Gravity Beam

General Questions
#1

Lets say you took quantity of mass equivalent to the earths mass and through some kind of unbelievable technology formed it into a solid shaft ten miles in diameter and X miles long. How big is X?

Now, once you have that shaft, is the gravity felt at the ends stronger than if you were standing on the side in the middle? It seems like you would have all that mass lined up below you when you’re at the end whereas you’d only have ten miles of mass below you at the middle. So it seems like the shaft is generating a gravity “beam” from each end. Or does the shaft as a whole still retain the 1G level of attraction in all directions?

#2

X is simple math, but I’d have to look up the volume of the Earth.

The gravity in different positions is trickier, but solved problems. If you stand at the end with the whole rod under you you’d feel the attraction of the whole Earth, but a lot of it would be much further away in this configuration that in the ball configuration. On the other hand less would be at an angle. My wag would be that you’d experience less than 1G in that position.

#3

The Earth has a volume of about 260,000,000,000 cubic miles. Using that figure, I get a length of about 3,250,000,000 miles (35 au, so about the distance from the Sun to Pluto) - At the ends, gravity would be pretty weak (compared to Earth’s usual gravity) - most of the mass would be a lot further away than the mass ever gets from someone on the Earth’s surface. At the middle of the cylinder, I’d use the approximation here Gauss's law for gravity - Wikipedia which amounts to 2 G / r times the linear density of the cylinder - and r would be 5 miles - which suggests that the gravity on the surface of the cylinder would be quite a bit higher than gravity at the ends (if I feel less lazy later , I’d do the calculation).

A long way off from the mis-shapen Earth, the effects of shape would “smooth out” so the gravity felt from Earth at Alpha Centurii would be pretty much the same as what Alpha C currently feels from Earth

#4

It seems my plans are once again thwarted.

#5

Ignoring edge effects acceleration could calculated it simply by integrating -2⋅G⋅m(X)/X3 over the length of the shaft, which is way easier than calculating the acceleration of a spherical or shell distribution. The ‘surface gravity’ at the end will depend on the length of the shaft and the mass distribution throughout the rod (e.g. whether it is all high density, like the iron/nickel core of the Earth, or lower density, like the mantle and crust, or some mix thereof), though obviously the the mass closer to the surface has greater contribution; at some point you could just ignore the portion of the length at a certain distance as a rounding error.

Stranger

#6

I presumed a homogenous mass.

#7

Back of the envelope calculations suggest that the gravity at the surface of the middle of the cylinder would be on the order of 1/800 g

#8

Would the magnetic field of such a cylinder affect the gravitational field?

~Max

#9

Using the Gaussian result I mentioned above (the same method described by @Stranger_On_A_Train, I think), I get the more exact result of 0.002 g (or 1/500th of a g)

#10

Would a uniform homogeneous cylinder have a self-generated magnetic field? Earth’s current one is a function of its nickle-iron core rotating.

#11

I had assumed there would still be a molten core, but doh! - 10mi in diameter isn’t going to be enough for that.

~Max

#12

The way the OP is written, are we assuming the “mass” has the same density as the Earth?

#13

Sure. If that helps calculations.

#14

That’s the assumption I used for my calculations.

#15

I should point out that the material that make up Earth are no where near strong enough to form a cylinder billions of miles long and only 10 miles across - it will start to collapse back to a sphere (or several spheres) in short order. Again, I haven’t done the calculations, but I suspect there’s no material strong enough for that.

#16

Thus, the unbelievable technology.

#17

Scrith! Ringworld - Wikipedia Scrith | Larry Niven Wiki | Fandom

#18

Scrith…the material that requires a tensile strength of approximately the residual strong force.

Stranger

#19

Would that be neutronium (or, as the video calls it, nuclear pasta, a quintillion times stronger than steel)? [queued to the relevant bit]

Obviously impossible to access or use though.

#20

If you are ok with science-fiction then gravitic whips have been contemplated. It’s all hocus-pocus but sci-fi will allow it.