(To be clear, I’m more interested in the reach and scaling of gravity.)
If we were to suddenly double earth’s mass, but not it’s size, would it’s field double in size (I assume strength as well), to the extent that if I were to measure this increased gravity at the same place I measured earth’s normal gravity, it would simply be double? If so, the least measurable point of both gravitates should also be the same?
Just wondering if there’s diminishing returns or if mass and density affects a gravity field the same regardless of whether it’s an asteroid or a billion Solar Masses.
Feel free to share any views I’m not taking into account in regards to gravity fields.
Thanks
Assuming a spherical earth, if you doubled its mass but kept the radius the same then the gravitational force on the earths surface would be twice that of the current earth.
As long as you keep the earths mass reasonable, you’re in the realm of Newtonian gravitation. Newton’s law of gravitation depends linearly on the mass of the attracting source. So doubling the mass doubles the gravitational force.
At 1 billion solar masses (firmly in the not-reasonable mass range for the earth), you’d need to consider the formation of a black hole. The Schwarzschild Radius for a 1 billion solar mass black hole (aka the event horizon) is almost 20 astronomical units or 2 billion miles. So in that case you wouldn’t be able to measure the change in gravity as you’d be within the event horizon of a black hole.
At an intermediate mass there might be some general relativity effects that could alter the linear relationship between earth mass and gravitational force as measured on the earths surface, but I’m not sure what that would be. If you were to measure earths mass from a large distance, then it should follow Newtonian dynamics and behave linearly with mass.
Thank you so much for the reply.
My understanding is that most(all?) force fields are made up of waves (as is everything?), so  hypothetically, a Gravity field should be as well? 
No.
A field is a value assigned at every point in space. It is not “made of waves”. But if the field is perturbed by an acceleration, then the perturbation is propagated as a wave.
Simple analogy: every point in the sea has a “depth”. That’s like a field. If a motorboat creates a wake, the “depth” changes temporarily. You see that change as a wave.
The other answer is correct, it’s not really accurate to say that gravity is made of waves.
In physics, a field is a physical quantity that has a value for each point in space and time The most accurate model for the gravitational field is general relativity, however for many cases it’s sufficient to just use Newtonian Dynamics. In GR, changes to the gravitational field propagate at the speed of light in a vacuum, c. It’s possible to create gravitational waves by rapidly accelerating a massive object, which occurs in inspiralling black holes or neutron stars. But the gravitational force pulling the pair of black holes together isn’t made of waves; the black holes are minimizing their gravitational potential energy as defined by the gravitational field.
force fields are made up of waves (as is everything?),
I wanted to address this since I think you might have a common misconception. Particles (photons, electrons, quarks, protons, neutrons, etc) are described in quantum mechanics using a wavefunction. But this doesn’t make these particles “waves”, they are still quantum mechanical particles. They simply don’t have a defined location (if using a spatial wavefunction, you can also work in an alternative basis like energy or momentum). If the particle interacts with something on the classical scale, it’s wavefunction will collapse to a single point where the location is defined.
If you try to model a quantum mechanical particle as either a classical point-like particle (single point in space) or a classical wave you will fail. Before quantum mechanics was discovered lots of very smart people tried and failed to use just waves or point-like particles. Quantum mechanics, using wavefunctions, is consistent with the fundamental nature of reality as far as we can tell.
You’ve gotten good answers about how much the gravitational force would be at the same position, so I want to answer your 2nd question.
The strength of the gravitational field decreases as the inverse of the distance squared. So double the distance away and the force is one quarter.
That also means it never really ends. Technically you are feeling the effects of gravity of everything in the known universe (as we currently see it). Obviously since most of those things are so far away you don’t really notice, but it’s still happening.
But let’s assume there’s some minimum amount we can detect. If you put you’re probe at the spot where it was measuring the minimum and earth’s gravity doubles, then you would measure double the minimum. You could then move your probe farther away by a factor of the square root is 2 and you would again measure the minimum value.
This seems interesting.
Gravity fields are just potentials, as gravity requires at least two bodies, right?
If the universe only contained one body of irrelevant mass, without anything else to interact with it would just sit there. Further there would be no time, as there would be no change.
If Earth’s mass were to double, all gravitational relations, including potentials, would also increased but it’s not exactly double as the equation should also account for the other body or bodies masses.
I’m not a scientist, I’m just smoking weed on a sunday. I’m hoping some actual smart people can explain this like I’m high.
If Earth’s mass were to double, all gravitational relations, including potentials, would also increased but it’s not exactly double as the equation should also account for the other body or bodies masses.
I think the simple Newtonian version is: Break down each gravitational relation (A and B pull on each other) in it’s components: A pulls on B, and B pulls on A. If you double the mass of A, this has two effects:
- A pulls on B twice as much
- B pulls on A the same, but needs twice as much force to achieve the same acceleration (a = F / m)
From my layman perspective, yes the measured gravity would be double it’s original value if measured from the same place.
Gravity is an [edit: inverse squared] function, so it gets weaker at an exponential rate as you move away from the source. But even if it’s a value of 1.0 at Earth’s surface and .02 at some distant point from Earth, doubling Earth’s gravity would double both values to 2.0 and .04, respectively.
Gravity is not exponential. It is linear with mass and inverse square with distance.
Thank you for the clarification. Best way to get the right answer is to post the wrong one.
This one, @[email protected], has the correct answer…
So, don’t believe in the crackpot idea that it would be exponential
Whoops! I said exponential instead of inverse squared. What a crackpot I am.
I did makeI have made many mistakes, much worse than this one and on many occasions. I would say : don’t be so hard on yourself since it’s important to forgive ourselves.I do believe the following correction should be made again to your text though :
Gravity is an [edit: inverse squared] function, so it gets weaker at
an exponentiala squared rate as you move away from the source.
This was the answer I was after, thank you!
Additional question that’s related, if you’d like to try it: I’ve read about vacuum energy/zero point energy - hawking radiation exists because of those theories. From what I’ve read, vacuum energy has the potential for any form of matter but because of the uncertainty principle, less likely to produce higher forms of energy, and thus why most fluctuations produce only virtual particles. My main question then is: so no matter what, all of space ether has matter or potential for matter? If so, should a photon actually collide with a virtual particle it would actually stay in physical existence)?
Thank again
I would assume I’m not interested in any of the associated crackpot ideas some have.
If a photon collides with a particle, virtual or not, then the particle will eventually emit another photon.
