What the Sun will do with long rod?

[Eagle9]

Here we have got a very long rod (100 million km length or so) built/placed on the lunar surface (near equator) and directed towards the Sun (imagine that Moon is in synchronous rotation with Sun and therefore that point (where this rod is built) is always looking at the Sun). The distance between these two celestial bodies is equal to 150 million km, so the rod covers two thirds of this distance.
I would like to know-will the Sun’s gravity destroy this rod? Rod’s closest point (relative to the sun) is 3 times closer than the points on the Moon, so the Sun will attract this end 9 times more than other end (according to Newton's law of universal gravitation its value is inversely proportional to the square of the distance between them)….so, what will happen? Will the Sun’s gravity be able to break the rod and attract broken piece(s)? Will the rod withstand Sun’s tidal forces? We know rod’s length, width, mass and all kind of property of the material that was used for manufacturing the rod (its specific strength for example), then what? :shy:

 

[mathman]

Mercury is about 60 million km from the sun and it is still there. If the rod is made up of strong enough material it should hold up.

 

[AlephZero]

Mercury is about 60 million km from the sun and it is still there. If the rod is made up of strong enough material it should hold up.

The reason Mercury and the other planets are "still there" is because their diameter is small in relation to their distance from the sun. Tidal forces are caused by the difference in gravitational attraction between the nearest and farthest points from the sun.

If your rod extended between the Earth's orbit and mercury's, the "natural" orbital period of one end of the rod would be 365 days, but 88 days for the other end. Very large forces in the rod are required to make everything orbit at the same rate.

Making the rod thicker doesn't work, because twice the mass needs twice the force to create the required accelerations.

 

[Vanadium 50]

I am certain the rod won't remain intact. Let's ask a simpler question - on a "flat earth" (i.e. gravitational strength is independent of height), what is the longest rod that you can stand on its end?

The compressive strength of steel is something like 400 MN/m². So we have:

\sigma > \frac{F}{A}

\sigma > \frac{mg}{A}

\sigma > \frac{\rho Ahg}{A}

h < \frac{\sigma}{\rho g}

Putting the numbers in, you see that there's a lint of 5 or 6 km. A rod that tall, standing straight up, will be under so much compression that it will yield. Since it will not yield in a perfectly axially symmetric fashion, when it yields, the top part will tip in some direction or other and the whole thing will come crashing down.

(Of course, a real rod with imperfections will yield even sooner)

You're talking about something 20 million times longer, albeit in a varying gravitational field. No way will it survive.

 

[DaveC426913]

If your rod extended between the Earth's orbit and mercury's, the "natural" orbital period of one end of the rod would be 365 days, but 88 days for the other end.

This is a very backwards way of examining the problem.

Parts of the rod (or anything else at a given point) do not have some sort of natural tendency to orbit. A rod based in the Moon will not have its tip "try to orbit".

It's not completely invalid - you can look at the forces needed to keep the end in freefall as if they are forces needs to keep it in orbit. It's just very strange.

Very large forces in the rod are required to make everything orbit at the same rate.

OK well this is definitely wrong, since the forces you're talking about would have to be large enough to streeeeetch the rod by a large fraction of it own length if its tips were going to remain in Earth and Mercury orbit!

 

[DaveC426913]

I am certain the rod won't remain intact. Let's ask a simpler question - on a "flat earth" (i.e. gravitational strength is independent of height), what is the longest rod that you can stand on its end?

The compressive strength of steel is something like 400 MN/m². So we have:

\sigma > \frac{F}{A}

\sigma > \frac{mg}{A}

\sigma > \frac{\rho Ahg}{A}

h < \frac{\sigma}{\rho g}

Putting the numbers in, you see that there's a lint of 5 or 6 km. A rod that tall, standing straight up, will be under so much compression that it will yield. Since it will not yield in a perfectly axially symmetric fashion, when it yields, the top part will tip in some direction or other and the whole thing will come crashing down.

(Of course, a real rod with imperfections will yield even sooner)

You're talking about something 20 million times longer, albeit in a varying gravitational field. No way will it survive.

But you're not taking into account that g drops off to virtually nothing within the first fractions of the rod's length.

Well, no matter - your point is made. If a rod cannot stand 6km high, then it doesn't matter whether it can stand 6000km high...

 

[AlephZero]

This is a very backwards way of examining the problem.
...
It's not completely invalid - you can look at the forces needed to keep the end in freefall as if they are forces needs to keep it in orbit. It's just very strange.

I don't agree. It seems entirely sensible to me to consider what is happening to a smal piece of the inner end of the rod. If it was in a circular orbit around the sun with no internal forces acting in the rod, it would have an orbital period one quarter of what the OP wants it to have. Therefore, there must be a tensile stress in the rod, such that the stress gradient along the length is enough to keep the inner part of the rod from falling into the sun. Integrating that gradient over 10^8 km is going to give one very large stress.

From your other post,

But you're not taking into account that g drops off to virtually nothing within the first fractions of the rod's length.

The moon's gravity has a negligible effect on the situation, except locally at the outer end of the rod. The whole of the rod would be in tension resisting the sun's gravity, right down to lunar ground level.

OK well this is definitely wrong, since the forces you're talking about would have to be large enough to streeeeetch the rod by a large fraction of it own length if its tips were going to remain in Earth and Mercury orbit!

It's not wrong if "the same rate" means "the same angular velocity", which is the situation the OP was talking about.

 

[DaveC426913]

I don't agree. It seems entirely sensible to me to consider what is happening to a smal piece of the inner end of the rod. If it was in a circular orbit around the sun with no internal forces acting in the rod, it would have an orbital period one quarter of what the OP wants it to have.

What does orbital velocity have to do with anything? The OP never mentions anything about it. It's simply hanging down from the Moon.

It's not wrong if "the same rate" means "the same angular velocity", which is the situation the OP was talking about.

The OP never mentions "the same rate". Are we having the same conversation?

 

[mistergrinch]

I just did a quick calculation where I end up with this:

G/D * [M_s(1/y + y^2/2) + M_p/(1-y)]

being the quantity to compare to the specific strength, which I took to be T/(A*ro) = T/mu
where ro and mu are density and linear density, A is cross sectional area of the rod, D = distance between bodies, y = (distance from the sun) / D.

So I think if you plug in y=1/3 and compare this to specific strength you will know if it breaks.

I did this quickly so you should double check this!

 

[Eagle9]

mathman

Mercury is about 60 million km from the sun and it is still there. If the rod is made up of strong enough material it should hold up.

Yes and I would like to know is it possible to calculate this

AlephZero

The reason Mercury and the other planets are "still there" is because their diameter is small in relation to their distance from the sun. Tidal forces are caused by the difference in gravitational attraction between the nearest and farthest points from the sun.

If your rod extended between the Earth's orbit and mercury's, the "natural" orbital period of one end of the rod would be 365 days, but 88 days for the other end. Very large forces in the rod are required to make everything orbit at the same rate.

Well…could you please tell me how this will happen? The rod is not a planet to orbit around the Sun…….actually it will but with Earth, so it will have 365 day period of orbiting, I think so

there must be a tensile stress in the rod, such that the stress gradient along the length is enough to keep the inner part of the rod from falling into the sun

I think that we should compare tensile strength (probably multiplied to rod’s cross-section’s are) with difference of Sun’s gravity at the end of the rod and its beginning….

mistergrinch

I just did a quick calculation where I end up with this:

G/D * [M_s(1/y + y^2/2) + M_p/(1-y)]

being the quantity to compare to the specific strength, which I took to be T/(A*ro) = T/mu
where ro and mu are density and linear density, A is cross sectional area of the rod, D = distance between bodies, y = (distance from the sun) / D.

So I think if you plug in y=1/3 and compare this to specific strength you will know if it breaks.

I did this quickly so you should double check this!

Until I check this I want to know something
First of all what is G is that formula? Gravitational constant? What is M_s and M_p? :shy:


Vanadium 50
Well, I guess we should use the lightest and the most strength materials for this purpose, for example Carbon nanotubes
Could you please tell me the meaning of these symbols? F, A, Mg….:shy:

 

[Vanadium 50]

F = force
A = area
M = mass
g = acceleration due to gravity

Note that you need something several million times stronger than steel under compression. Carbon nanotubes are about 100x as steel under tension (nowhere near a million) but poor under compression.

 

[Eagle9]

Vanadium 50

F = force
A = area
M = mass
g = acceleration due to gravity

Ok, but what about others? \sigma, p, h, what do they mean? I would like to know each one of them to calculate independently

Note that you need something several million times stronger than steel under compression. Carbon nanotubes are about 100x as steel under tension (nowhere near a million) but poor under compression.

And could you please tell me which material is the best for this task? I mean the material with the lowest possible density and with the highest value of strength under compression
P.S. Do you mean area of rod's cross-section (the rod is simply log cylinder) under A?

 

[Vanadium 50]

What do you think h means? Recall how I used it. Context is important.

And there is no material that you can buy that is a million times stronger than steel.