…from asteroids:
To properly explain the crater distribution, Ito and Malhotra say some other factor must have been involved. One possibility is that we simply haven’t seen all the craters yet: the ongoing lunar mapping missions may help on that score.
Another idea is that the Earth’s tidal forces tear Earth-crossing asteroids apart, creating a higher number of impacts than might otherwise be expected.
But the most exciting and potentially worrying possibility is that there exists a previously unseen population of near Earth asteroids that orbit the Sun at approximately the same distance as the Earth. These have gone unnoticed because they are smaller or darker than other asteroids, say Ito and Malhotra.
“More complete observational surveys of the near-Earth asteroids can test our prediction,” they say.
And let’s not waste too much time about it. By some reckonings, asteroid impacts represent the greatest threat to humankind that we are able to calculate.
Even more to the point, it’s not only the greatest one we can calculate, it’s probably the greatest one that we can actually mitigate (short of colonizing the galaxy).
“earth’s tidal forces” illustrates a very common misconception about gravity. They don’t exist, or are at most not measurable in any known situation. The exception is theoretically in the vicinity of a “black hole” which is also only theoretical.
Think about it. What’s the pressure at the center of the Earth or the Sun? Zero.
What’s the pressure at the center of the Earth or the Sun? Zero.
How do you figure?
Think about it. What’s the pressure at the center of the Earth or the Sun? Zero.
/facepalm
Naw, let’s wait ’til after the planet is dead to worry about it.
Pressure is the result of gravity. Atmospheric pressure is the “weight” of the atmosphere per unit area. The pressure in the ocean is depth dependent and is the “weight” of the water above. We calculate the force of gravity assuming point masses. Thus, the gravity at the surface of the earth is calculated by placing its full mass at the center. Simple calculation.
But actually it is the summation of the gravitational pull of each individual atom of the earth – the ones on the far side obviously being far weaker than the nearer. When you reach the center of the earth the gravitational force of the individual atoms is the same in all directions, thus giving a net zero.
As to gravitational tidal force, it is far more realistic to analyze them using the principles of General Relativity rather than Newtonian physics. In GR bodies, including photons, follow geodesics which are least action paths through the 4 dimensional continuum. This means no acceleration – the old falling elevator trick. Of course we, as observers, perceive an acceleration but the body doesn’t.
This leads to another glaring puzzle in astrophysics. The Sun is thought to be a thermonuclear furnace. The initial trigger for the reaction was supposedly the tremendous gravitation pressure at the center. No pressure, no trigger.
We calculate the force of gravity assuming point masses. Thus, the gravity at the surface of the earth is calculated by placing its full mass at the center. Simple calculation.
It can only be treated as such from a position on or outside of the sphere.
When you reach the center of the earth the gravitational force of the individual atoms is the same in all directions, thus giving a net zero.
That doesn’t mean that the pressure is zero. All of those gravity force vectors are converging on a single point from all directions, and yet you imagine that this will result in zero pressure?
The pressure increases with depth as you approach the center, due to all the weight above. Are you postulating some kind of magical singularity at the center where it suddenly goes from humungous to zero?
Most references calculate the pressure at the terracenter as on the order of 400 GPa.
Rand,
The gravity force vectors are not converging on a point, they are pulling on a point. As to the monotonically increasing pressure as you approach the center of the earth – that is conjecture, mistaken in my view. No measurements.
Off hand I can’t visualize the curve. I guess I could whip up model but I don’t care to spend that much time and besides I have to go to the store. Just realized I’m out of gin.
Snark aside, this is something I’ve thought about off and on for 40 years. My calculus prof considered it, shook his head and said he had some tests to grade. No modeling back then. Didn’t I say think about it?
Regards,
Roy
It’s not hard to model, mathematically. Those who have “thought about it” and done so come up with the pressure estimate that I described. I’m unaware of any who have come up with a different result, until you. If you want to convince us, you’ll have to show an actual analysis that conforms with math and known physics (e.g., no discontinuities at the center). The analysis would have to show the pressure curve as a function of distance from the center (assuming homogeneity as a function of depth for simplicity and preferably normalized, where the radius has a unit of one).
It is true that the inside of a hollow sphere has no net gravity force field. It is not true of the inside of a solid one, even at the center.
Interesting bit of applied mathematics.
A solid sphere is a hollow sphere where the size of the hollow is zero.
I’ll have to think about this.
Hokey Dokey.
Assume five units masses spaced one unit apart:
a….b….c….d….e
The calculation of the gravitational forces:
a: 1 + 1/4 + 1/9 +1/16
—b—–c——d—–e
b: -1 +1 + 1/4 + 1/9
……a…c……d……..e
c: -1/4 – 1 + 1 +1/4
—–a—–b—d—-e
I don’t know how to make big Sigmas or integrations symbols – not on my keyboard.
Make sense?
Sorry for the confusing diagram. Typing area is fixed font. comment is proportional.
You’re not showing the calculation. You’re just stating what you imagine the results are.
Rand,
a…..b…..c…..d…..e
F=GMm/d^2
f(a) = f(b) + f(c) + f(d) + f(e) = 1 + 1/4 + 1/9 + 1/16
f(b) = f(a) + f(c) + f(d) + f(e) = -1 + 1 + 1/4 + 1/9
f(c) = f(a) + f(b) + f(d) +f(e) = -1/4 – 1 + 1 + 1/4 = 0
Sorry, that’s not rigorous math.
Rand,
Tough to express using ASCII character set. I’ll try words.
The force on a equals to sum of the forces of b, c, d and e. Since the distances are 1, 2, 3 and 4 respectively those forces are 1/1^2, 1/2^2, 1/3^2 and 1/4^2 = 1 + 1/4 + 1/9 + 1/16.
For b the distances are -1, 1, 2 and 3 = (-1) + 1 + 1/4 + 1/9.
For c they are -2, -1, 1, 2 = (-1/4) + (-1) + 1 + 1/4 = 0.
That is indeed rigorous even if the notation looks funny.
I’m not sure what any of that has to do with the question of tidal forces. The real point is that in order for tidal forces to affect these near-earth asteroids they need to be either (1) extremely large or (b) extremely close to Earth’s surface. Math? Why?
I suspect the forces that would actually affect the asteroids in question would be Earth’s gravity per se, altering orbits and potentially flinging a larger number at the Moon than would otherwise hit it.
You’re talking about two different things. Both of these statements are true.
The pressure increases with depth as you approach the center, due to all the weight above.
When you reach the center of the earth the gravitational force of the individual atoms is the same in all directions, thus giving a net zero.
High pressure and zero gravity are both true at the center.
Ken,
Pressure = weight = gravitational force.
If gravitational force = 0 then whence cometh the pressure? Numbers please?
BTW, Roy, this solves your glaring puzzle in astrophysics. Although the mass near the center does indeed contribute little to pressure, the pressure itself builds all the way to the center.
Pressure and tension may come from gravitational forces, but are independent (they could exist even if gravity was not the source.)
If instead of using a weight on a string I could apply tension by pulling on both ends.
Roy, you seem to have a sign problem. Here’s a hint: gravity is a vector, but pressure is scalar…
When you reach the center of the earth the gravitational force of the individual atoms is the same in all directions, thus giving a net zero.
The problem is, you’re calculating gravity, not pressure. The two are not equvalent.
If you go from sea level to 300 feet below sea level, there is no significant change in gravity but the pressure goes up tremendously. Pressure is not proportional to gravity, it is proportional to the weight of the column of water above you.
Similarly, at the center of the Earth, pressure is proportional to the weight of the rock above you. Only an infinitesimal portion of that rock is located at the exact center of the Earth. Therefore, you cannot assume zero net gravitational pull on the entire column of rock. Instead, you have a gravitational pull that varies between zero at the center and 1.0 at the surface. When you integrate over that, the result does not turn out to be zero.
Pressure = weight = gravitational force.
No, if that were true, a vacuum bottle would be an antigravity machine.
Roy, you seem to have a sign problem.
Rand, I don’t see any sign problem. The examples he gave use the proper signs. The problem is assuming gravity and pressure are equivalent.
Rand, I don’t see any sign problem.
Then you, like Roy, don’t understand the difference between vector and scalar values.
Edward is quite right. Pressure is not weight. The thing to remember here is that every point inside the Earth has to support the weight of whatever is above it. Pressure is the resisting force per unit area to counter the summed force of that overburden per unit area.
Again, once one understands the difference between vector values (gravity force) and scalars (pressure), all becomes clear.
Gentlemen,
Thank you for a lively discussion. Like I said, think about it. There is no practical experiment that can resolve this question. I know that it has disconcerted profs that I posed it to. I’ll continue to think about it too, though I doubt that I have 40 more years to ponder.
Regards,
Roy
Then you, like Roy, don’t understand the difference between vector and scalar values.
I know the difference. I’m saying his signs are correct for a WRONG assumption. Both vectors and scalars can have negative values.
Roy, are you getting the point that pressure may be caused by gravity but is not the same as gravity?
There is no practical experiment that can resolve this question.
No, but theory can do so, with little effort.
And in fact, an experiment could do so, in weightlessness. Put in pressure transducers in a sphere suspended in weightlessness. It could probably be done on ISS.
Not that it is necessary, because there is no uncertainty in the theory.
I know that it has disconcerted profs that I posed it to.
Then they understood neither the math, or the physics.
Gentlemen,
I defer to your greater knowledge. My last real taste of physics was 40 years ago. Since then it’s been bits and bytes.
Thank you again,
Roy
Also tidal forces are legitimate forces in certain local frames of reference. They’re just local variations in global force fields that objects of extended size experience. Any mass, not just black holes generates tidal forces.
Would the tidal forces caused by passing near the Earth across the diameter of an asteroid be sufficient to fracture it into multiple objects? Especially if the asteroid were a loose aggregate and not a solid object to start with.
Would the tidal forces caused by passing near the Earth across the diameter of an asteroid be sufficient to fracture it into multiple objects? Especially if the asteroid were a loose aggregate and not a solid object to start with.
Not unless it were an asteroid of comparable mass to the earth.
In other words…no.
Ken is correct.
It depends on what “it” means. I assumed that it was referring to the earth. If it is referring to the asteroid, then yes, there is a good chance that it would break up from tidal forces, depeneing on size, composition and closeness of path to the earth.
“The problem is, you’re calculating gravity, not pressure. The two are not equvalent.
If you go from sea level to 300 feet below sea level, there is no significant change in gravity but the pressure goes up tremendously. Pressure is not proportional to gravity, it is proportional to the weight of the column of water above you.”
Well reasoned point. But there are some folks around here not receptive to reason no matter how clearly it is presented.
My poor grammar… I meant ‘it’ to refer to the asteroid.
As an interesting aside: if you dig a hole on Earth, does the acceleration of gravity increase or decrease as you descend into the hole?
Think carefully.
It may do either depending on the density distribution. My intuition says it would initially go up since the heavier elements are at the core and acceleration goes up with the square of distance, but it eventually must decrease toward zero as you approach the center.
Ken: that’s correct. As I understand it, the acceleration would reach a maximum at the mantle-core boundary, then decrease to zero at the center.
But to come back to the article:
“But the most exciting and potentially worrying possibility…”
Still more exciting than worring to me, because I’m thinking that such an unnoticed population of NEAs would have very low delta-V’s for round trips from Earth (although I expect that launch windows for any one target would be astonishingly infrequent).
Them possibly being small isn’t any disadvantage from the point of resource retrieval to High Earth Orbit for use in space manufacturing, either. Smaller helps us get started with more modest requirements and capabilities.
I don’t buy that there’s a missing population of NEAs that we haven’t seen yet. The thing is that there’s plenty of evidence in the geological record of multiple asteroid impacts over short periods of time. For example, a number of craters (and suspected craters) date from 65 million years ago (see here for a somewhat confusing discussion of five possible impacts with the margin of error containing the KT boundary). They might be spread out over a few hundred thousand years or they might be a large asteroid that, due to tidal forces, broke up prior to impact with Earth.