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83

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 ...


31

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

As you probably know, Newton thought that energy is linearly proportional to velocity. The second law's original formulation reads: "Mutationem motus proportionalem esse vi motrici impressae" = "any change of motion (velocity) is proportional to the motive force impressed". This law, which nowadays is wrongly interpreted as: $F = ma$ (there is no ...


13

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 ...


8

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

The answer is yes. The de Broglie wavelengths of freely propagating particles (i.e. forget the influence of interactions and gravity perturbations, just consider the Universe as a whole) are redshifted by the expansion of the universe. Another way of saying this is that their peculiar momenta with respect to a co-moving local volume decrease as the inverse ...


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 ...


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

"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 ...


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

I believe the difference comes from the fact that forces can do different amounts of work in different reference frames. In particular, the normal force by the ramp does no work in the "lab" frame, but does do work in the moving frame (since there is a component of velocity that is now parallel to the normal force). I don't think you accounted for this work ...


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 ...


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

You've already got some answers, but nobody mentioned Noether's Theorem yet. Noether's theorem maps a conserved quantity to each continuous symmetry. The relevant continuous symmetry needed to prove the conservation of energy is the one that leaves the laws of nature invariant, meaning the laws of physics don't change with time. Each continuous symmetry ...


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 ...


1

Work done on a system increases the energy of the system. But you have to be clear about what system you are talking about. For the system consisting of the Earth and your other object, there is no external force, and no work done on the system. The total energy is constant; it just changes from potential energy to kinetic energy. If your system is the ...


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 $10^{-9}$ ...


1

I don't think energy conservation and momentum conservation can explain charge conservation in particle physics.


1

No. There are more conservation laws: conservation of angular momentum, electric charge, color charge, weak isospin. Without the charge conservation you cannot explain why electrons do not decay into neutrinos.


1

When we lift those three weights from Y to X, we can use the reversbile machine. So, we ease the machine B, because machine A takes those weights. Is that a right picture? There are two weights with mass $M$ and $3M$. Initially, both weights are at the same height $h_0$ which we can freely set to zero: $h_0 = 0$. Now, machine B lowers mass $M$ to ...


1

Suppose you have a constant angle $\theta$ slope in the original frame (we suppose that the transitions from horizontal movements to the slope is quasi-instantaneous). Call $x$ and $x'$ the horizontal displacements in the original and moving frame. Call $T$ the total time for going to $z=h$ to $z=0$. Then you have $x'= x- v_0 T$ With $ x= h \cot \theta$, ...


1

Let's look at what happens if we put numbers into your equations to see if there actually is a violation of the conservation of energy. Using numbers for Earth from wikipedia we have a velocity of $29.78*10^3\frac{m}{s}$ for orbital velocity and a mass of $5.972*10^{24}kg$. For say the Voyager 2 we have mass of about 730 kg and a speed of ...


1

Some conservation laws are related to conservation of angular momentum. There is a famous example (from Feynman if I recall correctly), where you assume an infinite flat space and conservation of angular momentum about any point, and then you get conservation of linear momentum for free. Intuitively, to get say the $x$ component of linear momentum is ...



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