4 ms·
Yes, but a better clock means more precise measurements, means we can locate smaller masses to higher precision.
by btilly 3y ago
Yes, but a better clock means more precise measurements, means we can locate smaller masses to higher precision.
- defrost 3y agoDoes it? Inversion is rarely unique, and it's not due to the precision with which the field is measured. https://earthsciences.anu.edu.au/study/student-projects/novel-approaches-geophysical-inversion https://earthsciences.anu.edu.au/study/student-projects/nove... https://inside.mines.edu/~rsnieder/snieder_trampert_00.pdf https://inside.mines.edu/~rsnieder/snieder_trampert_00.pdf Epilogue: Linear inverse problem theory is an extremely powerful tool for solving inverse problems. Much of the information that we currently have on the Earth’s interior is based on linear inverse problems Despite the success of linear inverse theory, one should be aware that for many practical problems our ability to solve inverse problems is largely confined to the estimation problem.
- btilly 3y agoYes, inverse problems are hard. And not always possible in practice. See, for example, https://www.ams.org/publicoutreach/feature-column/fcarc-199706 https://www.ams.org/publicoutreach/feature-column/fcarc-1997... for a case where one isn't possible. That said, the gravity technique is one that actually gets used today. With better precision, it can be even more useful than it already is.
- JanisErdmanis 3y agoThe problem is that the planet could be hollow and produce the same gravitational measurements on the surface and outside. It needs to be coupled with a model that introduces constraints for the inverse problem to be defined.
- blincoln 3y agoSince mining is only concerned with material that's within maybe 0.1% of the distance from the surface to the core, seems like you'd just need to move the sensor around and make sure the signal changes about where you'd expect for a mass of X Kg at a depth of Y meters instead of a supermassive chunk of dense material much deeper. Or, to put it another way, build a grid map of the area and subtract any background signal. Would that not work for some reason?
- btilly 3y agoConsider the special case of a spherical deposit. You can find the center and mass of the deposit, but not its volume or density. But now that you know it is there, you can use other techniques, like seismic measurements, to nail that down.
- JanisErdmanis 3y agoIn practice, that's what would happen. Move around until seeing some larger gravitational pull, likely indicating some deposit. However, formally, this is not correct due to the mere fact that the gravitational force is proportional to 1/R^2, just like a Columb force. Thus, there are infinite numbers of mass distributions that produce the exact same gravitational field on the surface. The planet could be hollow, and we would not know it only from the field measurements. A practical constraint is mass density, which has maximum and minimum values. We can make a crude approximation that the planet's density is constant, evaluate the field on the surface from the planet's shape and compare it with measurement. This would be more useful, but still, it wouldn't tell us whether there is a combo of water reservoir and a large massive deposit below it.
- cyanydeez 2y agoThats why you generate typical geologic formations and add a few drillhole constraints. Sure this isnt going to be a star trek scanner but for practical purposes theres a bunch of other techniques to constrain the results