# Time Dilation and Redshift for a Static Black Hole

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The following is an overview of the time dilation and gravitational redshift effects of a static (Schwarzschild) black hole. In accordance with general relativity, a strong gravitational field can slow down time. The closer you get to the event horizon of a black hole (if you can survive the gravity gradients, g-forces and have some means of propulsion that will keep you stable in the overwhelming gravity), there will appear no difference to yourself but time will pass slower for you and quicker for people outside the field (with no apparent difference to them either). On returning from the edge of the black hole, depending on how close you were to the event horizon and how long you were there for, you would have aged less than the people outside the black holes strong gravitational field. This can be calculated using the following equation-

$$\tau=t\sqrt{1-\frac{2Gm}{rc^2}}$$

which can be rewritten as

$$\tau=t\sqrt{1-\frac{r_s}{r}}$$

where $\tau$- actual (proper) time experienced within black holes gravitational field, $t$- time observed from infinity, $r_s$- Schwarzschild radius of black hole (where $r_s=2Gm/c^2$), $r$- the coordinate radius time was spent at ($G$- gravitational constant, $m$- mass, $c$- speed of light).

For example, if an observer was to spend an ‘observed’ 3 days at 1 km from the event horizon of a 12,000* solar mass black hole, they would only experience-

$$\tau=3\sqrt{1-\frac{35,444,396}{35,445,396}}$$

$$\tau=0.01594\ \text{days = 23 minutes}^{**}$$

where t=3 days, $r_s=2Gm/c^2$ which for a 12,000 solar mass black hole equals 35,444,396 m and to be 1 km from the event horizon would mean r=rs+1000 m=35,445,396 m

*A 12,000 solar mass BH was chosen due to the tolerable tidal forces (dg) near the event horizon, $dg=(2Gm/r^3)dr$ where dr=2 (approx. height of a person), which means dg=71.5 g which is 7.3 earth g from head to toe which is just about manageable.

**That is, if they instantaneously travel to the mark and travel instantaneously back again, otherwise time dilation would need to be taken into account for the various stages of the trip.

Likewise, due to the immense gravity, the electromagnetic information being sent out by person would be extremely redshifted. This can be calculated using the following equation-

$$z=\frac{1}{\sqrt{1-\frac{2Gm}{rc^2}}}-1$$

(black hole event horizon = surface of infinite redshift)

and to calculate the actual change in wavelength-

$$z=\frac{\lambda_o – \lambda_e}{\lambda_e}$$

which becomes

$$\lambda_o=(z \cdot \lambda_e) + \lambda_e$$

where $\lambda_o$- wavelength observed from infinity and $\lambda_e$- wavelength emitted from the object.

Therefore, the redshift at 1 km from the event horizon would result in z=187.27. Information at a wavelength of 550 nm (the colour green in the visible spectrum) $(\lambda_e)$ would be stretched to a wavelength of 103,549 nm (0.10355 mm or 103.55 um) $(\lambda_o)$ which would put it in the FIR wavelength (far infrared light waves) (in comparison, the sun has a surface gravitational redshift of z=0.0000021 which has near to zero effect on the electromagnetic information. A white dwarf with a 1.14 sol mass and a 4.5×10^6 m radius has a gravitational redshift of z=0.00037. For a neutron star with a 2 sol mass and a 12 km radius, z=0.4034.). The strong gravitational field would have the opposite effect on light going the other way as it was pulled in towards the event horizon. Viewed from outside the strong gravitational field, the individual would appear to slow down and disappear overtime as the light coming from them was redshifted towards radio waves, but from the inside, the observer might see the outside 3 days pass in the 23 minutes they spent at 1 km from the event horizon. It’s very likely that this light would be blue shifted into the x-ray and gamma ray spectrum which would no doubt have fairly drastic consequences for the observer.

Imagine a spaceship is at 1,000,000 km from a 12,000 solar mass black hole (even here, the craft would have to have powerful engines in order for it not to be pulled in by the black holes gravity which would still be 1,511,711 m/s^2*** at this point). Gravitational redshift would be minimal (550 nm $(\lambda_e)$ would be 560 nm $(\lambda_o)$, green would still be green). Here, time dilation would already be beginning to take effect (~59 minutes would pass compared to every hour in regular space outside the black holes gravitational field). A scout ship with the capacity to travel over 1,000,000 km almost instantaneously (an average velocity of 30,000 km/s, 0.1c, covering the 1 million km in ~35 seconds. Time dilation due to this velocity is virtually undetectable) is sent out towards the black hole. Due to the velocity of the craft, it would appear orange to the main ship for the majority of the journey but as it reached the 100,000 km mark from the event horizon, it would begin to turn deep red (light wavelengths from the scout ship would be stretched from 600 nm to 650 nm due to gravitational redshifting). At around the 50,000 km mark, it would disappear to the naked eye and the observer ship would have to use infrared equipment in order to monitor the scout ships progress. As the scout ship passed the 10,000 km mark it would appear to slow down as time dilation meant that 30 seconds for the scout ship would appear as 1 minute for the main ship (the scout ship would appear to half its speed). By the time the ship reached the 1 km mark from the event horizon, it would be in far infrared. It would be seen to hover for 3 days, visible only to infrared equipment on the main ship, the crew having plenty of time to monitor the scout ship’s position. After a full 3 days at the 1 km mark, the scout ship would at first appear to slowly make its way back to the ship, picking up speed as it went through far, mid and near infrared, back into visible light, 100,000 km from the event horizon, appearing nearly full speed now as the time dilation here is only 0.85 (1 minute for the main ship would be ~52 seconds for the scout ship). Over the remaining 900,000 km, the ship would appear blue due to its velocity. Once the scout ship had returned, it would inform the crew of the main ship that it had been travelling at maximum speed all the way, taking only 35 seconds to arrive at the 1 km mark where it hovered for just 23 minutes before accelerating away at full power to arrive back at the main ship.

***Gravity for a static black hole is $g=Gm/(r^2\sqrt(1-r_s/r))$.

Time dilation and gravitational redshift become more extreme the closer you get to the black hole, at 1 m from the event horizon (r=rs+1 m), you would only experience-

$$\tau=3\sqrt{1-\frac{35,444,396}{35,444,397}}$$

$$\tau=0.0005\ \text{days = 43 seconds}$$

The redshift would be z=5,957.83 (550 nm (green) $(\lambda_e)$ would be at 3,277,358 nm or 3.277 mm $(\lambda_o)$, which would put it in EHF, Extremely high frequency radio waves, detectable only by radio receivers)

and at 5 mm from the event horizon (r=rs+0.005 m)-

$$\tau=3\sqrt{1-\frac{35,444,396}{35,444,396.005}}$$

$$\tau=0.000036\ \text{days = 3 seconds}$$

The redshift would be z=104,962.52 (550 nm (green) would be at 57,729,938 nm or 57.730 mm, which would put it in SHF, Super high frequency radio waves).

Within the last 5 mm as you reached the event horizon, you would shift through medium to very low frequency to infinite wave length and time dilation would go from 3 days passing in 3 seconds to centuries passing in mere fractions of a second. A life time would pass for anyone observing you from the outside with sensitive radio equipment.

At exactly on the event horizon, where gravity is considered to be infinite, if a person was to look outwards from the black hole, they might see some portion of the future of the universe flash before them before they free fall at close to the speed of light towards the singularity, the time of this fall (for a static Schwarzschild black hole) can be calculated by the following (which seems derivative of the antipode distance of a hypersphere based on the gravitational radius $(M=Gm/c^2)$)-

$M \pi/c = Gm \pi/c^3 = 1.548×10^{-5}\text{ x solar mass = time in seconds to singularity}$

For a 12,000 solar mass black hole, this would be a full 0.1857 seconds to admire the view before reaching the singularity.

Note: As stated, the spaceship would experience a slight time dilation of its own. If the spaceship was hovering at r=10^6 km, then 3 days at infinity would be 2 days, 22 hours, 45 minutes on the spaceship. If it was in orbit at r=10^6 km, then the time dilation would be 2 days, 22 hours, 7 minutes where-

$$\tau_o=t\sqrt{1-\frac{r_s}{r}}\cdot\sqrt{1-v_s^2}$$

where $\tau_o$ represents the time dilation for an object in stable orbit and $v_s$ is velocity required for stable orbit where $v_s=\sqrt(M/(r(1-r_s/r)))$.

The above equation is equivalent to

$$\tau_o=t\sqrt{1-\frac{3}{2}\cdot\frac{r_s}{r}}$$

where $\frac{3}{2}r_s$ is considered the absolute last stable orbit, it also defines the photon sphere.

Another interesting equation is the proper distance to the event horizon for a hovering observer-

$$\Delta r’=\frac{\Delta r}{\sqrt{1-\frac{r_s}{r_s+\Delta r}}}$$

where $\Delta r'$ is the proper distance as observed by the hovering observer and $\Delta r$ is the coordinate distance as observed from infinity.

At 1 km from the EH (as observed from infinity), the hovering observer will see himself to be 188 km from the EH, at 1 m, the hovering observer will see himself to be 5.954 km from the EH and at 5 mm, 421 m from the EH.

Also note, the equations shown for time dilation are for an object hovering at a specific r, for an object that has fallen from rest at infinity, the time dilation would be-

$$\tau=t\left(1-\frac{r_s}{r}\right)$$

where $\tau=t\sqrt(1-r_s/r)\cdot\sqrt(1-v^2)$ where for an object that has fallen from rest at infinity, $v=\sqrt(r_s/r)$

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42 replies
1. SlowThinker says:

Nice article, with some interesting and hard to find formulas.Do I understand it correctly that at the horizon, gravity acceleration is infinite, while the tidal forces are manageable/finite?Second question: I'm in my ship, hovering 10 meters above horizon. In that situation, I turn the engines off. Will the bottom of my ship turn to dust before I hit the horizon?I'm trying to figure out how gravitational time dilation works on fast moving objects.

2. SlowThinker says:

I was, uh, thinking some more.1. The tidal forces are a difference between gravity at my head, minus gravity at my toes, right? (Lets define gravity as the acceleration necessary to stay in place relative to a distant observer)2. If my ship goes lower so that my head is where my toes were, the difference is nearly the same as before.3. But when I reach the horizon, the acceleration needed is infinite.Does that mean that, as long as my engines can keep the ship in place, I can go down and down and never reach the horizon? I think this must be the case, and something funny must be happening with the height of my ship.Sadly this is not related to the topic any more, is there a chance for a follow-up article discussing the horizon thoroughly? Or is there such an article already?

3. Edriven says:

Loved the post,  but is respectfully disagree with use of time dilation.  Doesn't gravity affect light and matter?  Is it not having an affect on our clock?  Clock was designed to work on Earths surface. A change in function of this machine is only a loss of calibration to a different environment. Thus this time piece can work correctly in a stable environment, if it is recalibrated to that environment. So it would be closer to the truth if you said time dilation is only a relationship of the rest of our universe using Earths time.  But time is a constant and can't be bent or altered.

4. Edriven says:

In addition.  I think of it like this. Light traveling from a distant star can be used as a historical record of time. Much like archeologists date objects based on depth of dirt. But no one is saying we can bend the dirt-time continuum

5. Bernard McBryan says:

So, outside the event horizon in the three cases:1) hovering (engines accelerating fighting 1,511,711 m/s^2 ):  experience listed gravitational time dilation.2) engines just turned off, same height but now in free fall but still at slow velocity: experience same listed gravitational time dilation.3) in circular orbit (free fall at same height but with high tangent orbit velocity): experience slightly higher gravitational time dilation (perhaps due to the same listed gravitational time dilation but with extra velocity time dilation due to the orbital tangent velocity.

6. Mike Holland says:

SlowThinker has commented a couple of times that the gravitational acceleration at the horizon is infinite! This is surely incorrect. The escape velocity at that point is c. Approaching the singularity the gravitational force would tend to infinity, but it becomes meaningless at the singularity..

7. PeterDonis says:

Good post in general and good choice of subject, this is often a source of confusion on PF threads.

I have a few comments, though. First, it’s important to be very careful to specify the state of motion of the object that appears to be time dilated, or which is emitting radiation that appears redshifted from far away. The equations you give are for a static object–i.e., an object which is “hovering” at a fixed altitude above the horizon and has no tangential motion at all–and the radiation it emits. However, later in the article, you discuss a spaceship in orbit about the hole at some altitude, and a scout ship free-falling into the hole. The “time dilation” equations for these are different from the one you give.

For someone in a free-fall orbit about the hole, the equation is

$$tau = t sqrt{1 – frac{3GM}{rc^2}} = t sqrt{1 – frac{3}{2} frac{r_s}{r}}$$

There is also a correction to the redshift observed in light emitted by an orbiting observer, but IIRC it is somewhat different (it’s similar to the transverse Doppler effect in SR). (Also, if a spaceship is in orbit about the hole, it doesn’t “have to have powerful engines to resist the hole’s gravity”; a free-fall orbit requires no engine power at all. If you meant that the ship was “hovering”, maintaining a constant altitude while having no tangential velocity at all, then “orbit” is not a good word to use.)

For someone free-falling into the hole along a radial trajectory, the “time dilation” factor isn’t really well-defined, because this observer is not at rest relative to an observer at infinity, so they don’t have a common standard of simultaneity. However, the “redshift factor” for light emitted by the free-falling observer and received by an observer at infinity is well-defined, and can be calculated as the redshift factor for a static observer at altitude ##r##, which is what you wrote down, combined with the Doppler redshift for an observer falling inward at velocity ##v = sqrt{2GM/r}## relative to the static observer.

Also, at the end of the article, there is an incorrect statement:

“At exactly on the event horizon, where gravity is considered to be infinite, if a person was to look outwards from the black hole, they might see some portion of the future of the universe flash before them”

The “future of the universe” is not in the person’s past light cone when they cross the horizon (or even when they reach the singularity). In fact, a person free-falling into the hole does not even see light from the outside universe blueshifted; he sees it redshifted (by contrast with a person “hovering” just above the horizon, who does see light from the outside universe highly blueshifted). It’s important to be very careful about describing what happens near, at, and inside the horizon, because of the many common confusions people have, based on pop science literature.

8. PeterDonis says:

Do I understand it correctly that at the horizon, gravity acceleration is infinite, while the tidal forces are manageable/finite?

Yes. The interpretation of tidal forces as “the difference in acceleration between two neighboring points” only works when you are very far above the horizon; it breaks down close to the horizon (and obviously doesn’t work at and below the horizon).

I’m in my ship, hovering 10 meters above horizon. In that situation, I turn the engines off. Will the bottom of my ship turn to dust before I hit the horizon?

It depends on how large the hole is. For a hole of one solar mass, your ship might well be dust even before you get 10 meters above the horizon. But for a hole of a billion solar masses (such as probably exist in the centers of some quasars), you won’t even feel the tidal gravity at the horizon, and your ship will be just fine.

9. SlowThinker says:

It depends on how large the hole is.

Sorry for the misunderstaning. I was referring to aging of the ship, due to infinite(?) time dilation.
The mechanical stresses, radiation etc. are a story for another time.

10. PeterDonis says:

I was referring to aging of the ship, due to infinite(?) time dilation.

The ship itself doesn’t experience any time dilation; to someone on the ship, everything seems perfectly normal, clocks tick at one second per second, etc. This is true even if the ship falls through the hole’s horizon. If the ship starts falling 10 meters above the horizon, the time it takes to fall those 10 meters, by the ship’s clocks, will be short–the sort of time you would expect an object to take to free-fall 10 meters in a strong gravity field.

Time dilation is something an observer very far away from the hole will observe if they watch what is happening on the ship.

11. stevebd1 says:

The figures provided at a distance of 1km, 1m and 5mm from the EH of the black hole are for an object (i.e. the scout ship) hovering at these particular radii. The statement about seeing ‘some portion of the future of the universe’ is based on the notion of the scout ship trying to hover closer to the EH within the last 5mm but as it reaches the event horizon, were there is no stable r, being pulled into the BH. Within those last 5mm, just before being pulled into the BH, the occupants would see the outside universe speed up significantly, albeit via immensely blueshifted and highly radiative light. As they were pulled across the event horizon, I imagine that this image of the outside universe would begin to redshift as they fell towards the singularity. I’ve also added some footnotes and supplementary equations.

12. PeterDonis says:

The figures provided at a distance of 1km, 1m and 5mm from the EH of the black hole are for an object (i.e. the scout ship) hovering at these particular radii.

Ah, ok. That wasn’t really clear to me from the description; you might want to emphasize it more at the start of the scout ship portion of the article.

Within those last 5mm, just before being pulled into the BH, the occupants would see the outside universe speed up significantly, albeit via immensely blueshifted and highly radiative light.

Yes, but the term “future of the universe” is still misleading. It’s not the “future” that is being seen from the viewpoint of the occupants; it’s still the past. Being close to a black hole’s horizon doesn’t allow you to see things outside your past light cone.

As they were pulled across the event horizon, I imagine that this image of the outside universe would begin to redshift as they fell towards the singularity.

Yes.

13. jartsa says:

Nice article, with some interesting and hard to find formulas.
Do I understand it correctly that at the horizon, gravity acceleration is infinite, while the tidal forces are manageable/finite?

Consider a fleet of spaceships landing on a planet that has a frictionless surface, very deep gravity well, and a few moons.

The spaceships will slide around into different directions pulled by the gravity of the moons, slowly according to far away observers, at great speed according to surface observer.

14. Boing3000 says:

I have also got a question regarding two free falling ship case:
1) vertically “falling”
2) orbiting “falling”

In both those frame there is no local gravity/acceleration, but I think the gravitational potential (hence time ratio) is somewhat different along the vertical axis in both ship (and rapidly changing in ship 1
When both ship are at the same altitude, do they experience (locally) the same time dilatation ? Or does the “horizontal” movement of the orbiting ship 2 change it slightly compare to ship 1 ?

From the article, it is not evident to guess the difference in time dilatation for a 12,000 solar mass black, between the foot and the head (let’s say 2 meter) of someone free falling, and whose foot are at 1 meter of the event horizon. If the tidal force (the radial angle difference) is manageable, I suppose the time dilation must be quite weak too.

I think SlowThinker has inverted the gravitational time dilatation effect, because his toes are not supposed to turn to dust, but his head is:nb)

15. SlowThinker says:

I have also got a question regarding two free falling ship case:
1) vertically “falling”
2) orbiting “falling”

In both those frame there is no local gravity/acceleration, but I think the gravitational potential (hence time ratio) is somewhat different along the vertical axis in both ship

I’m pretty sure that for an object in free fall, the time dilation is the same throughout the volume, or at least nearly the same – it follows from the equivalence principle. Further, a free falling object is gaining speed, which, due to Lorentzian time dilation, is exactly equal to the gravitational dilation at any lower height (or at least I am sure it is), so no paradox there.
For ship 2, you just need to add “standard” Lorentzian time dilation due to orbital speed.

Yes you are right, my head would age fast while my legs are OK.

16. PeterDonis says:

The tidal forces are a difference between gravity at my head, minus gravity at my toes, right?

Only in the Newtonian approximation, where you can view “gravity” as a “force”. This approximation is certainly not valid near the horizon of a black hole.

The more general definition of tidal gravity involves geodesic deviation: how fast a pair of freely falling objects that start out at rest relative to each other separate or converge with time (where “time” here means proper time along either geodesic). It is quite possible for tidal gravity, by this definition, to be negligible at the horizon of a black hole, if the hole is massive enough. But the proper acceleration required to “hover” goes to infinity at the horizon regardless of the hole’s mass. This makes it obvious that tidal gravity and proper acceleration are not related in the general case.

Does that mean that, as long as my engines can keep the ship in place, I can go down and down and never reach the horizon?

No. See above.

17. PeterDonis says:

In both those frame there is no local gravity/acceleration, but I think the gravitational potential (hence time ratio) is somewhat different along the vertical axis in both ship (and rapidly changing in ship 1

Strictly speaking, “gravitational potential” is a property of the spacetime, not the object, and it is only well-defined to begin with in a special class of spacetimes, called “stationary spacetimes”. In such spacetimes, any two objects at the same spatial location have the same gravitational potential.

However, this does not mean they necessarily have the same time dilation factor. See below.

When both ship are at the same altitude, do they experience (locally) the same time dilatation ?

No. First let’s take the simplest case, where the vertically free-falling ship has just started falling from rest at that altitude (it has just turned off its rockets and stopped “hovering”, or it was free-falling upward and has just reached its maximum altitude and is starting to fall back down again). In this case, the (instantaneous) time dilation factor of the vertically free-falling ship will be the same as that of a “hovering” ship at the same altitude, i.e., ##sqrt{1 – 2M / r}## (in units where ##G = c = 1##). But the orbiting ship will have a time dilation factor of ##sqrt{1 – 3M / r}##, i.e., a larger one.

The extra time dilation is due to the orbiting ship’s motion; in general, in a stationary spacetime, we can split up any object’s time dilation factor into two pieces, a gravitational piece (which depends only on its position, as above), and a piece due to velocity relative to a static observer (i.e., an observer whose position is not changing) at the same position. The general formula for the time dilation factor is then ##sqrt{1 + 2 phi – v^2}##, where ##phi## is the gravitational potential. In the case we’re discussing, where the gravitating mass is spherically symmetric and static, we have ##phi = – M / r##, and an object in a free-fall circular orbit has ##v = sqrt{M / r}##; plugging these in will give you the specific formulas I quoted above.

If the tidal force (the radial angle difference) is manageable, I suppose the time dilation must be quite weak too.

No. The two are not related, for much the same reasons that the tidal gravity and proper acceleration are not related. See my previous post.

I think SlowThinker has inverted the gravitational time dilatation effect, because his toes are not supposed to turn to dust, but his head is

No. Tidal gravity, when it gets strong enough, will affect both his head and his feet; it will “try” to make them separate rapidly, and at some point the structural strength of his body will be overcome and his whole body will turn to dust. Assuming his body is small compared to the size of the black hole, one end won’t be affected significantly before the other end is.

I’m pretty sure that for an object in free fall, the time dilation is the same throughout the volume, or at least nearly the same – it follows from the equivalence principle.

Yes; more precisely, it follows from the EP plus the assumption that the object’s size is very small compared to the size of the black hole (roughly speaking, to the radius ##r = 2M## of its horizon).

Further, a free falling object is gaining speed, which, due to Lorentzian time dilation, is exactly equal to the gravitational dilation at any lower height (or at least I am sure it is), so no paradox there.

I’m not sure what you mean by this. Gravitational time dilation and “Lorentzian” time dilation (by which I assume you mean time dilation due to velocity, which I explained above) are two different things.

18. SlowThinker says:

The more general definition of tidal gravity involves geodesic deviation

Are you 100% sure that the height contraction does not play a crucial role? I still don’t see a flaw in my reasoning:
1. The acceleration needed to keep my head in place is ##g_{head}##. Similarly define ##g_{feet}##.
2. The difference ##g_{head}-g_{feet}## seems to fit your definition of tidal effects, and so is finite everywhere except singularity.
3. If the ship is descending slowly, ##g_{feet}## eventually reaches infinite value.
4. Therefore, there has to be an infinite number of heights before horizon.
Which of these steps are wrong?

I’m not sure what you mean by this. Gravitational time dilation and “Lorentzian” time dilation (by which I assume you mean time dilation due to velocity, which I explained above) are two different things.

I was imagining this scenario: Let’s have 2 ships, one hovering low, another high. The high ship turns engines off. By the time it meets the lower ship, what are their relative time speeds, and tallness? I thought they are the same, but now I’m not so sure…

19. PeterDonis says:

Are you 100% sure that the height contraction does not play a crucial role?

What is “height contraction?”

The acceleration needed to keep my head in place is ##g_{head}##. Similarly define ##g_{feet}##.

Yes, these can be defined. But they aren’t the same as tidal gravity. Read my description of tidal gravity again, carefully. Note that I specifically talk about freely falling objects, and how they separate or converge. If you are accelerating to stay in the same place, you are not in free fall, and the difference in acceleration across your body is not the same as tidal gravity.

No, it doesn’t. See above.

and so is finite everywhere except singularity.

No; the acceleration needed to “stay in the same place” goes to infinity at the horizon. Once more: the change in the acceleration needed to stay in the same place, and tidal gravity, are two separate things. They don’t go together.

Let’s have 2 ships, one hovering low, another high. The high ship turns engines off. By the time it meets the lower ship, what are their relative time speeds

Since they are at the same altitude, the contribution of gravitational potential to their “time speeds” is the same in both cases. So the only difference is their relative velocity. Each will measure the other’s clock to be running slow, just as two observers in relative motion in SR do. An observer at infinity, watching them both, will see the falling ship’s clock running slower than the stationary ship’s clock.

and tallness

I’m not sure what you mean by “tallness”.

20. jartsa says:

Consider a fleet of spaceships landing on a planet that has a frictionless surface, very deep gravity well, and a few moons.

The spaceships will slide around into different directions pulled by the gravity of the moons, slowly according to far away observers, at great speed according to surface observer.

So clearly on the surface the situation is such that huge tidal forces are mangling the fleet. If the spaceships are chained together the chains will break.