# Thread: Gravitational time dilation, and weight.

1. Order of Kilopi
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## Gravitational time dilation, and weight.

Say you could lower a 10kg piece of metal from a spaceship in geostationary orbit*, onto some scales on the ground. The whole weight isn't supported by the scales, only around half of it. The rest of the weight is carried, and measured, by the spaceship.

So my question is would the two measurements add up to 10kg, or would it be less, due to gravitational time dilation(ignoring the weight of the supporting wire)?

*
edit: or more simply, hovering over one of the poles, at a great distance.
Last edited by WaxRubiks; 2018-Aug-08 at 02:06 AM.

2. First things first, before delving into relativity, you must separate mass and weight. Weight is a force and its unit is Newtons if you want to use kg masses. If you want to accelerate a mass you need a force and to stop a mass freefalling in a gravity field you need a force. Geostationery satellites are in orbit, cunningly at an orbit period that matches Earth’s rotation. The mass there is floating about. Lowering it involves changing the radius and therefore sideways forces to keep it alomg a radius, Coriolis force. At the pole the hovering craft has to exert a thrust to stay there, to balance its weight. So lowering the mass does not require the side force. The wire “sees” the force all the way down, it’s increasing due to the changing distance using just Newton. So if you set up your polar experiment ignoring the wire mass and weight, the hovering craft would measure 98 Newtons approx or whatever the g value is at that point times 10 kg. The hoverer-is not moving relative to Earth, despite or thanks to, its thrusters, so there is no relativity calculation to do.
Last edited by profloater; 2018-Aug-08 at 08:46 AM.

3. Weight is a force. There is no time component to the units of force. Therefore unaffected by time dilation. Time dilation comes in when you generate acceleration using your force, or calculate a force from acceleration, observing from a different reference frame.

Grant Hutchison

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Originally Posted by profloater
First things first, before delving into relativity, you must separate mass and weight. Weight is a force and its unit is Newtons if you want to use kg masses. If you want to accelerate a mass you need a force and to stop a mass freefalling in a gravity field you need a force. Geostationery satellites are in orbit, cunningly at an orbit period that matches Earth’s rotation. The mass there is floating about. Lowering it involves changing the radius and therefore sideways forces to keep it alomg a radius, Coriolis force. At the pole the hovering craft has to exert a thrust to stay there, to balance its weight. So lowering the mass does not require the side force. The wire “sees” the force all the way down, it’s increasing due to the changing distance using just Newton. So if you set up your polar experiment ignoring the wire mass and weight, the hovering craft would measure 98 Newtons approx or whatever the g value is at that point times 10 kg. The hoverer-is not moving relative to Earth, despite or thanks to, its thrusters, so there is no relativity calculation to do.
yes, I know that mass and weight are different things.

In both cases the ship would have to maintain altitude, and if the wire hooked over some scales, it would measure weight, wouldn't it, with the ship being a pretend ground..?

...

Yes, I see Grant....that makes sense.

5. no they are very different.
The orbiter is in free fall so the mass there is weightless. To make it drop to Earth requires a tangential deceleration. Then it would adopt a lower and actually slower orbit but from Earth view it would look faster. To go all the way it would require continuous tangential deceleration and if following a vertical path, it would at each point be in unstable orbit. The vertical force would increase from weightless to full weight as seen on the orbiting scales.
The polar hoverer would measure the weight as local g times 10 kg, and it could drop the mass toward the pole seeing its weight increase toward the normal surface value. The hoverer is not moving nor accelerating relative to Earth in the Earth frame so no relatavistic effects arise. Given the continous tangential thrust to the mass to get it to fall straight down, it arrives with no relative velocity to Earth and no radial velocity at the end, so no relativity to calculate.

6. Originally Posted by grant hutchison
Weight is a force. There is no time component to the units of force. Therefore unaffected by time dilation. Time dilation comes in when you generate acceleration using your force, or calculate a force from acceleration, observing from a different reference frame.

Grant Hutchison
actually the dimensions of force are ML/T^2 ie mass times acceleration but it still the case that time dilation is not relevant to the OP thought experiment.

7. Originally Posted by profloater
actually the dimensions of force are ML/T^2 ie mass times acceleration ...
Yeah. But in the case of balanced forces, as with the conventional definition of gravitational weight used in the OP, there's no acceleration, so you ending up putting it in only to cancel it out. I sort of tried to distinguish between the static case and the second-law case, but I think I made a hash of it.
However, since the question invokes GR, I guess there "really" is an acceleration going on, as the cable from the spacecraft continuously accelerates the weight relative to a succession of instantaneous local free-fall reference frames.

Grant Hutchison

8. Originally Posted by grant hutchison
Yeah. But in the case of balanced forces, as with the conventional definition of gravitational weight used in the OP, there's no acceleration, so you ending up putting it in only to cancel it out. I sort of tried to distinguish between the static case and the second-law case, but I think I made a hash of it.
However, since the question invokes GR, I guess there "really" is an acceleration going on, as the cable from the spacecraft continuously accelerates the weight relative to a succession of instantaneous local free-fall reference frames.

Grant Hutchison
that's interesting but my take is that the specific question involves the space station being stationary relative to the earth surface. The centripetal of the orbit case surely does not invoke relativity since its distance remains the same. If we were to estimate its size with a telescope its speed relative to the speed of light would cancel if we view both sides at once in an image.

So to answer what weight the space station measures in both cases, ignoring the wire, is actually simple and does not involve SR nor GR in the static case. If we have to look at a falling accelerating weight, it's more complicated.

9. As soon as the OP invokes "gravitational time dilation" it involves general relativity, though. There's a difference in proper time between the space station and the ground, which doesn't occur in Newtonian gravity.

Grant Hutchison

10. Originally Posted by grant hutchison
As soon as the OP invokes "gravitational time dilation" it involves general relativity, though. There's a difference in proper time between the space station and the ground, which doesn't occur in Newtonian gravity.

Grant Hutchison
Ah, now i see, the proper time difference would affect the reading of weight from the distant station? Not much in this case, but i may not be clear about the question, since it talked about shared weight. If we assume the wire is arranged to be static, then the force in the wire is the same both ends, and there is no change to reveal a proer time difference. Is that right?

11. Originally Posted by profloater
no they are very different.
The orbiter is in free fall so the mass there is weightless. To make it drop to Earth requires a tangential deceleration. Then it would adopt a lower and actually slower orbit but from Earth view it would look faster.
I'm probably misunderstanding, but this seems impossible to me. If you are closer to an object but going slower, then I would guess you are not in an orbit anymore but would have to fall further to gain the speed to be in actual orbit, since the force of gravity grows as you get closer. But maybe you are talking about the process of getting to a lower orbit?

12. Originally Posted by Jens
I'm probably misunderstanding, but this seems impossible to me. If you are closer to an object but going slower, then I would guess you are not in an orbit anymore but would have to fall further to gain the speed to be in actual orbit, since the force of gravity grows as you get closer. But maybe you are talking about the process of getting to a lower orbit?
Maybe i did not put it well. The thought experiment requires an orbiting satellite to drop a mass on a wire while weighing it. If you shot it out radially towards Earth it would still have its tangential velocity so it would be in a too low orbit and would climb. To make it stay on the vertical line you have to decelerate it tangentially, so if at some point on the descent you let it go it is now going too slow to be in orbit at that height, so it spirals down. That complicates the experiment, so the OP offered a polar version but that needs the craft to be hovering using thrust to stay up.

13. Maybe i should use the opposite, if you want to launch a geostatioary satellite, you start out thrusting up and away and then tangentially climbing away from Earth and going ever faster. An Earth observer sees the satellite moving but eventually it is going so fast it reaches an orbit where its orbit period matches Earth spin. Now it appears staionary but it’s whizzing through space. To drop down vertically along a radius is an unlikely plan but it needs a constant reverse thrust. It would then fall down the radius.
Last edited by profloater; 2018-Aug-09 at 09:10 AM.

14. Originally Posted by profloater
Maybe i did not put it well. The thought experiment requires an orbiting satellite to drop a mass on a wire while weighing it. If you shot it out radially towards Earth it would still have its tangential velocity so it would be in a too low orbit and would climb. To make it stay on the vertical line you have to decelerate it tangentially, so if at some point on the descent you let it go it is now going too slow to be in orbit at that height, so it spirals down.
To be clear, in the absence of atmospheric drag, just newtonian gravity, there is no spiraling, it's either in orbit or it collides. Of course, there could be other complications, but those aren't really in consideration
That complicates the experiment, so the OP offered a polar version but that needs the craft to be hovering using thrust to stay up.
Yeah, so that there is no relative movement between the two platforms, with a connecting cable so that some of the weight of the 10kg mass is supported by both platforms, in some fashion. That does require sensitive and accurate station keeping--the effect is bound to be extremely tiny anyway, but this is just a thought experiment, we don't actually have to build it.

The OP becomes a question of how the tools on each platform determine mass. And if the gravitational time dilation affects that determination somehow, however subtly.

15. Ok so i considered the whole weight for simplicity, ignoring how tiny the time dilation is , and assuming the set up can measure a static state, does time dilation affect weight? I thought not. But then you say it does? OK more thought required.

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