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84

Cory, here's a different way of thinking about gravity assists that may help: First is my short answer for readers in a hurry: What is really going on is a giant game of pool, with fast-moving planets acting as massive cue balls that impart some of their energy when they whack into tiny spacecraft. Since you can't bounce a spacecraft directly off the ...


33

Energy is in fact conserved, even in gravitational slingshots. After the slingshot, the velocity of the spacecraft may indeed change, which means its kinetic energy will also change. If this happens, the energy increase (or decrease) will be made up by a commensurate decrease (or increase) in the kinetic energy of the planet. In plain English: The planet ...


14

1. If the universe is expanding, what is it expanding into? The universe isn't expanding into anything. Space-time isn't curving into a higher-dimensional space. So what do we mean by "curved" and "expanding", words usually having a meaning only for objects in space? The answer is it is just an analogy. Mathematicians have found properties of space an ant ...


9

A simple counterexample: Imagine two particles with opposite direction and equal speed. The center of mass does not move, yet the kinetic energy of the system is non-zero. Now let both particles come to rest (by friction, hitting a wall, whatever). The kinetic energy is now zero, and total momentum has been conserved, while energy is not. The crucial ...


7

Gravitational potential energy is usually measured as a negative value. We do this because an object that is so far away from a gravity well that it practically is unaware of it shouldn't be considered as having any potential energy. So as $r\to\infty$, $PE\to0$. As an object falls into a gravity well, it loses potential energy, so gravitational PE is a ...


7

Gravity assists don't change speed in the two body problem. An object approaching a lone gravitating body will enter and leave the vicinity of that body with exactly the same speed. All that a lone gravitating body can do is change the direction in which the object is heading. The body that provides the assist needs to be moving with respect to the target ...


7

Noether's theorem states that to every continuous symmetry of a physical system there is an associated, conserved quantity. The conserved quantity associated with time translation invariance (i.e. it doesn't matter if you perform an experiment now or tomorrow, provided you set it up the same way) is what we call energy. Therefore, somewhat tautologically, ...


5

Good question! Many astronomers think that the moons of Mars, Phobos and Deimos, are captured asteroids. Others object precisely because of the issues that you raised. Capture is not easy. Sans a collision, capture is impossible in the Newtonian two body problem. A hyperbolic trajectory stays hyperbolic. On the other hand capture in the multi body problem ...


4

@soumyadeep is on the right track, but wrong. The object starts at rest. When the "external force" is removed, it starts to slide down the slope, picking up speed. When it reaches the point where the force of friction plus the force of the spring equal the force of gravity, it doesn't just stop - it stops accelerating. Thus, to get equality you need to ...


4

Potential energy for a point mass (and also for a sphere) is not $PE = mgh $ (this is special case for a uniform field) but rather: $$ PE = - \frac{GMm}{r}$$ where G is the gravitational constant, M and m are both masses and r is the distance between the masses. (in the case of a sphere, the distance is to the centre of the sphere) Can you see the answer ...


4

If the magnet is strong enough to pull the ball up the bottom slope, it will be too strong to let the ball fall. Even worse, as you have drawn the diagram the magnet is pulling the ball down the lower slope when it is toward the left end. Anywhere to the left of where the perpendicular from the magnet to the ramp, the magnet is pulling more right than up. ...


4

The key here is the antimagnetic strip, quite aside from whether or not such a device can be built. When you insert the anti-magnetic strip, you must change the shape of the magnetic field. You must force the magnetic field to "leave" the high permeability ball. The same magnetic induction $|\vec{B}|$ in a high permeability $\mu$ material represents a lower ...


4

"applying a constant force, F with a rocket engine that I supply a constant amount of energy to" Heere's your problem! Rockets are not as simple as you think. Yes, you could apply a constant force (thrust). And yes, this would consume fuel at a constant rate, but as @garyp points out, this does not mean you are adding to the kinetic energy of the ...


4

One mistake you are making is equating a constant power $\frac{\delta E}{\delta t}$ with a constant force. One does not imply the other. I'm going to switch to a more standard notation and use $K$ for kinetic energy. And let's say $U=\text{const}$ to keep things simple in our system. Finally, let's imagine there's only this one force $F$ on this ...


3

You might be able to get it to work for quite a while, depending on your skill as an engineer. But there is a critical difference between a well-engineered machine that runs for a while, and a perpetual motion machine that runs forever without input. The latter is impossible.


3

We don't know! If we were to observe a situation where energy conservation did not appear to work, that would be a major puzzle. As you say, either we would have to discover some alternative contribution to the energy that we had been neglecting, or we would have to give up on energy conservation. A priori it is not obvious which one of those two resolutions ...


3

The "missing" kinetic energy is still there ... it's now in the rotation of the yo-yo.


3

But once it's moving in the vacuum of space, with no gravity or magnetic field nearby, could it spin nearly forever (aka billions of years) producing a magnetic field, No. A rotating magnet creates a changing magnetic field. Similar to an oscillating electric field, it will radiate electromagnetic energy. This energy will come from the rotational ...


3

Here's a general overview of how to approach this: Since the only external forces are vertical (gravity pulling the balls down, normal force of the surface holding the balls up), we can use conservation of momentum in the plane. Similarly, there is no external torque rotating things in the plane, so that component of the angular momentum is conserved. And ...


2

According to the question, distance $d$ is measured from the pivot down: [...] a peg located a distance $d$ straight down from the pivot point You have computed it from the bottom of the swing up. I expect that $d = \frac{2L}{3}$ is one of the answers in your list... Your analysis is otherwise correct. One word of advice: ALWAYS draw a diagram. If you ...


2

Conservation of energy only applies to closed (isolated) systems, and your example, with an external source of energy, is definitely not of that kind.


2

Energy is conserved so it can't be created or destroyed. All we can do is change energy from one form to another. In your example we are changing the potential energy of the mass $m$ into kinetic energy. The increase in kinetic energy must be equal to the decrease otherwise energy wouldn't have been conserved. By an external force I assume you mean some ...


2

To calculate for a situation like this, consider the Law of Conservation of Momentum: Pi = Pf In the case of the billiards: KEi = 1/2 mu1^2 + 1/2 Mu2 (u1, u2 = initial velocity) KEf = 1/2 mv1^2 + 1/2 Mv2^2 (v1, v2 = final velocity) Based on this law, initial kinetic energy and final kinetic energy in an elastic collision are equal. KEi = KEf Hope ...


1

If you impact the second body its axis of percussion it will purely rotate. By carefully choosing the inertial properties of the two objects you can make the first object stop translating in the process. See this post for more details on a particle to rod impact.


1

There is a confusion between the terminology "perpetual" , which means "continuously", devices that almost move forever, and a machine that can produce energy. As the other answers point out energy is conserved and if it looks as if energy is provided from nothing a closer analysis shows the mistake, as in the drinking bird perpetual setup. In the case of ...


1

For what it is worth, your formulation of the spherical wave business is flawed. See here for example. These have the form $$u(r, t) = \frac{A}{r} e^{i(\omega t \pm k r)},$$ where the symbols have their usual meaning. So, intensity, which goes as the amplitude squared, will go as $$I = \left(\frac{A}{r}\right)^2$$ falling off as the square of the distance, ...


1

A "collision course" is a very fuzzy concept: if you are "barely going to hit" you are on a collision course but don't need a lot of deflection. However, let's assume for a moment a stationary earth, a meteorite of mass $m$ at distance $D$, heading for earth of radius $R$ with velocity $v$. The equations you need are conservation of angular momentum and ...


1

The modern mind picture of electrons bound to a nucleus is that of an orbital rather than of an orbit as you seem to be thinking. However, your ideas are still somewhat meaningful in this modern picture in that region of electron delocalisation is more tightly confined around the nucleus for lower energy orbitals than it is for higher energy ones. You might ...


1

BECAUSE WE MEASURE WITH ATOMS (or atomic properties) a change in the measured value can be either a true change of the observed quantity or a change in the properties of the ruler (the atoms). I'll revert the paradigm the space expands to this one: the atoms are shrinking THEN I have simple answers : If the universe is expanding, what is it ...


1

Both are correct. The elastic energy of solid is stored in the chemical bound. On the other hand, it also contributes, although by very very tinny amount, to the total mass of the system due to $E=mc^2$. So combining both facts, we arrived at the conclusion that the chemical bound has mass, even though the mass of the chemical bound is about only $10^{-9}$ ...



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