How Strong Does Superman Need to Be to Pull Our Sun?

[PeterDonis]

The result of that calculation (which I'll let the OP discover for himself) does not actually give a tensile strength that violates the known laws of physics--it still implies a sound speed in the material that is less than the speed of light.

Actually, I should clarify this. A better way of stating the limit on tensile strength for this problem is that the tensile strength in pressure units, or equivalently energy density units (which are the usual units for tensile strength) must be less than $1/3$ of the energy density of the material. This allows one to explore a range of parameters for the chain, balancing the energy density against the cross sectional area (basically, the smaller you want the cross sectional area to be, the greater the energy density has to be to stay within the limit, since smaller cross sectional area means larger tensile strength required for a given force).

Given the above, there is a range of parameters for the chain that allow a tensile strength that does not violate the known laws of physics. But those parameters are pretty extreme.

 

[RosutoTakeshi]

Once you have the acceleration, you know the force required to impose that acceleration on an object with the mass of the Sun. Then you can just use standard formulas for the breaking strength of a cable to find what tensile strength would be required for a cable of your given cross sectional area to sustain that force.

Alright let me take a shot at this. I need to figure out the tensile strength of a cable (4in. in diameter) pulling an object with the mass of our sun (1.989e30kg) going from 0 to 0.99c in 1 minute

Acceleration:
a = (v-u)/t
a = (296344604m/s - 0m/s) / 60s
a = 296344604 / 60
a = 4,939,076 m/s^2

Force:
F = ma
F = 1.989e30 * 4939076m/s^2
F = 9.82e36 N

Cross Sectional Area (Cylinder):
πr^2
3.14(2in^2)
3.14 * 4
12.56in^2
Conversion to mm2:
8103mm^2

Ultimate Tensile Stress:
T = F/A
T = 9.82e36/8103mm^2
T = 1.21e33 Mpa

... I was struggling all night too. How does it look?

 

[PeterDonis]

How does it look?

The general form of your calculations looks fine. Now you should calculate what the minimum energy density of the chain material would need to be in order to meet the condition that the tensile strength must be less than 1/3 of the energy density. How does this energy density compare to, for example, the energy density of a neutron star?

 

[RosutoTakeshi]

The general form of your calculations looks fine. Now you should calculate what the minimum energy density of the chain material would need to be in order to meet the condition that the tensile strength must be less than 1/3 of the energy density. How does this energy density compare to, for example, the energy density of a neutron star?

I haven't been able to find anything relatable. All information on energy density is regarding electric fields, magnetic fields and electromagnetic waves

 

[PeterDonis]

I haven't been able to find anything relatable.

There are plenty of places online that will tell you the mass density of a neutron star. The energy density is just the mass density times $c^2$.

 

[RosutoTakeshi]

There are plenty of places online that will tell you the mass density of a neutron star. The energy density is just the mass density times $c^2$.

Right, information on neutron stars is available, but energy density related to the chain you wanted me to figure out... I can't seem to find

 

[PeterDonis]

energy density related to the chain you wanted me to figure out... I can't seem to find.

You don't need to find anything. You have all the information you need to calculate what I asked for.

You calculated a tensile strength in MPa for the chain. MPa (megapascals) is the same as MJ/m^3 (megajoules per cubic meter) in SI units, i.e., pressure and energy density have the same units. So your calculated tensile strength equates to an energy density. By the rule I gave before, to avoid violating the limits imposed by relativity, the tensile strength equated to an energy density must be less than 1/3 the energy density of the material that has that tensile strength. So the energy density of the chain has to be at least 3 times the tensile strength you calculated. How does that compare with the energy density of a neutron star?

 

[DaveC426913]

If the neutronium chain is strong enough to withstand that acceleration, can we also assume the mass (the neutron star itself) is rigid enough to be accelerated that quickly? Or will it get ... cut in half?

 

[PeterDonis]

If the neutronium chain is strong enough to withstand that acceleration, can we also assume the mass (the neutron star itself) is rigid enough to be accelerated that quickly?

I have not made any assumptions about rigidity of the neutron star or the chain in what I've said in this thread. The calculations I have described do not tell you any details about how the shape of either one changes during the acceleration. They are just simple calculations of what could be viewed as reasonable average values.

Or will it get ... cut in half?

I have assumed that it wouldn't. However, that assumption might be rendered implausible if it turns out that the energy density of the chain is substantially higher than the energy density of a neutron star (that's meant to be a hint ). That's not a matter of rigidity so much as hardness or sharpness--is the chain effectively "sharp" enough to cut into the neutron star instead of just transmitting a force to it? Obviously the narrower the chain, the more likely this is to happen; but I have not discussed how that might be estimated beyond the rough heuristic I just gave.

 

[RosutoTakeshi]

MPa (megapascals) is the same as MJ/m^3 (megajoules per cubic meter) in SI units, i.e., pressure and energy density have the same units. So your calculated tensile strength equates to an energy density.

So...

1.21e33 Mpa = 1.21e33 MJ/m^3

So the energy density of the chain has to be at least 3 times the tensile strength you calculated

1.21e33 * 3 = 3.63e33 MJ/m^3

How does that compare with the energy density of a neutron star?

Well I'm still trying to figure that one out. Haven't found anything that offers a straight answer. As of now, I need to find its Electric and Magnetic field in order to see. I'll keep looking

 

[RosutoTakeshi]

Well I'm still trying to figure that one out. Haven't found anything that offers a straight answer. As of now, I need to find its Electric and Magnetic field in order to see. I'll keep looking

Scratch that. I think I'm looking at it the wrong way

 

[PeterDonis]

I think I'm looking at it the wrong way

Yes, you are. EM fields do not provide any significant amount of energy density to a neutron star. All you need is the mass density, as I pointed out in post #35.

 

[some bloke]

It's probably a safe bet that the chains are simply required to be strong enough to accelerate the mass of the star, and that there may be some sort of cradle or dampening field or some other scifi handwavium to prevent the start from getting cut in half - after all, the chains are used to move stars, not to cut them!

So essentially Superman would also need to be strong enough to move said star in the same manner to be equally as strong as the chain, right?

There is also the physical properties to consider - tensile strength is one aspect, but brittleness is another - I can snap a hardened steel file in half over my knee, but I couldn't pull one so hard that it snap like a christmas cracker! If the chains are specifically designed with tensile strength in mind, and were not expected to have to be malleable, then they may be easier to break through a concerted effort to break them than they are when being used for their designed purpose.

 

[essenmein]

Not entirely sure how a rope would even "hold" a ball of plasma, its not particularly erm "solid".

Let alone plasma that is in fact a raging ball of nuclear fire.

 

[PeterDonis]

Not entirely sure how a rope would even "hold" a ball of plasma, its not particularly erm "solid".

Let alone plasma that is in fact a raging ball of nuclear fire.

This was already pointed out in post #23. The discussion has more or less shifted to an alternative scenario where the object to be pulled is a neutron star of the same mass as our Sun.

 

[some bloke]

I feel we're getting caught up on the wrong part of this issue. The issue is the strength of chains used to haul a mass (which we assume to be that of the sun) in space. I don't know the equations, but I would expect that you would be looking at the minimal forces needed to overcome inertia in order to move the star - I doubt that this is going to be about making that star move quickly (after all, they will have to stop it somewhere too!).

The logistics of moving a star don't factor into how strong superman would have to be to break these chains.

 

[DaveC426913]

I would expect that you would be looking at the minimal forces needed to overcome inertia in order to move the star

You still have two things to factor:
how strong Superman is, and how much thrust he can produce in flight.

With insufficient flight power, all he's doing is hauling himself toward the star.

 

[PeterDonis]

I feel we're getting caught up on the wrong part of this issue. The issue is the strength of chains used to haul a mass (which we assume to be that of the sun) in space.

If you think we haven't discussed what you say the issue is, I think you haven't read the thread. The strength of the chain is exactly what we've been discussing.

I don't know the equations

A number of formulas have already been given in the thread.

 

[jackwhirl]

You still have two things to factor:
how strong Superman is, and how much thrust he can produce in flight.

With insufficient flight power, all he's doing is hauling himself toward the star.

Superman, in flight, can accelerate himself to speeds greater than the speed of light. His flight thrust is infinite. Maybe more than infinite? If infinite thrust is required to reach 1c, how much gets you to 1.1c? ...That's meaningless, isn't it?

 

[PeterDonis]

That's meaningless, isn't it?

As far as trying to analyze using the laws of physics, yes.

 

[DaveC426913]

Superman, in flight, can accelerate himself to speeds greater than the speed of light. His flight thrust is infinite.

Perfect.

Now all we have to do is establish how indestructible he is.

If his flight thrust has no upper limit and he is indestructible, his strength could be zero, and he could still move a star.

Of course, then he would look like this - using his flight but not his strength:

 

[gmax137]

Getting back to the strength of the chain. If it turns out the neutronium energy density is insufficient, Superman could use chain woven from Chuck Norris's belts!

 

[DaveC426913]

I may be attempting to lighten the thread, but I'm not attempting to derail it. When examining the question it he title of the thread, one cannot ignore all physics.

Yes, you can ignore inconsequential things, like "the target doesn't collapse" but you can't ignore Superman's flight when asking if he's strong enough to move a star. They're inseparable.

 

[Jarvis323]

Given that superman can achieve infinite acceleration, it might be worth considering another explanation beyond physical strength for his chain breaking ability. Sure, he is depicted with muscles and his effort looks strenuous. But, take Keanu Reeves as an example. At first he became a kung fu expert. Then he became a super fast punch blocker, and then he transcended into something else altogether, beyond physical. The truth is that his fighting power was an illusion. All of those super human abilities were never shows of strength, reflex, or even skill.

Superman may still be at a stage where he believes he is somehow using his muscles. The truth is, maybe, there is no chain.



In fact, his muscles may be doing no more work to break those chains, than Mr. Mxyzptlk's muscles are when he pulls universes out of his hat.

 

[MathematicalPhysicist]

I don't know how much superman is strong but God-Man can do anything!

 

[Uberhulk]

You still have two things to factor:
how strong Superman is, and how much thrust he can produce in flight.

With insufficient flight power, all he's doing is hauling himself toward the star.

Superman's strength tops out at bench pressing the equivalent mass of the Earth. When he needed to move an object larger than the Earth (mass unknown), he needed help.

 

[Uberhulk]

[Moderator's note: Unnecessary introductory statement deleted.]

There's a comic where Superman breaks out/shatters chains that were designed to haul stars across the galaxy
- Let's say these chains were made to pull (our) sun
What tensile strength would a chain need to help pull our sun without breaking? And if I'm not asking the right questions, please let me know what other information you need to solve thisView attachment 261146

You missed the previous scan, which shows this was a dream. Superman is in bed with Lois, starts to dream and suddenly he's chained up. Nowhere in the story did we see him captured and chained up. He closes his eyes, then he's chained up. It's obvious we're shown a dream.

 

[Whipley Snidelash]

Supermans strength, or any superhero for that matter, is directly dependent on the writer god in control and whether or not he wants superman to win or lose in any particular situation. This is all contrived because the way he is defined is that he is completely powerful and invulnerable to everything except kryptonite and magic. So when a writer is in control he does get hurt sometimes he gets killed sometimes he gets beat up sometimes and it’s not always by kryptonite or magic If I recall correctly I’m not a big superman fan

 

[h1a8]

I have no idea how the discussion became about Superman moving stars with flight. Superman appears to break the chains by pulling them apart with his arms. So the question is: "What is the minimum force of tension the chains can be subjected to without breaking?"

Assumptions:
1. The star is replaced with an indestructible solid spherical object with the sun's mass and having a diameter of 4in.
2. The object is accelerated by the chain from 0 to 184,410 mi/s in exactly 60 seconds.
3. Relativity applies
4. 1 of the 2 following cases apply (to simply things):
case1: Chain supplies a constant pulling force (implies a decreasing acceleration)
or
case 2: Chain accelerates object at a constant rate (implies an increasing force).

The relativistic force equation is
$F=\frac{d}{dt} \left[\frac{Mv}{\sqrt{1-\frac{v^2}{c^2}}}\right]$
After doing some Calculus and Algebra
$$ \rightarrow 1) \space F={\frac{M}{\left(1-\frac{v^2}{c^2}\right)^{3/2} } \frac{dv}{dt}}$$

$$case1:\space a=\frac{dv}{dt}\text { is constant}$$
$$\rightarrow v=at$$
$$\rightarrow 1*) \space F=\frac{Ma}{\left[1-(\frac{at}{c})^2\right]^{3/2}}$$

Step1: Calculate the acceleration
You already did.
$a = 4.94463e6 \frac{m}{s^2}$

Step 2: Calculate the mass of the object
You already did.
$M = \text{1 solar mass} = 1.989e30kg$

Step 3: Calculate the maximum force of tension (or the maximum force the chain exerts on the object) during the time interval [0,60].

Since the acceleration is constant (case1:), the force of tension would be increasing on $[0,60]$.
Use eqn 1*) and plug in both the acceleration and the value of t in [0,60] that gives the maximum force of tension in that interval.
$F_{max} =$

Step 4: Calculate the minimum tensile strength (optional)
$P = \frac {F_{max}} {A} = \frac {F_{max}}{\pi \cdot 4in^2} =$$$case2:\space \text {F is constant}$$
$$\rightarrow \frac{dv}{dt}>0 \space\text{and decreasing on} \space[0,60]$$
$$\rightarrow 1**)\space \int_0^{t_f} \frac {F}{M}dt=\int_0^{v_f}\frac{dv}{(1-\frac{v^2}{c^2})^{3/2}}$$
$$\rightarrow \space F=\frac{Mv_f}{t_f \sqrt{1-\frac{{v_f}^2}{c^2}}}=$$

Then you can calculate both the maximum force of tension (using the appropriate t and v values) and then the minimum tensile strength (optional) like in steps 3 and 4 above.