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I suspect this will be closed as "opinion based". I don't believe there is a canonical answer. Usually microscopic scale relates to phenomena that occur on a level much smaller than the system under consideration (atoms in a crystal when you are thinking about the crystal, for example). There is an analogy with micro- and macro-economics. Micro-economics ...

4

The #1 cause of low-earth-orbit decay is atmospheric drag. There is just enough air up there to cause a tiny amount of drag, slowing the satellite down just like an aircraft without engines. End result: everything in LEO will return to the surface rather soon. Things that are higher up, like geosynchronous satellites or things in high inclination orbits ...

3

You are partially correct. If you have two objects with moment of inertia $I_1$ and $I_2$ then if one applies a torque to the other, they will start rotating in opposite directions. So if one object is the reaction wheel and the other is the satellite, the satellite will indeed rotate (while the reaction wheel, internally, is rotating in the opposite ...

3

According to Newton's second law, $\overrightarrow F=m\overrightarrow a$ which means an object accelerates in the direction of the applied force. Acceleration means change in velocity is in the direction of the applied force. The final velocity will not necessarily be in the direction of the applied force, except when the initial velocity is zero.

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The parallel axis theorem $I_B = I_A + md^2$ only applies when $A$ is the center of mass.

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There is a sense in which the first and third laws are the same, because the are both just saying that momentum is conserved. The first law tells us that the momentum of an isolated object remains constant, and the third law tells us that the net change of momentum in any interaction between two bodies is zero i.e. momentum is conserved. So both laws can be ...

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The first law is really just the statement that intertial frames exist.

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This answer is based on $\phi = \phi_0 \cos(\omega_0 t)$ being true. EXPRESSION OF VELOCITY BY KINEMATICS It is important to note that $l\dot\phi \neq l\omega_0 \phi$. You find this to be the case whenever you find $\dot\phi$ by differentiation. $$\dot\phi = \frac{d}{dt}(\phi)=\frac{d}{dt}(\phi_0\cos(\omega_0t))$$ $$= -\phi_0 \omega_0 \sin(\omega_0 t)$$ ...

2

I believe you want to replace mass with charge and angular velocity with the magnetic induction. The Coriolis effect is an apparent force due to the fact that the observer is measuring with respect to a rotating frame of reference. There is no actual force acting on the body, so this can be made to disappear by changing the frame of reference. Classical ...

2

Energy is relative. You allow the potential energy to become negative (since you set it at zero at the top), so the sum of kinetic plus potential remains zero. If you had set potential energy = 0 at the bottom of the trajectory, you would have had a constant positive energy at all time. You are not really "learning" anything about the energy of the rope - ...

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Because this radius $a$ is the semi-major axis of the elliptical orbit. It is not the instantaneous distance.

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Classical mechanics are very important for everyday physics. For the energy scales, relative velocity differences, and mass scales that we experience is our everyday lives, Newtonian physics provide us with an extremely valuable tool of predicting outcomes of events. In other words Newtonian physics are an accurate enough approximation to the more precise ...

2

Newton's describes his notion of absolute time and space his Scholium on Time, Space, Place and Motion. In Newton's time, civil time was still measured by the motion of the Sun. Newton needed to distinguish time as measured by a sundial from the time as measured by a clock (or by the motions of the planets, or of Jupiter's moons). Scientists in Newton's day ...

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Reaction wheels, momentum wheels, and control moment gyros are three somewhat distinct ways of controlling the rotation and orientation of a spacecraft. Reaction wheels are the easiest to understand, at least in their simplest form. Consider a spacecraft such as a space telescope that is nominally not rotating with respect to inertial space. The reaction ...

2

Newton's first law says that unless an object is applied a force it will continue its motion with the same velocity (or linear momentum). This law expresses the homogeneity of the Euclidian space: nothing changes from one place in the space to another. Newton's third law says that if a body A acts on a body B by a force $\vec F$, then the body B acts also ...

1

I can't quite fathom the source of your confusion (I think it might have something to do with a focus on the notion of rotation here---angular momentum does not require rotational motion), so I'm having trouble writing a really clear response. For the moment I would rather offer a program for practicing the right skills rather than reinforcing the mistaken ...

1

If you apply a force on an object the component of velocity in the direction of force is only get changed. for example in projectile motion we take the horizontal component of velocity to be constant since there is no force in the horizontal direction. We only calculate how the component of velocity in the direction of the force(gravity) ...

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Two reasons. First, I strongly recommend Asimov's essay Relativity of Wrong that explains very concisely and clearly why questions such as this one are more wrong than Newtonian physics. Second, the concepts of Newtonian physics are necessary for explaining pretty much anything in physics, including general relativity and quantum mechanics. These ...

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https://www.youtube.com/watch?v=dmnmuTv4pGE I think this is what you are looking for.

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Solar flares etc can heat the highest reaches of the upper atmosphere and make it expand outwards, thereby increasing the drag on satellites. Using an artificial heating via HAARP was explored as a way of altering the trajectory of incoming ballistic missiles across the pole. Allegedly without much effect (so the US govt claims)

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Historically, I'm not really sure what prompted Newton to write down his third law. Physically, however, it is just a statement of momentum conservation. Say object 1 pushes on object 2 with force $F_{12}$.Then by the third law object 2 pushes on object 1 with force $F_{21}=-F_{12}$ Rearranging and using Newton's second law: $F_{12}+F_{21} = \frac{d}{dt} ... 1 Here is how I would do it: From conservation of energy you find $$mgh = \frac12 m v^2 + \frac12 I \omega^2$$ And the relationship between$v$and$\omega$:$v=\omega R$. The result then follows. 1 I see your update but I don't understand the equality, more exactly the right hand side. I get there$2(x_1 −x_0) +(x_3 −x_2 )$. This equality is also directly obvious. From it one gets indeed your relation between accelerations$2a_2 +a_1 −a_{P_{smallest}} =0$1 Because you dont need general relativistic (tensors ,differential geometry etc) calculation to send a rocket to the moon. 1 That's a very common misunderstanding of classical mechanics. The theory doesn't make ANY statement about the microscopic structure of space or time, it only makes a number of symmetry statements. The mechanistic view of what the theory "means" is a philosophical construct completely external to the theory. It's metaphysics, not physics. Time, in the ... 1 The first phenomenon you are stating is a "local" one, whilst the second is a "global" one. The radius you speak of is not the norm of the position vector from the sun to the planet, which would be a function of time, but rather the length of the semi-major axis of the ellipsis, which is a constant. The period time is also a constant, but the planet can ... 1 I am not convinced that answer (B) is correct. It is not enough to say "the liquid would spill out of the top if the top wasn't there" - because the middle of the liquid is lower, and you can't decide just from that piece of information whether those two effects balance out, or whether one is bigger than the other. In the steady state scenario, viscous ... 1 Eccentricity plays no role in determining the period of a Keplerian orbit. The period is given by$T = {2 \pi}\sqrt{\frac {a^3}{G(M+m)}}$where$T$is the orbital period,$a$is the semi-major axis of the orbit,$M$is the mass of the central body, and$m\$ is the mass of the orbiting body. Eccentricity plays a definite role in determining the instantaneous ...

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In linear control systems theory one way control systems engineers model physical systems is by transforming the set of linear differential equations describing the system into the frequency domain. So using the LaPlace transform one winds up with a rational polynomial function that related some desired sets of system inputs and outputs. The benefit of using ...

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I will argue that A,B, and C, are all wrong on the grounds of logic (and remark on pedagogy). Firstly, A is wrong because the first law does not follow from the second law. The first law makes the specific testable predictions that in the absence of a net force, that the motion not change. But the second law makes no such claim because it is possible (see ...

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