Does observer's sense of gravity change?

[tanus5]

I know this is a very newbie question but I'm a newbie when it comes to the mathematics involved in GR. My question is if gravity is the curvature of space-time then would an observer on a larger gravity source, such as Jupiter actually experience more gravity? Since space-time itself is compressed wouldn't Jupiter be much larger to a local observer than we have measured since they are also compressed more by space-time? I would think this would mean the gravity experienced by the observer would end up the same since their space-time compressed self would see a smaller portion of the gravity due to the fact that the observer and gravity source are both space-time compressed. I don't yet have the mathematical skills to try to do calculations on tensors to answer this question myself. I don't think lower "weight" experienced on the moon is good enough proof for this since a portion of the gravity of the moon is actually canceled, on the light side of the moon, by the gravity of the earth.

 

[Nugatory]

I don't yet have the mathematical skills to try to do calculations on tensors to answer this question myself. I don't think lower "weight" experienced on the moon is good enough proof for this since a portion of the gravity of the moon is actually canceled, on the light side of the moon, by the gravity of the earth.

Just for grins, you might want to try calculating how much that "cancelling" effect is, compare with the actual known difference between the surface gravity of the Earth and the surface gravity of the moon... No tensors needed, just a calculator, wikipedia to look up the values of a few constants, and the formula for the gravitational force between two objects:
$$F=\frac{Gm_{1}m_{2}}{r^2}$$ where G is the gravitational constant, the two m's are the masses of the two objects, and r is the distance between their centers.

 

[tanus5]

Thanks for the reply, but I can't make any sense of the answer.

Distance between Earth and moon = 3.844 x 10^8 m
Mass of Earth = 5.97219 × 10^24 Kg
Mass of moon = 7.34767309 × 10^22 Kg
Gravitational constant = 6.67300 × 10^-11 m^3 /kg s^2

If my calculations are correct this brings me to an answer of

1.9817 x 10^21 m kg/s^2

Earth gravity is listed as 9.78 m/s^2

Assuming I can ignore the kg in my answer (cringe) the number seems much larger than the gravitational pull of the Earth so this answer doesn't make any sense to me. Here are my intermediate results...

2.9282258 x 10^37 m^3 kg s^-2 / 1.4776336 10^16 m^2

Either way... something went wrong here. Can you fix my mistake?

 

[tanus5]

If my calculations are correct this brings me to an answer of

1.9817 x 10^21 m kg/s^2

As an afterthought to get rid of the kg I divided by the mass of the moon which gives me

3.3318 x 10^-2 m/s^2 which is very small compared to Earth's gravity. To be completely honest this exercise has simply confused me more about GR. If these calculations are true than the curvature and compression of space-time doesn't seem to fully account for gravitational effects since the observer is still getting crushed (to death) even with the compression of space-time.

 

[Nugatory]

Either way... something went wrong here. Can you fix my mistake?

You're calculating the force between the Earth and the moon and getting a huge number, not so surprising when you consider that that the moon is a huge chunk of rock and that force is holding it in its orbit.

To see whether the Earth's gravity can counteract any significant fraction of the moon's gravity you'll have to assume a mass (say 1kg to make the math easy) sitting on the surface of the moon, on the side facing us. Calculate the force of the Earth's gravity on that mass, calculate the force of the moon's gravity, compare.

 

[tanus5]

Nugatory,

Thank you for your assistance. I may have the issue clearing up a bit. By dividing by the mass of the moon I believe I calculated the average gravity experienced by the moon from the Earth which should be equal to the negating effect. On realizing this I considered the astronauts jumping around, seemingly with less weight, but in reality while they may visually see themselves as being able to jump higher with less effort, in reality, based on GR, they are actually traveling through the same amount of space as they would if they jumped on earth. Distorted because the space is less compressed on the moon. While I can't use the tensors yet to figure out an exact answer it seems that the observer would feel the same amount of gravity, but on Jupiter the amount of mass weighing down on their heads would be much greater, due to both the compression and the larger amount of atmosphere the increased curvature would allow. No different than a submarine going too deep in the ocean, it would get crushed, but those inside a submarine on the surface of Jupiter would experience more gravity because it will take more effort to move the same amount of distance due to the increased amount of space they must travel through to move the same distance. In such an environment I don't think the weight/kg would be increased, so if we were in a submarine we wouldn't be crushed by the weight of our head, but we'd barely be able to move due to how compressed the space is.