4 ms·

In the interior of a sphere of uniform density, the gravitational strength is proportional to the radius, exactly like Hooke's law for an ideal spring. That's w

In the interior of a sphere of uniform density, the gravitational strength is proportional to the radius, exactly like Hooke's law for an ideal spring. That's why the object in the bore hole undergoes harmonic motion. Why is the graviational strength proportional to the radius? Firstly, you have to know that the field strength is zero inside a hollow sphere. This is part of that shell theorem. So for a point at a given depth inside the sphere, we can divide the sphere into a hollow sphere consisting of everything less deep, and a solid sphere consisting of everything deeper. Only the deeper sphere matters; we can ignore the hollow sphere. So as we progress toward the centre, the attraction is due to a smaller and smaller sphere, whose mass is proportional to r^3. The radius is shrinking though, which has the effect of increasing gravity: the gravitational field strength is proportional to 1/r^2. Wen we combine these factors, we get r.

FredPret 1y ago
It’s really odd how often and where damped spring motion comes up
- C-x_C-f 1y ago
  Spring motion is the motion of systems where the force is proportional to the distance. Many interesting systems (like springs) are near equilibrium, which means that the potential energy is at a local minimum. A spring is an example, but also a pendulum. When the potential is at a local minimum, its gradient is zero. So if you Taylor expand it you only get second-order contributions. For a spring, the potential energy looks like V(x) = V(0) + k * x * 2 where x is the displacement and k is a constant. Differentiating, you get harmonic motion: F(x) = k * x Broadly speaking, this applies to all systems near equilibrium, simply from Taylor expanding the energy. And it's not only in classical mechanics, but in all branches of physics. Sydney Coleman [0] is often quoted as saying something like "QFT is simple harmonic motion taken to increasing levels of abstraction." [1] [0] https://en.wikipedia.org/wiki/Sidney_Coleman https://en.wikipedia.org/wiki/Sidney_Coleman [1] https://physics.stackexchange.com/questions/355487/qft-is-simple-harmonic-motion-taken-to-increasing-levels-of-abstraction https://physics.stackexchange.com/questions/355487/qft-is-si...
- snickerbockers 1y ago
  Isn't that just a second order differential equation?

FredPret 1y ago

It’s really odd how often and where damped spring motion comes up

C-x_C-f 1y ago
Spring motion is the motion of systems where the force is proportional to the distance. Many interesting systems (like springs) are near equilibrium, which means that the potential energy is at a local minimum. A spring is an example, but also a pendulum. When the potential is at a local minimum, its gradient is zero. So if you Taylor expand it you only get second-order contributions. For a spring, the potential energy looks like V(x) = V(0) + k * x * 2 where x is the displacement and k is a constant. Differentiating, you get harmonic motion: F(x) = k * x Broadly speaking, this applies to all systems near equilibrium, simply from Taylor expanding the energy. And it's not only in classical mechanics, but in all branches of physics. Sydney Coleman [0] is often quoted as saying something like "QFT is simple harmonic motion taken to increasing levels of abstraction." [1] [0] https://en.wikipedia.org/wiki/Sidney_Coleman https://en.wikipedia.org/wiki/Sidney_Coleman [1] https://physics.stackexchange.com/questions/355487/qft-is-simple-harmonic-motion-taken-to-increasing-levels-of-abstraction https://physics.stackexchange.com/questions/355487/qft-is-si...
snickerbockers 1y ago
Isn't that just a second order differential equation?