 6.1 ms[sup]-1[/sup]

In the extreme case, when there’s no gravity, you get

v[sup]2[/sup] = 0[sup]2[/sup] + 2 x (2 - 0.0) x 9.8 x 1

v = 6.3 ms[sup]-1[/sup]

That’s the fastest you can jump (given zut’s initial assumption that you can jump 1 m in earth’s gravity). Not very much greater than zut figure of 4.4 ms[sup]-1[/sup].

Now, working out the size of the planetoid, and using zut’s assumptions about density, you get

6.3 = sqrt (2 x 6.67x10[sup]-11[/sup] x r[sup]2[/sup] x 23000).

r = 3600 m, which isn’t very far off zut’s figure of 2500 m. The initial estimation of how high you (or an olympic athlete) can jump on earth is the critical thing to be determined.

First, a nitpick: What we’re interested in here is escape speed, not escape velocity. Velocity has a direction, but if you launch a projectile in any direction at all at escape speed, it’ll escape (so long as it doesn’t hit something first, like the planet).

Now, then, for the golf ball. I understand that on Earth, the longest drives are on the order of 300 meters. If we neglect air resistance (which is admittedly probably not justified), the maximum range for a projectile, launched at an optimal 45 degree angle above the horizontal, is given by R = v[sub]0[/sub][sup]2[/sup]/g . This gives us an initial speed for the ball of 54 m/s, which is probably somewhat low (see above about air resistance).

So, we want a planetesimal with the density of the Earth and an escape speed of 54 m/s. Unless my calculations are off, this would imply a radius of about 33 km. I also find that for a constant density, escape speed at the surface should be directly proportional to radius, so if my golf ball speed were low by a factor of two (that is to say, if the correct speed is 108 m/s), then the radius required would be 66 km.

This site says 140mph (63 m/s), so your 54 m/s ain’t bad. Dunno if that’s an “average” or “maximum” drive, but I suspect it might not matter, as I would think that drive length on Earth is more a function of how well you hit it that how hard. I’m not a golfer, though, so I’ll defer to Sam Snead.

The image of standing on some distant planet(oid) and driving golf-balls into the great unknown is very nice. Thank you for brightening my day :slight_smile:

The best part, IMO, is that if you get exactly the right diameter (mass) of the planetoid, the golf balls go into orbit. You can stand there and drive all day and they don’t come down but they don’t excape either. Eventually you have this (metaphorically) buzzing mass of 4 cm satellites whizzing around your planetiod. Now THAT’s a mental picture I LIKE!

escape…really, I don’t say or think “excape”–it’s just a typo. The X is right below the S and my finger slipped. Really.

According to the Discovery Channel show 95 Worlds And Counting, an Olympic high jumper on Phobos (maybe it was Deimos) could launch himself into space under his own power. That’s assuming he’s not wearing a spacesuit, of course…

would it be possible to put a golf ball into a stable orbit from the surface of a small planetoid? It doesn’t seem like it should be possible because you’re not able to correct the trajectory at any point after driving the ball - I think the resulting orbit would eventually intersect with the planetoid’s surface.

I have this wonderful image of the golfer smacking a couple of shots off the tee, standing there having a good chuckle with his buddies and then a little while later…

WHACK! His low-planetoid-orbit ball comes sailing up over the horizon behind him and pops him in the helmet.

I think Kim Stanley Robinson made the same point in his Martian trilogy. I know that’s science fiction, but he seems to have done his science homework before writing the books.

You could throw the golf ball (better yet, a baseball) parallel to the local surface of the planetoid (tangent) at just the right speed so it could stay the same height from the ground as it circles the planetoid, so you can grab a bat and wait for your pitch to come around…

Just tee it up, and then use a reeeeeally low-loft driver to hit it nice and flat. By the time the ball comes around to the exact same position, the planet will have rotated the tee out of the way. :smiley:

And also Heinlein, in The Rolling Stones.

Physics tidbit, by the way: For two spherical bodies of the same density, the surface orbital period is the same. That is to say, an object orbiting just above the surface of the Earth takes about an hour and a half to make a complete orbit, and an object orbiting just above the surface of an asteroid with the same density as the Earth will take the same amount of time.

The dimples on the golf ball, combined with the spin of th ball, create a lot of lift (assuming you hit the ball here on earth). This is a large part of the reason the ball flies so far. It’s not just the initial kinetic enery imparted to the ball. Also, a typical drive leaves the club head at a much lower angle than 45 degrees.

No, the initial kinetic energy is ALL that drives the ball. It does not gain more energy on its way from nothing.
Rotation may create a lift as you described, but rotation is just another form of kinetic energy.
Now I don’t know whether the kinetic energy is devided into forward movement and rotation right at the beginning, or whether air resistance turns forward movement into rotation later on, but be sure that all the energy is there right from the start.