G-forces during gravitational slingshot maneuvers

[Yora]

The issue came up in the comments on the latest Because Science video. If you slingshot around a binary neutron star to a quarter of the speed of light, would the g-forces kill you?

And as I've been thinking about it, I've come to the conclusion that you would not experience any g-forces at all.

When astronauts are experiencing high g-forces during launch, their bodies are feeling compression because the bodies are stationary object and the seats are pushing into them. It's not really the acceleration that you feel but the seat wanting to go faster than your body currently does. And then your body wanting to go faster than your blood currently does. With constant acceleration we don't feel that as a momentary impact but as getting compressed.

But when you are in a spaceship and you are getting accelerated by the gravity of a massive object in your path, isn't every atom in your body and in your spaceship getting gravitationally accelerated at the same speed? When you and a friend jump into a pool, you won't fall any faster or slower when your grabbing each other's hands (and effectively becoming a single falling object with twice the mass). To me this means that all atoms are being gravitationally pulled individually. And if your body, your blood, and your seat are all being accelerated at the same rate, there should not be any compression happening. So even if you get accelerated to a quarter the speed of light in seconds, there would be no g-forces on your body, right?

Where things would become a problem is tidal forces, where your feet facing towards a nearby neutron star would accelerate faster than your head, causing the body (and also the spaceship) getting stretched. But that would be a completely different effect from g-forces, right?

[Tyndmyr]

Slingshot maneuvers are usually coupled with engine burns, so yeah, you'll feel the g force as normal for that engine burn.

How much engine burn varies depending on the specifics. A final speed isn't enough to calculate out acceleration without other variables.

[Aeson]

I'll just add that an essentially-uniform acceleration field over a given volume isn't necessary "safe" even though everything is accelerating together - your body is meant to function in an environment with an essentially-uniform acceleration field of the strength of 1 standard (Earth) gravity. It will not function nearly as well in an environment with an essentially-uniform acceleration of the strength of, say, 5 standard gravities, and being exposed to such conditions - especially for a prolonged period of time - could do you significant harm; a sufficiently strong gravitational field could potentially kill you outright without any need for tidal forces to rip you apart.

[Khedrac]

With a slingshot maneuver round a neutron star it's not the acceleration that kills you - it's the tides!
Well, actually it depends how close to the neutron star you get, but Larry Niven's short story Neutron Star covers the problem quite well (though he admits he forgot that the ship would have been spinning).

[Radar]

I'll just add that an essentially-uniform acceleration field over a given volume isn't necessary "safe" even though everything is accelerating together - your body is meant to function in an environment with an essentially-uniform acceleration field of the strength of 1 standard (Earth) gravity. It will not function nearly as well in an environment with an essentially-uniform acceleration of the strength of, say, 5 standard gravities, and being exposed to such conditions - especially for a prolonged period of time - could do you significant harm; a sufficiently strong gravitational field could potentially kill you outright without any need for tidal forces to rip you apart.

I think you are confusing our situation at the surface of the Earth and essentially a free fall in gravitational field. Only in the latter case we truly feel a uniform acceleration, which means we do not feel anything at all. In the former case the forces we feel are very much nonuniform, since the ground is pushing us up, so we do not fall further down.

This means that aside from tidal forces (which are proportional to 1/r^3, where r is the distance to a center of a massive object) you only feel the thrust of your spaceship through the floor you are standing on. If you turned off the engines and the ship was not spinning, then you would not feel any acceleration whatsoever.

[shawnhcorey]

You got most of the physics correct except for one problem. A slingshot around a star will not increase your speed. Slingshots only change your speed if there are two other bodies, say the Earth and the Moon. By slingshotting around the Moon, you are very slightly decreasing the Moon's orbit speed to increase your own. Without a third body to have an orbit, you cannot change your speed.

[Strigon]

I'll just add that an essentially-uniform acceleration field over a given volume isn't necessary "safe" even though everything is accelerating together - your body is meant to function in an environment with an essentially-uniform acceleration field of the strength of 1 standard (Earth) gravity. It will not function nearly as well in an environment with an essentially-uniform acceleration of the strength of, say, 5 standard gravities, and being exposed to such conditions - especially for a prolonged period of time - could do you significant harm; a sufficiently strong gravitational field could potentially kill you outright without any need for tidal forces to rip you apart.

If you're talking about in orbit - as in, not on the ground, so not being pushed into anything - you're completely wrong about that.
One could experience literally any gravitational force and survive, as long as it was evenly distributed and not pushing you into the ground. A gravitational field wouldn't do any harm to you on its own.

In fact - and I could be talking complete rubbish here, or at least an extremely simplified version of the truth - but I seem to recall that, in truth, you wouldn't even be accelerating at all. I mean, you are, in that your velocity is changing relative to everything around you, but my understanding is that gravity bends spacetime, so that what the universe considers to be a straight and level path actually involves curving and accelerating from an outside perspective. For all intents and purposes, though, until you hit something else, you're just sitting there.


On this topic, though, can someone please explain why the term "microgravity" is considered more technically correct than "zero-G"? I mean, I've heard the explanations, and I'm entirely unconvinced. After all, "G" forces are used to measure acceleration that's felt, generally; we say a person standing on the surface of the earth is under 1 G, even though they aren't actually accelerating. By that logic, anyone in orbit should be experiencing 0 G.
But microgravity makes absolutely no sense, because it implies that the gravitational force has just been greatly diminished, and that's why things behave as though they're weightless. However, that's demonstrably less true than the whole argument about G-forces, because the gravitational force only decreases a tiny amount in our typical orbits, and in some cases is actually higher.
Is there a compelling reason that I'm missing why zero-G is considered a misnomer, and microgravity is considered correct?
Is that even the correct terminology, or was I taught incorrectly?

Edit:

You got most of the physics correct except for one problem. A slingshot around a star will not increase your speed. Slingshots only change your speed if there are two other bodies, say the Earth and the Moon. By slingshotting around the Moon, you are very slightly decreasing the Moon's orbit speed to increase your own. Without a third body to have an orbit, you cannot change your speed.

This is true, but if an object is slingshoting around a star, it seems likely that it's an interstellar traveler. One could easily imagine an object leaving one star system, orbiting around the galactic center, and getting a gravity assist from another star system. It would take a long, long, long time, especially since 0.25c was only achieved after the slingshot, but there's no reason it couldn't be done.

[Fat Rooster]

You got most of the physics correct except for one problem. A slingshot around a star will not increase your speed. Slingshots only change your speed if there are two other bodies, say the Earth and the Moon. By slingshotting around the Moon, you are very slightly decreasing the Moon's orbit speed to increase your own. Without a third body to have an orbit, you cannot change your speed.

The OP did say binary.

I don't think you would be able to get close enough for tides to be a major problem. Neutron stars are several hundred thousand degrees hot, and have magnetic fields that have higher energy density than solid matter. You run into other problems first, and unless your binary system is extremely tight you will be able to get a fairly extreme velocity off it anyway.

[gomipile]

You got most of the physics correct except for one problem. A slingshot around a star will not increase your speed. Slingshots only change your speed if there are two other bodies, say the Earth and the Moon. By slingshotting around the Moon, you are very slightly decreasing the Moon's orbit speed to increase your own. Without a third body to have an orbit, you cannot change your speed.

Others have pointed out that the OP was referring to a binary. However, even in the case of a non-binary neutron star, your statement is not correct in practice.

Your statement seems to have a hidden assumption that the neutron star is stationary with respect to some background reference frame. There does not seem to be such a reference frame in our universe. All of the stars are moving with their own velocities.

Yes, there is a general average instantaneous velocity that most stars near us are close to, in their orbits within our galaxy. However, there are exceptions which have noticeably different velocities to their neighbors.

If the neutron star does not have a velocity in the general direction of the maneuver, then your statement would be correct in spirit, if still not in fact.

But, assume there was a neutron star, or any star, moving relatively fast relative to the neighborhood in a direction a spacecraft wanted to accelerate more or less towards. Then a slingshot could be plotted using that star's momentum. Assuming it was in a convenient position, etc.

Since the star would be in orbit around the galaxy, in practice this would still be approximately like the 3 body slingshots we're used to in the solar system. But you just said "a star." Since most stars we care about are in galaxies, and they all have velocities, your statement was still wrong in practice.

[factotum]

On this topic, though, can someone please explain why the term "microgravity" is considered more technically correct than "zero-G"? I mean, I've heard the explanations, and I'm entirely unconvinced.

Because there's no such thing as zero gravity, really. The actual gravitational attraction of Earth at the altitude of the ISS is still around 90% of what it is at ground level. The astronauts inside don't feel it because they're essentially falling constantly around the Earth. Also, because the ISS is not a microscopically small object, the things inside it are mostly not orbiting at the correct velocity for their altitude, so there will be a tiny residual gravity field due to that--it would probably not even be sufficient to overcome air resistance to actually make any object move, but it's nonetheless there, so "microgravity" is truer than "zero g".

[shawnhcorey]

Your statement seems to have a hidden assumption that the neutron star is stationary with respect to some background reference frame. There does not seem to be such a reference frame in our universe. All of the stars are moving with their own velocities.

No, I haven't. In order for you to change speed, you must change your kinetic energy. That energy has to come from somewhere. If an observer changes their frame of reference, the kinetic energy of all bodies must change is proportion.

If there are only two bodies involved, it can be solved using the reduced mass . This means each body's motion can be calculated as if it were a single body in a centralized field. That means the sum of its kinetic energy and its potential energy is always the same.

But if there are three or more bodies, reduced mass can no longer be used. The sum of the KE and PE of a single body can change. It is this property that leads to the slingshot effect.

[Strigon]

Because there's no such thing as zero gravity, really. The actual gravitational attraction of Earth at the altitude of the ISS is still around 90% of what it is at ground level. The astronauts inside don't feel it because they're essentially falling constantly around the Earth. Also, because the ISS is not a microscopically small object, the things inside it are mostly not orbiting at the correct velocity for their altitude, so there will be a tiny residual gravity field due to that--it would probably not even be sufficient to overcome air resistance to actually make any object move, but it's nonetheless there, so "microgravity" is truer than "zero g".

But 'G' and 'gravity' aren't the same.
A fighter pilot pulling an eight-G maneuver isn't suddenly under eight times the gravity of a person on the ground. Likewise, saying one's in a zero-G environment doesn't mean there's no force of gravity.

[factotum]

But 'G' and 'gravity' aren't the same.
A fighter pilot pulling an eight-G maneuver isn't suddenly under eight times the gravity of a person on the ground. Likewise, saying one's in a zero-G environment doesn't mean there's no force of gravity.

Er, that's exactly what an 8-G manoeuvre means? The pilot is experiencing 8x the gravity that he would be if sat stationary in his cockpit. Per Einstein, if the pilot was in an enclosed space with no instruments he wouldn't actually be able to tell the difference between being under 8G due to acceleration, and 8G due to being on the surface of a high gravity planet.

[Yora]

Or he could be facing backwards and is decelerating. Also indistinguishable.

[Radar]

I would like to add one more thing about gravitational slingshots: aside from gaining speed from a moving celestial body there is another advantege, since the faster you go, the more efficient your thrusters are (up to exhaust velocity obviously), so they are at their best when you swing down the gravity well. So when you are saving up your fuel, which is always, you schedule your burns for when you travel close to a celestial body.

[Strigon]

Er, that's exactly what an 8-G manoeuvre means? The pilot is experiencing 8x the gravity that he would be if sat stationary in his cockpit. Per Einstein, if the pilot was in an enclosed space with no instruments he wouldn't actually be able to tell the difference between being under 8G due to acceleration, and 8G due to being on the surface of a high gravity planet.

No, G is a measure of acceleration, usually acceleration felt. Gravity is a measure only of... well, the force of gravity. The pilot isn't experiencing any change in gravity. He can tell the difference between one of them results in his spine being compressed and blood rushing from his brain, and one doesn't.
To be more clear, Standing on the surface of earth, you feel 1G. You aren't accelerating, relative to everything else on the planet. The net force acting on your body - gravity combined with the normal force of the earth - is zero.
When in freefall, or orbit, you feel nothing. According to your body, you're under 0G. And you are accelerating - the net force acting on you is ~9.8 m/s^2.
The two aren't interchangeable.

Which of those is 1G? Is a person standing still under 1G, or is the person in free fall under 1G? If the answer is only the person standing still, then someone is in freefall would be under 0G. If the answer is only the person in freefall, then what is the person standing still under? 0G, because he's stationary? And if the answer is both, then clearly G is purely a reference to gravitational force, not acceleration, and a person in a rocket accelerating upwards at 9.8 m/s^2 must also be under 1G.

[Radar]

No, G is a measure of acceleration, usually acceleration felt. Gravity is a measure only of... well, the force of gravity. The pilot isn't experiencing any change in gravity. He can tell the difference between one of them results in his spine being compressed and blood rushing from his brain, and one doesn't.
To be more clear, Standing on the surface of earth, you feel 1G. You aren't accelerating, relative to everything else on the planet. The net force acting on your body - gravity combined with the normal force of the earth - is zero.
When in freefall, or orbit, you feel nothing. According to your body, you're under 0G. And you are accelerating - the net force acting on you is ~9.8 m/s^2.
The two aren't interchangeable.

Which of those is 1G? Is a person standing still under 1G, or is the person in free fall under 1G? If the answer is only the person standing still, then someone is in freefall would be under 0G. If the answer is only the person in freefall, then what is the person standing still under? 0G, because he's stationary? And if the answer is both, then clearly G is purely a reference to gravitational force, not acceleration, and a person in a rocket accelerating upwards at 9.8 m/s^2 must also be under 1G.

What factotum was talking about is that acceleration and standing still in a gravity field are indistinguishable situations - this is the cornerstone of general relativity. So anyone in a freefall is not experiencing any force - just the definition of a straight line gets to be redefined. I am not sure how relevant it is to the discussion, but it is the current understanding of gravity that it is not a force, but just a consequence of non-Euclidean geometry of space-time.

[gomipile]

That means the sum of its kinetic energy and its potential energy is always the same.

The sum of the total kinetic and potential energy of the system is constant, but that is not true of each body separately.

A straight-on zero-spin collision between two ideal billiard balls of equal mass can also be solved using the reduced mass and a potential function representing electron shell repulsion instead of gravity. But in that case, in the reference frame of one of the balls, it starts off with zero speed and ends with the same speed as the other ball initially had. Energy is conserved for the system as a whole, but not individual elements.

[shawnhcorey]

The sum of the total kinetic and potential energy of the system is constant, but that is not true of each body separately.

A straight-on zero-spin collision between two ideal billiard balls of equal mass can also be solved using the reduced mass and a potential function representing electron shell repulsion instead of gravity. But in that case, in the reference frame of one of the balls, it starts off with zero speed and ends with the same speed as the other ball initially had. Energy is conserved for the system as a whole, but not individual elements.

But when two bodies are influenced by a central field, their individual total energy is preserved. There is a difference between this type of field and a collision.

[gomipile]

But when two bodies are influenced by a central field, their individual total energy is preserved. There is a difference between this type of field and a collision.

This isn't a true central field, since the force on one object depends on both objects' distance from the barycenter, not just on that one's distance.

[shawnhcorey]

This isn't a true central field, since the force on one object depends on both objects' distance from the barycenter, not just on that one's distance.

True but in the two-body problem, their mutual field can be replaced with a central field with reduced mass. Then their motion can be determined independently of the other as though they were in a single field.

[Radar]

True but in the two-body problem, their mutual field can be replaced with a central field with reduced mass. Then their motion can be determined independently of the other as though they were in a single field.

Yes, but this is just one of many possible reference frames. Gravitational slingshot makes sense, if the object you use is actually moving with respect to your point of origin and/or destination and moving in the right direction is also important.

Also important to stress: the conservation of individual energies is also happening only in the center of mass reference frame. Any other will show transfer of energy from one object to the other.

[shawnhcorey]

Yes, but this is just one of many possible reference frames. Gravitational slingshot makes sense, if the object you use is actually moving with respect to your point of origin and/or destination and moving in the right direction is also important.

Also important to stress: the conservation of individual energies is also happening only in the center of mass reference frame. Any other will show transfer of energy from one object to the other.

You're missing the point. The two-body problem is a special case where each body preserves its total energy regardless of the frame of reference. Each body orbits the centre of mass as though the other did not exist. I'm not going to repeat the math here but this is how it works out.

[Radar]

You're missing the point. The two-body problem is a special case where each body preserves its total energy regardless of the frame of reference. Each body orbits the centre of mass as though the other did not exist. I'm not going to repeat the math here but this is how it works out.

No, this only works in the center of mass frame of reference. Add a constant velocity vector in whichever direction to both bodies and you will readily see that their energies will change with time. Still, total energy of the two body system will be conserved.

The sitaution can be simplified a lot, if you think of it as a collision: two bodies move with constant velocities toward each other, interact and go away with some new velocities (value and direction). You can easily show even in a simplistic one dimensional case that there is an exchange of energy and momentum if the center of mass is moving. It is all relative and the only frames of reference that have special meaning in case of travel are the origin and destination frames of reference.

Also relevant: Voyager probes used gravity assist from effectively single bodies - around Jupiter or Saturn there is no other object of comparable gravitational influence.

[shawnhcorey]

No, this only works in the center of mass frame of reference.

There is no no about it. The math is clear. Each body's total energy does not change in the two-body problem. That is the fact.

[Radar]

There is no no about it. The math is clear. Each body's total energy does not change in the two-body problem. That is the fact.

Kinetic energy is dependent on the reference frame and does not scale in a linear fashion with velocity, so it will not remain constant. Center of mass is used so often as a reference frame, because the special quality of perserving individual energies simplifies the calculations greatly, but it is just a single, very special reference frame.

Quick example: you have a body on a circular orbit around something with speed v and period T. In the frame of reference of the center of the orbit, the kinetic energy of that body is simply

E_k = m v^2/2

Now make the frame of reference move with constant speed u in any direction let's say in the plane of orbit. Now the percieved velocity is a sum of both v and u vectors and the direction of v changes all the time, so the total speed is

sqrt(v^2+u^2+2uv cos(alpha(t)))

where alpha(t) = 2pi/T+alpha₀ is the angle between v and u. If we now plug that into expression for kinetic energy, we get

E_k = m(v^2+u^2+2uv cos(alpha(t)))/2

which clearly is time-dependent.

[shawnhcorey]

Kinetic energy is dependent on the reference frame and does not scale in a linear fashion with velocity, so it will not remain constant. Center of mass is used so often as a reference frame, because the special quality of perserving individual energies simplifies the calculations greatly, but it is just a single, very special reference frame.

Quick example: you have a body on a circular orbit around something with speed v and period T. In the frame of reference of the center of the orbit, the kinetic energy of that body is simply

E_k = m v^2/2

Now make the frame of reference move with constant speed u in any direction let's say in the plane of orbit. Now the percieved velocity is a sum of both v and u vectors and the direction of v changes all the time, so the total speed is

sqrt(v^2+u^2+2uv cos(alpha(t)))

where alpha(t) = 2pi/T+alpha₀ is the angle between v and u. If we now plug that into expression for kinetic energy, we get

E_k = m(v^2+u^2+2uv cos(alpha(t)))/2

which clearly is time-dependent.

Any change in energy is independent of the reference frame. In the two-body problem, the total energy of a body does not change. This is a fact. And since you don't want to learn about the two-body problem, this conversation is over.

[Fat Rooster]

Any change in energy is independent of the reference frame.

Demonstrably false, assuming you are talking about sub parts of a system, such as one body in a two body system. If we start a non-trivial interaction in a frame of reference where a body starts stationary, it clearly starts with 0 kinetic energy and ends with some. It has gained kinetic energy. Compare this to the frame of reference where the body ends stationary. It ends with 0 kinetic energy, so it has lost some.
Your confusion seems to stem from misunderstanding the limitations of the statement "Any change in energy in a closed system is independent of the reference frame".

In the two-body problem, the total energy of a body does not change.

The total energy of the system does not change, and we can solve the two body system by carefully choosing the reference frame so that the total energy of a body does not change. It is a special property of a particular reference frame, not a general rule. It disappears if we change reference frame.

This is a fact. And since you don't want to learn about the two-body problem, this conversation is over.

There is very little to learn about the two body problem. The existence of a frame of reference where a closed solution is easily found means we can solve it in any reference frame by translating into and out of that frame. We need to be aware what properties we are gaining and losing when we do these transformations though, which you clearly are misunderstanding.

[Radar]

(...) And since you don't want to learn about the two-body problem, this conversation is over.

I find this remark somewhat condescending - especially right after I backed up my claim with specific calculations. If you do not wish to continue the converstation, it is perfectly fine, but there was no reason to say things like that.

Two body problem is just that: a specific problem in the whole wide subject of classical mechanics. In order to properly understand the statements made for this particular case it is useful to first understand the context, so it is important to take time and go through everything from the basics up. In this case, understanding how mechanical quantities depend on reference frame is at the core of our dispute and the easiest way to sort it out is sit and personally do the calculations. The most basic case would be a an elastic collision between two bodies of even mass in one dimension. Center of mass reference frame helps to solve the problem easily, but then the point was about seeing the same process from a different perspective using Galilean transformation to change the frame of reference.

What is sometimes difficult to grasp is that energy is not absolute. Within a given inertial reference frame it is conserved (as in sum of energies of all the bodies - not individual energies), but each reference frame can give different values thereof precisely due to the connection between velocity and energy.

[wumpus]

But when two bodies are influenced by a central field, their individual total energy is preserved. There is a difference between this type of field and a collision.

If two bodies are influenced by a third field, you no longer have a two body problem and have a three body problem.

Even if there isn't another planet/star in the neutron star's system, it can still gain momentum wrt to some distant star, thus making it not a two body problem.

There are two issues due to G-forces in such maneuvers:
Tidal Forces. remember F=G*m1*m2/r^2? If the distance from the center of mass of the neutron star to your head and feet are sufficiently different (assuming you are standing upright) then you will experience higher gravity on your feet than your head even in "zero gravity". Being racked is not pleasant at all. This also applies to any two points in your spacecraft, so it better be strong enough to survive.

Actual acceleration (or more accurately, dp/dt): While I'm pretty sure you can't exceed a change in velocity greater than the difference in velocities between the body you are slingshoting from and the body of reference you will be traveling to/through, I'd expect that your slingshot orbit can be pretty small (assuming zero atmosphere) and the acceleration can be pretty high. You probably don't have to worry about the difference between classical acceleration and proper (relativistic dp/dt), but it should work out the same.

PS: you will still get the tidal effects in a single body system (although why you would be there is an open question). I'm not entirely sure the "acceleration" in an "n body" slingshot would be directly measurable onboard ship, although coriolis effects might be serious (I can't say I've really studied them).