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I wanted to elaborate in Qmechanic's answer. The angular average for $f(\theta)$ is $\langle f(\theta)\rangle_\theta= \frac{1}{2\pi}\int_0^{2\pi} f(\theta) d\theta$. The time average of $f(\theta)$ is $\langle f(\theta)\rangle_t=\frac{1}{\tau}\int_0^{2\pi} \frac{f(\theta)}{\dot{\theta}} d\theta$. Where $1/\dot{\theta}=\frac{m}{L(A+B\cos{\theta})^2}$ and $... 4 The virial theorem$\langle U\rangle_t =-2\langle T\rangle_t$is for time-averages$\langle f\rangle_t=\frac{1}{T}\int_0^T\! \mathrm{d}t~f(t)$, while OP considers angular averages$\langle f\rangle_{\theta}=\frac{1}{2\pi}\int_0^{2\pi}\! \mathrm{d}\theta~f(\theta)$. These averages will in general be different because of the larger (smaller) angular velocity ... 0 Is an object that is neutral buoyant the equivalent of the same object in the microgravity (free fall) of the International Space Station? In some ways they are very similar, in other ways, not so much. In either case, the net weight is (approximately)$0$. This is the major similarity. Some differences would depend on the medium that you are buoyant in. ... 0 One obvious difference is that the neutral buoyancy state of the ping pong ball on water is broken if it is moved vertically. If it is lifted and released it will fall back to the water surface; if it it pushed down and released it will rise back to the water surface. Whereas the ping pong ball in the ISS can be moved in any direction. Another difference is ... 1 For an object to be in free fall there must not be any apparent weight.What means is that if you stand on a weighing scale it would give a reading of zero given that it has zero velocity relative to you. In free fall the reading will be zero, you must have seen astronauts floating in space? They would be weight less as they would not "push" the weighing ... 2 This question is answered really well by the answers above. Of course, as explained in all answers the statement quoted from your textbook is incorrect. The phenomenon of weightlessness could be explained in two different ways: The object orbiting the planet is in free fall (as described in other answers). The centrifugal force counteracts the earth's ... 3 A freely falling body is a body which moves in the presence of gravity without any other external force acting on it. A ball dropped on the surface of the earth is the example that you have in mind. Now let's give some initial tangential velocity to the object (wrt the direction of free fall). Now is this case any different than the ball falling without ... 5 When people talk about "free fall," what they are referring to is the absence of other forces besides gravity. For example, there are no "normal forces" from the ground holding the object up. This is a correct description of how satellites move. It is a confusing term for the reasons you bring up. "Free fall" does give a sense that something is moving ... 7 You are absolutely right, and the caption conveys a common error. The textbook is wrong! Congratulations on discovering this error. Indeed, gravity at typical spacewalk height is not all that different from gravity at the surface. Because we have that $$g \propto r^{-2}$$ where$r$is the distance from the center of the planet, we can estimate (using an ... 1 A satellite is falling, it's just also moving forward fast enough that it is always falling around the curve of the Earth. So it is in freefall. If you throw a ball straight out in front of you, you will see it follow a curve as it goes forwards and falls downwards. If you could throw it fast enough, and there were no air resistance, it's forward speed could ... 12 May be Newton's cannonball-thought-experiment can convince your intuition that there is no principal difference between a body falling down to the earth and a satellite circulating the earth. Consider a cannon located on top of a high mountain, shooting in horizontal direction. And let's neglegt air resistance. (image from Wikipedia - Newton's cannonball) ... 5 Yes it is doing free fall. Consider a satellite at some moment in time. Say its velocity at that moment happens to be in the direction of the pole star. If the satellite were not being attracted to the Earth, it would move in a straight line and go off in that direction, moving further and further away from Earth. But, because of the attraction to the ... 4 Any idea whether this is correct and if so how it is calculated? That's about correct. An object released from a space elevator at an altitude less than about 14500 miles above the surface of the Earth will impact the Earth in less than half an orbit. A release altitude of 15000 miles will result in an orbit with a perigee altitude of about 473 miles. There'... 6 The astronaut can stay afloat because he/she is in free-fall around the earth. Picture a cannon on a cliff shooting a projectile with increasing speed. Eventually, with enough speed, the projectile will make a whole round trip. In space, with less air-resistance and friction, this process can go on for a much longer (in theory infinite, in reality finite, ... 8 Whatever written in the book is wrong. But let us assume it is correct then had it been the case then the moon would have flown away as well as the ISS along with the astronaut. 58 The statement in the image is shockingly wrong. Given that the Earth is visible behind the astronaut, the reduction of the gravitational force due to distance must be fairly small; on the ISS for example the force of gravity is about 88% the value at the Earth's surface - very far from being negligible! The reason why the astronaut appears weightless is that ... 8 You are correct that the caption is wrong. The cause of weightlessness is that gravitational and inertial forces cancel, exactly according to the equivalence principle. 5 I'm afraid your book is wrong. At distances from the Earth where it's still as big as in this picture, the gravitational field is still very relevant. As you said yourself, the astronaut is in orbit around the Earth, and thus effectively in free fall, leading to the illusion of an absence of gravity. 0 Obviously, at geostationary orbit a dropped object will just stay where it is: the orbital velocity is exactly matched to the velocity it has, so it is in a circular orbit. Equally obviously, an object dropped at ground height will go into a very eccentric, Earth-intersecting orbit (it is not falling entirely vertically when you look at it in a non-rotating ... 0 The gravitational force of the Earth on an object distance$R$from the center is given by$F_g = \frac{GMm}{R^2}$, where$M$is the mass of the Earth, and$m$is the mass of the object. The centripetal force required to keep an object in an orbit of radius$R$, with an angular velocity of$\omega$is$F_c = m \omega^2 R$. Assuming the space elevator is at ... 0 Roll a wheel down the road. If it’s really rolling, the place where the wheel touches the road doesn’t move relative to the road: it’s just sets down as the wheel gets to a spot, then lifts up after. On the other hand, the top of the wheel is moving twice as fast as the axle. The skyhook is like that. It’s engineered to be rotating and orbiting at just the ... 1 the moon orbits both Earth and the Sun. Since the Earth orbits the Sun, and the moon orbits the Earth, it necessarily also orbits the Sun. These two motions of the moon almost do not interfere with one another, because its orbit about the Sun is almost identical to the Earth's orbit around the Sun (because they are approximately the same distance from the ... 0 The null point... ... it is unstable i.e. the force on that point is indeed zero but what if we displace the object by$\mathbf{\delta x}$? What I mean to say is that the system is in unstable equilibrium. The force on null point$Nmay be zero but the potential energy at the point is maximum suggesting unstable equilibrium. What is unstable ... 0 In this situation, the minimum speed is a limiting value. Anything above that will carry the projectile beyond the null point. 0 The same reason you can jump up in a bus and not be slammed into the back of it. The balloon already has the same velocity as the earths rotation. As does the air around it (ignoring air currents). Lets take the bus example again. If you are in a bus, and it goes 50km/h, everybody else in that bus wouldnt seem to be moving at all. Because the bus is the ... 5 A hot air balloon floats immersed in the air, similar to the way a fish might float immersed in the ocean. The fish doesn't see the seafloor below it zoom off at 1000 miles per hour because it is inside the ocean which is rotating with the Earth. Similarly, a hot air balloon rests inside the atmosphere which is rotating with the Earth. The ocean and the ... 0 I think this is your problem?. so try to calculate the intersection points. so the result will be: $$\phi_2-\phi_1=\text{const}\,\Delta T$$ 0 L1 is a saddle point in the potential: (From here) As such, a body at L1 feels tidal forces. An extended body will tend to align due to these tidal forces. This is similar to behavior near just one planet, where extended bodies align up and down. In close orbit, like ISS, the stable orientation is to rotate once per orbit, so that “up and down” are ... 0 No, it cannot happen because of a natural instability in the gravitational forces. Gravitational forces are inversely proportional to the distance between the objects squared. If that distance decreases, then the forces will increase. There will be a key balancing point where the forces of gravity are exactly equalized. At this point the object is in "... 1 A body in orbit may lose energy due to friction, such as caused by tidal forces. The loss is higher at high speed so the the speed tends to average out. This implies a relaxation to circular orbits, which have constant orbital speed. 4 Circles often are seen as a symbolism of stability and symmetry. A lot of physics simplified models use to consider circles as the trajectory of some motions because it's way simpler to work with them. Constant radius, which makes the measures involved with the inverse square laws (really present in nature) the same for every point around the circle. There ... 1 Orbits are elliptical. An orbit that’s just a little bit elliptical, like the one perturbed by 1 m/s here, looks like bobbing up and down around the original circular orbit. And the period of that bobbing oscillation is equal to the orbital period: the motion goes down once and up once before returning to the same point and repeating. 0 1 m/s is a small change to modify too much an orbit. In general, the elliptic orbit would change. If it was in a perfectly circular orbit, it would become slightly elliptic. 0 This is an answer for why there is a constant n involved in the angular momentum classically. $$F_c=\frac{1}{r}mv^2=k\frac{q^2}{r^2}=F_p$$ $$L=rmv$$ $$r=k\frac{q^2}{mv^2}=k\frac{q^2}{m(\frac{L}{mr})^2}$$ $$r=\frac{L^2}{kq^2m}$$ $$\frac{1}{2}mv^2=\frac{1}{2}m\omega^2r^2=\frac{1}{2}m\omega^2(\frac{L^2}{kq^2m})^2=\frac{1}{2}k\frac{q^2}{r}$$ $$\omega^2rL^... 1 The answer is any mass. The speed v of a circular orbit between the two particles of the same mass is v=\sqrt{Gm/2R}. You can have an orbit of any desired radius. v will be slow because m is very small. You can increase it by reducing the distance as much as you want. 0 According to Newton: a = \frac{GM}{r^2} a = "planet's" acceleration... for Earth a=9.8 meters/second^2 on the surface. M = mass of "planet" = 1 atomic mass unit for a neutron = 1.6603 \times 10^{-27}~\text{kilograms} r = radius from the planet center = let's pick 1 centimeter just as an example. G = the universal gravitational constant = 6.6726 \... 2 So roughly speaking this is the contents of Newton's first law. Newton was living in an era when people wanted to think that the natural state was things standing still, and that all motion required a force to explain it. But but during Newton's life, Galileo had observed something different: when he rolled a smooth metal ball over a large flat floor, it ... 1 "Forces" don't keep things in motion - that was the old idea of Aristotelian mechanics, and it is not a good principle for understanding the rest of the Universe, already dated by the time of Sir Isaac Newton. "Forces" influence motion, but "motion" exists of its own accord. So what I suppose you are really asking is "what force confines the electrons to ... 0 Classical Picture There is no need for a force to cause motion, it's the presence of them that disturbs motion! (Cause: Inertia) But if you are asking for considering the circular motion of electron (note that this model was used by Bohr but isn't currently in use) then yes the Electromagnetic Force is the cause of it. 0 Galaxies do collide due to gravity. They don't stick together exactly, because they're not solid bodies; but they do often merge. See this article on the NASA website, for example. 2 I believe this problem can be tackled using polar coordinates. The equation of an conic in these coordinate is:$$r(\theta) = \frac{r_0}{1+e \cos (\theta + \phi)},$where$r$is the distance to one of the foci of the ellipse (the center of mass of the system, or the center of the star if it is much more massive than the orbiting object),$e$is the ... 1 I think the issue you're running into here is that there isn't a single vector describing the inertial path between different points in space, but infinitely many. The reason for this is that you can approach this point at any time you'd like. To simplify, consider the case of a baseball you want to throw from point A to B in a linearly uniform ... 0 What general relativity describes as curved is the space-time continuum. Objects travel through this continuum on geodesics. Like a line in flat 3D space, geodesics have a direction. You don't expect two lines with a common point but different directions to continue together, do you? Same with the world lines of your particles: The direction of the geodesic ... 8 The relativistic way of looking at things is that objects don't have paths through space, they have paths through spacetime. These are called world-lines. Even in Galilean physics, the trajectory of a test particle through space is not really well-defined. For example, the planet earth doesn't have a well-defined trajectory through space that is an ellipse; ... 16 Light and matter both follow the curvature of spacetime when passing a massive object. The difference is that matter is ALWAYS slower than light, it will be in the more curved spacetime longer, so it's path is curved more than light. Compare it to throwing a ball, if you throw the ball very slowly, it will follow a sharp curve and fall down close to you. If ... 5 The orbital trajectory taken is dependent on the velocity of the body. Since light always travels at$c\$ and nothing with mass can, the orbits will always be different. Light does curve around the sun in a hyperbolic path. If there was something with a much more powerful gravitational field than the sun (like a black hole), light would orbit around it. This ...