Calculating tidal forces (Interstellar)

[SpaceXray]

So I know that i the movie Interstellar, Miller's planet was on the edge of the event horizon of Gargantua.

I read somewhere that technically, the tidal forces should've ripped it apart.

I want to calculate if that is true.

I already found out the mass of Miller(1.3xEarth mass), mass of Gargantua(100 million times of the Sun),

however I still need the distance between Gargantua's singularity to the surface of Miller's planet, as well as Miller's Radius.

Help me out please?

When done, I'll post my calculation.

As the 2 points of tidal pull, I took the opposite middles of Miller's radius.

After I find out the difference in the forces, I need to find out if Miller's would survive that difference elastically.

I need help on that too.

[*Aqua*]

You're asking for the Roche limit. It's the distance at which the tidal forces rips a satellite apart. But you'll need either the radius of the planet or the black hole to calculate it.

Let's try another way: The Schwarzschild-radius of the black hole.

The formula is r = 2 * G * M / c²

G - gravitational constant

c - speed of light

M = 1,989 * 10³⁸ kg (100 million sun masses)

r = 1,989 * 10³⁸ kg * 1.485 * 10^-27 m/kg

r = 295366500000 m = 1,97 AU

That means the planet must be at least 2 AU apart from the black hole because at that distance is the event horizon. Of course the distance must be greater than that by a lot.

Edited November 29, 2014 by *Aqua*

[Bill Phil]

OP, that's not how gravity works. Just because Miller has 1.3g at the surface doesn't mean it has 1.3 Earth Masses.

[SpaceXray]

OP, that's not how gravity works. Just because Miller has 1.3g at the surface doesn't mean it has 1.3 Earth Masses.

Please explain then. I am eager to know.

- - - Updated - - -

You're asking for the Roche limit. It's the distance at which the tidal forces rips a satellite apart. But you'll need either the radius of the planet or the black hole to calculate it.

Let's try another way: The Schwarzschild-radius of the black hole.

The formula is r = 2 * G * M / c²

G - gravitational constant

c - speed of light

M = 1,989 * 10³⁸ kg (100 million sun masses)

r = 1,989 * 10³⁸ kg * 1.485 * 10^-27 m/kg

r = 295366500000 m = 1,97 AU

That means the planet must be at least 2 AU apart from the black hole because at that distance is the event horizon. Of course the distance must be greater than that by a lot.

Thanks! But what distance is Miller's planet from the event horizon then? We know it is outside the accretion disc too.

[cantab]

M is primary mass, I'll use 100 million solar masses

m is satellite mass, I'll use Earth

r is satellite radius, I'll use Earth

Schwarschild radius = 2GM/c^2 = 295 Gigametres. The lower bound for a stable orbit is 1.5 times that.

Roche limit = 1.26r(M/m)^(1/3) = 258 Gigametres. This of course depends somewhat on the size of the satellite.

It seems reasonable to conclude that Miller's planet is well outside its Roche limit.

[TheShadow1138]

Please explain then. I am eager to know.

Surface gravity depends not only on the mass of the planet, but the radius of the planet. If it depended only on mass, then you would be correct, but to determine the mass of Miller you need to know the radius of the planet.

surface gravity = -(G*M_planet)/R_planet^2

The negative sign is there indicating that the acceleration is always towards the center of the planet. Surface gravity would be (1.3*-9.81m/s^2) = -10.1m/s^2 (to one significant figure). So in the equation you would have:

-10.1 = -(6.67E-11 * M_planet)/R_planet^2

Rearranging, the negative signs will cancel, and we will multiply both sides by R_planet^2 giving:

10.1*R_planet^2 = 6.67E-11*M_planet

Then divide both sides by the Universal Gravitational Constant (6.67E-11)

(10.1*R_planet^2)/6.67E-11 = M_planet

So, we are missing two pieces of information. If you know the radius of the planet, then you can easily find the mass of the planet. Having not seen the film, I'm not sure what information they gave, so you may not be able to determine the mass of the planet.

[cantab]

Well for a constant density, surface gravity is proportional to radius. We don't know the density of Miller's planet, but it's probably about the same as Earth, there's only a certain reasonable range for rocky planets. That would make Miller's planet around 1.3 Earth radii and two Earth masses.

Also, considering the work that was done in relation to Gargantua, I expect Kip Thorne checked that the planet would be safely outside the Roche limit.

[AngelLestat]

You are missing a big point about the distance of miller´s planet to gargantua and the tidal forces.

First, the massive black hole is super rotating (as all massive black should in theory), this drag the space time with it. So miller´s planet does not need to be so close to gargantua to have that time dilation.

In fact, miller´s planet only orbit to 0,45c due this effect.

Another difference that change all your math, if a planet is tidal locking (as the moon to the earth), there is no tidal forces, no matter how strong the gravity is..

In this case miller´s was tidal lock to gargantua but with a slight swing.

In case you dont know it, there is 400 pages book (The science of interstellar) explaning every aspect of the movie..

Made by Kip Thorne, the science designer behind the movie which is an eminence in "black holes-wormholes-quamtum gravity-string theory" fields.

[YNM]

Does anyone here know what does one use to calculate frame dragging ? I thought that the reason why there are several tides is due to frame dragging itself ?

Also, tidal force does affect an object no matter whether it's tidally locked or not. Consider very close binary stars - their shape aren't a ball anymore, but ellipsoid (maybe more akin to tri-axial ellipsoid if their intrinsic rotation are also considered). Tidal force are the difference of gravitational force within an object between the farthest portion and the nearest portion. This creates physical stress and tension within the object, causing the shape of an object to be more elongated along one axis. In the case of general relativity, it's due to the difference in curvature of spacetime. Correct me if i'm wrong, but that's what I get as long as this.

Maybe for tidally-locked objects it's that the tidal force no longer generating a torque wrt the rotation of the object, yes (ie. the axis no longer precess).

Edited November 30, 2014 by YNM

[magnemoe]

Mann's planet should get some serious tides too at Pe, as it look pretty low, this effect could be worse on the planet than the wobbling of Millers.

[AngelLestat]

Also, tidal force does affect an object no matter whether it's tidally locked or not.

Yeah, I express wrong, I mean to said that "tidal friction" on the planet only happens if is not tidal lock. Also in normal circustances (no BH) we would not see (to naked eye) the tidal effect (as the sea) if is tidal lock.

Correct me if i'm wrong, but that's what I get as long as this.

Yeah is due to curvatura of spacetime.

In the book is explained with much more detail how the "streching tendex lines and squeezing tendex lines work.

Maybe for tidally-locked objects it's that the tidal force no longer generating a torque wrt the rotation of the object, yes (ie. the axis no longer precess)

my english is bad. WRT??

Mann's planet should get some serious tides too at Pe, as it look pretty low, this effect could be worse on the planet than the wobbling of Millers.

I dint read the book part which talks about mann´s planet yet. But that orbit explain how they fall after achieve escape velocity from mann´s planet.

There is another neutron star orbiting the black hole in ahigher orbit than miller´s planet. This was placed to make more credible the delta V as gravity assist to leave or reach miller´s planet.

Edited November 30, 2014 by AngelLestat

[kerbiloid]

I may be wrong, but.

http://en.wikipedia.org/wiki/Gravitational_redshift

1 h on Miller = 7 years outside

Relativity factor = 7 * 365.25 * 24 / 1 ~= 60000.

Relativity factor = sqrt(1 - Rg/R) = 1/60000.

sqrt(1 - Rg/R) = 1/60000;

1 - Rg/R = 1/60000^2;

Rg/R = 1 - 1/60000^2 ~= 1 - 3e-10;

R/Rg = 1 + 3e-10;

R - Rg ~= 3e-10 Rg;

According to

http://en.wikipedia.org/wiki/List_of_most_massive_black_holes

The greatest known black hole mass = 40,000,000,000 Msun ~= 4e10 * 2e30 = 8e40 kg.

Radius ~= 2 * 6.67 e-11 * 8e40 / 3e8^2 = 1.2e14 m.

R - Rg = 1.2e14 * 3e-10 = 36000 m.

I.e. even if Gargantua would be the greatest black hole ever, Miller planet would be 36 km above its event horizon.

I.e. it would be a quark mess. not a planet.

[YNM]

my english is bad. WRT??

WRT is short for "With Respect To". Thought it's common...

I may be wrong, but.

http://en.wikipedia.org/wiki/Gravitational_redshift

1 h on Miller = 7 years outside

Relativity factor = 7 * 365.25 * 24 / 1 ~= 60000.

Relativity factor = sqrt(1 - Rg/R) = 1/60000.

Thought that z (redshift) and γ (lorentz factor; I don't know what the name would be for gravitational effects) is different ? But that's a fair approach (I think that γ and z should have connections somewhere, maybe within the metric below). A more correct way would be using the metric tensors below, but I don't fully understand it

[cantab]

kerbiloid, your oversight is that Gargantua is rotating quite fast. That makes the distortion of spacetime more complicated, and allows Miller's planet to experience that extreme time dilation while being rather further from the event horizon.

I don't know the maths myself, but I'll assume Kip Thorne does.

[kerbiloid]

Of course, Gargantua rotates fast, as almost any black hole.

But what would change except instead of 36 km there would be 100 km?

I don't have the movie at my place, so I can't estimate Gargantua mass more correctly and can't calculate immediately.

But I am absolutely sure that if a planet diameter is ~10 000 km and its day side is 100 km away from the event horizon, then relativistic factor on the opposite sides of the planet would differ thousands times.

So while an electron on one side makes 1 turn, and an electron on the opposite side makes 100 turns, we can't have any talk about a planet as a integral whole.

It would become a molecular mesh.

Also if an atom is a, say, centimeter far from event horizon, it would hardly stay as a whole.

And so on. That's, btw, about "soft" black holes whose horizon an astronaut would "pass and even don't notice". When one of his atom is a trillion years old, and for another atom just a second passed, I'm afraid, any astronaut would become a quark blot many times smeared along all the horizon.

[Kryten]

It doesn't really matter if the tidal effects are survivable, when blueshifted by 0.45c background radiation definitely won't be.

[AngelLestat]

@kerbiloid

You are not taking into account that is a spinning black hole.

See the image (in scale) from the first page of this topic.

Black hole mass: 100 millons suns. Similar to andromeda´s black hole.

Event Radius: 1 Au.

Miller´s orbit: 1.5 Au aprox from the Event.

Parking orbit: 5 Au aprox from the Event.

The black hole is spinning very close to the max possible.

Normal black holes would fine an equilibrium at 0.998 the max possible spin value (this max number and average equilibrium was in fact discovered by kip thorne), at these speeds the black hole finds harder to suck matter falling at the same equatorial plane and spin direction.

But can be different circustances where a black hole can increase its spin a lot closers to the max value.

In this case was 0,99....98 (1 trillon)

Black hole fussions, etc.

The only thing that is out of scale is Man´s planet orbit. In the graph seems like its apo is 100Au, but in fact is 600 Au.

SOF backward (photons orbit in opposite direction than spin)

SOF Forward (photons orbit in the same spin direction)

I mistake in the speed of Miller´s planet, is not 0.45c, is 0.55c

@Kryten

??? what? I think that health harm with that effect will be only notice at much higher speeds in a higher time frame (years in those conditions.)

PD: according to your numbers kerbiloid, the biggest black hole found, its horizon event is about 800 Au. Also if we take all the mass of the universe and we calculate the Schwarzschild radius, it will be 10000 Billions Ly, an amazing similarity with out universe it self.

More if we take into account other similarities as entropy, etc.

We are in a black hole.

Edited December 1, 2014 by AngelLestat

[kerbiloid]

Event Radius: 1 Au.

Miller´s orbit: 1.5 Au aprox from the Event.

Great.

Relativistic factor = 1 / sqrt(1 - 1/1.5) ~= 1.7

So, 1 hour outside = 1.7 hours on Miller. Not 7 years at all.

To be 7 years per hour this planet would be much closer to the Black Hollie.

[YNM]

Your relativistic factor is closer to the z (redshift or blueshift) of a light emmited near the blackhole, viewed by an observer at infinite (ie. no curvature). While time follows something akin to lorentz factor for general relativity (I don't know what's the name for it). Just the same that the observed z of a radially moving particle isn't affected by lorentz factor (it should but the connections are hard, not as easy as time dilation/length contraction).

[kerbiloid]

Your relativistic factor is closer to the z (redshift or blueshift) of a light emmited near the blackhole, viewed by an observer at infinite (ie. no curvature). While time follows something akin to lorentz factor for general relativity (I don't know what's the name for it). Just the same that the observed z of a radially moving particle isn't affected by lorentz factor (it should but the connections are hard, not as easy as time dilation/length contraction).

I mean a time dilation in a strong gravity field

http://en.wikipedia.org/wiki/Gravitational_time_dilation

I.e. the effect which is referenced by the movie authors to explain why 7 years per hour. The planet would be pretty close to the black hole to be dilated so much.

And also as I can see when from POV of an electron at one side of an atom another electron of this atom, but at the opposite side spends a trillion years while the first one spends a picosecond, they could hardly talk about some chemistry or at least as about an atom as an integral entity.

So, I think, any physical object which is not a material point would become a mess.

[AngelLestat]

Lol kerbiloid, you are ignoring many concepts.

First, time dilation is related to speed and gravity.

Second, Is a spinning black hole, It drags time and space with it.

Third, you know who Kim Thorne is? Or you guess that is just a another screenwriter from holywood.

http://en.wikipedia.org/wiki/Kip_Thorne

In his book it said that he did mostly all math calculations, some using special softwares to avoid any error.

Seeing from the endurance, miller´s orbit gargantua in 1,7 hours, now from miller´s planet surface point of view, gargantua seems frozen in time but if we look towards the endurance seems that miller´s is orbiting gargantua 10 times by second.

Now back to the endurance point of view, Man´s planet orbits gargantua in 44 days. Periapsis speed 0,5c, Apo speed: 0,2 (without any significative time dilation, the speed en periapsis can be similar to millers, but is not so deep in the gravity well)

I guess the movie is great, Is like the 2001 space odyssey of our time.

The movie it self does not explain the science at all, but mostly all we see has science and real math as base.

Many scientists reviews about the movie was wrong, that is because they only imagine some cases where it would not be possible. That is easy to do, but if someone wants to hear an expert opinion, he/she needs to tell us in what cases that may be possible.

I recommend the book. Is great.

Edited December 1, 2014 by AngelLestat

[kerbiloid]

First, time dilation is related to speed and gravity.

Second, Is a spinning black hole, It drags time and space with it.

Of course.

And as closer you get to the event horizon, time dilation for:

every pair of your atoms;

every pair of electrons in your atoms;

every couple of nucleons in your atom cores;

every couple of quarks in your nucleons

would dramatically increase to infinity.

Because any two particles always have different distance from the event horizon.

And this is hyberbolic function: 1 / sqrt(1 - Rg/R).

So, yes, tidal force would be weak as a feather weight, don't worry.

But as closer you are to horizon, so every particle of your body would watch any another particle living trillions times faster.

So, I presume that your body would finish its existance as a binded system of particles and would become smeared in time and space as a Planck-length -thick spot of quarks or so, with maximum enthropy value.

A movie  yes, nice: as a movie.

Yes, I know Kip Thorne books, thanks.

[AngelLestat]

I am trying to determine what is the cause of your mix-up concepts.

Something with your reasoning is not right... The fact that you still think that just quoting the time dilation formule (which you dont understand either) makes you able to disprove Kip Thorne´s math, is a bad sign.

Lets see the facts again:

The event horizon´s radius is 1 Au, the distance between the sun center and the earth. Now lets see the size difference between 1 person, and 1 Au.

1 au = 149.597.871.000 meters

1 person = 2 meters

Now tell me the time dilation and gravity difference between these 2 points: 149.597.871.000 and 149.597.871.002

Back to the spinning black hole case:

Spinning black holes are a lot more "soft", because they extend their relative effects much farther.

In this case miller´s planet is about 1.5 Au from the event horizon, this mean 2.5 Au from the singularity. So again, what is the gravity difference between one planet face and the other? Lets said that miller´s diameter is close to earth´s (10000 km)

Ah.. but you cant use that formule, you need to take into account the keer metrics.

http://en.wikipedia.org/wiki/Kerr_metric

This is what happens to a particle falling in different direction than the rotation spin:

So you see how strong this effect is, which you totally ignored.

Yes, I know Kip Thorne books, thanks.

It does not bite.. read it.

[kerbiloid]

...

Thank you, I will think about this on weekend.

[carrotSnack]

For very large black holes, the gradient of gravity is actually quite small at the event horizon, as it very far from the singularity. In fact its small enough to pass the event horizon and live until it inevitably will be strong enough to kill.

http://en.wikipedia.org/wiki/Spaghettification#Inside_or_outside_the_event_horizon

https://www.physicsforums.com/threads/how-is-it-possible-to-live-past-the-event-horizon.621428/#post-3999653

From http://en.wikipedia.org/wiki/Supermassive_black_hole

As with density, the tidal force on a body at the event horizon is inversely proportional to the square of the mass: a person on the surface of the Earth and one at the event horizon of a 10 million solar mass black hole experience about the same tidal force between their head and feet. Unlike with stellar mass black holes, one would not experience significant tidal force until very deep into the black hole.

Edited December 3, 2014 by carrotSnack