New answers tagged

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Taking towards the centre of the Earth as positive and applying Newton's second law to a system which consists of just mass $m$, $F = ma \Rightarrow mg-N = mr^2\omega$ where $N$ is the normal force on the mass due to the Earth. Again with the system being the mass but now considering the rest frame of the mass which is acceleration, applying Newton's second ...


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Imagine the transformation of a spherical Earth into a flat Earth. In what directions is the distortion taking place? What would it look like if the distortion continued in the same directions? That's right, the surface would fold upward into a bowl (with the original outer surface of the Earth on the inside) then a bottle, then a sphere -- the hollow Earth ...


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$$\omega = \frac{2\pi}{24\,{\rm h}} =\frac{\pi}{12}\,{\rm h^{-1}} $$ and you forgot the force with which the surface of the Earth pushes outward.


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The two bodies will actually revolve each other in an ellipse-shaped orbit. The only "real" force that applies to these two masses is actually just the a=Gm/r2 gravity acceleration. The centripetal force is apparent, like an illusion, but does not actually give any real acceleration to the objects. So the two bodies will keep on orbiting each other,...


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I will tell you what I feel should be the answer. See, in case two bodies, the gravitational force between them is alright. But the moment you consider their centre of mass, the situation changes entirely. For two individual bodies sitting at some distance from each other will feel the gravitational force. But centre of mass concept is different. This was ...


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There is a lot going on in the cloud as the raindrop forms. Firstly you have warm moist air rising, as this column of air rises it expands (pressure reduces as altitude gained) and cools eventually the moisture in the air will precipitate onto any aerosols in the rising air and form a cloud. The water droplets will coalesce and accumulate until the weight of ...


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Your calculus is right. What you've computed is the change in potential energy over a displacement $h$, not the total potential energy. Gravitational potential energy is negative. An object's potential energy is opposite the amount of energy it would take to move it from where it is to infinity. For a negative-valued force like gravity, closer to the bottom ...


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It's been mentioned in answers already, but I just want to reinforce that in terms of an opportunity for teaching a young child about physics, the following answer is actually a very good one: it would travel in a straight line if there were no gravity at all, for example if there were no Earth at all. In fact, free-falling objects in deep space, far away ...


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I like the shell answer, as it is not trivial. The only other answer is the trivial one: no shape, as in no Earth. Now if she wants a straight line with non-zero acceleration, then she is out of luck.


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If you mean straight lines on the spherical earth's surface, then objects in free-fall already follow straight line trajectories (if they are a given an initial velocity in the direction of the acceleration or just dropped from a height with no sideways components of velocity). Since the Earth is spherical, the strength of the gravitational force varies with ...


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Then she asked me what shape the Earth would have to be for an object in free fall to follow a straight line trajectory. Is it even possible? Yes, it is possible, under very strange (effectively purely hypothetical) circumstances. Suppose that the shape of the earth was a uniform-density hollow spherical shell and suppose that instead of living on the ...


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An object will follow a straight line trajectory if acceleration is in the direction of its velocity. In a gravitational field, acceleration is in a fixed direction. The trajectory will be straight only of velocity is in that direction. On a spherical planet, it will be straight if the initial velocity is straight up or down. There is no gravitational field ...


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An object in free fall is accelerated towards the centre of gravity of the Earth. That means that its trajectory will be a straight line if and only if , it has no component of velocity normal to the line between it and the centre of the Earth. Off the top of my head, I imagine that to hold true regardless of what shape the Earth takes.


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Dark matter is not distributed evenly across the universe so I don't think we could consider it coming from the centre black hole of galaxies. Physicists have used gravitational lensing to map it very clearly and it is very closely associated with particular galaxies. For example, they used gravitational lensing around the bullet cluster of galaxies as one ...


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"Why is this value independent of the central angle of the ring" It's because you're not integrating over an angle. You're integrating $\int_0^M dm$, wich yields M, plain and simple. If instead a angular mass density was givem, youi'd integrate over an angle instead, with some factor 2$\pi$ in the result. But that's not how it is stated.


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The force on a mass $m$ in a gravitational field $\vec g$ is $m\vec g$. A mass $m$ subjected to such a force and only that force will have an acceleration $a$ which can be found by using Newton’s second law, $m\vec g = m \vec a \Rightarrow \vec a = \vec g$. In other situations when a mass has other forces acting on it in addition to the gravitational force ...


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It seems to me that the theory that over 80 percent of our galaxy is composed of dark matter (as yet undetected) but deemed necessary hold the galaxy together is simply more of the same sort of guff in the same vein. Aether was one hypothesis introduced to explain one observation. Dark matter is one hypothesis introduced to explain almost a dozen ...


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First, there is nothing intuitive about Newton's Law of Gravity. No one on Earth had a clue that the same 'force' that caused an apple to fall to the Earth was the same 'force' that was holding the moon. This was an enormous leap away from intuition. Second, sure, maybe the gravitational constant isn't constant. What you need to do is what Newton did: ...


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The answer to the question lies in Newton's first law of motion: An object at rest remains at rest, and an object in motion remains in motion at constant speed and in a straight line unless acted on by an unbalanced force. https://www1.grc.nasa.gov/beginners-guide-to-aeronautics/newtons-laws-of-motion/


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Now the gravitational field on $P$ due to $dm$ is along $PA$ and it's magnitude is $$dE = \frac{Gdm}{z^2}$$ This is where you're wrong. Now do you understand why?


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If your mass is attached to the bottom of an un-stretched vertically hanging spring and released, the spring is initially exerting no force, and the mass will accelerate downward due to gravity. If you choose the zero point for gravitational energy at the equilibrium position where mg – kx = 0, it will start with gravitational energy mgx, and pass ...


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If the mass is released it will also have kinetic energy. You must include the kinetic energy in your expression.


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Now if the mass is released, the tension will be the restoring force of the spring and since the mass is neither going up nor going down But at that point the velocity (kinetic energy) is non-zero. So the claim that it neither goes up or down is false. In your energy conservation statement, you have zero kinetic energy and that happens only at maximum ...


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If you were initially standing in the elevator at rest, once the elevator started accelerating the floor would accelerate away from you and the ceiling would accelerate towards you. During this time, you would still be accelerating downwards with magnitude $g$ (relative to an external inertial observer). Once you hit the ceiling then you will be accelerating ...


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Otherwise, you can use $$T=\dfrac{2u\sin \left( \theta -\phi \right) }{g\cos \phi }$$ $$=\dfrac{2\cdot 10\sin \left( 45-30\right)^{\circ } }{9.8\cos 30^{\circ }}$$


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That's correct. If you start with the equation of motion of the Moon $$ m_M\ddot{\boldsymbol{r}}_M = - \frac{Gm_Mm_E}{|\boldsymbol{r}_M - \boldsymbol{r}_E|^3}\left(\boldsymbol{r}_M - \boldsymbol{r}_E\right), $$ and you take the center of mass as the origin: $$ m_M\boldsymbol{r}_M + m_E\boldsymbol{r}_E = \boldsymbol{0}, $$ then $$ \boldsymbol{r}_M - \...


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starting with two body system, Fig I, you have two equations: $$r_{{{\it cm}}}={\frac {m_{{1}}r_{{1}}+m_{{2}}r_{{2}}}{m_{{1}}+m_{{2}}} }=0 \\ r=r_{{2}}-r_{{1}} $$ from here you obtain that $$r_1=-{\frac {m_{{2}}}{m_{{1}}+m_{{2}}}}\,r\\ r_2={\frac {m_{{1}}}{m_{{1}}+m_{{2}}}}\,r$$ your Ansatz: (Fig II) you put the Mass $~M~$ at the center of mass and ignoring ...


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The best answer I found is based on the proof of Kepler's First Law. I am unsure if the force being proportional to the inverse square of the radius is a necessary requirement. See https://faculty.etsu.edu/gardnerr/2110/notes-12E/c13s6.pdf From equation (***) in page 8, $$ GM \left(\frac{\vec{r}}{r} + \vec{e} \right) = \vec{\dot{r}} \times \vec{C} $$ Recall ...


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It's sloppy phrasing. By "average radius" the question means semi-major axis. This is probably because, if you take a circle with radius $r$, and draw an ellipse with semi-major axis $a = r$ that shares a focus with the center of the circle, the circle will intersect the ellipse at the tips of the minor axis, dividing the perimeter of the ellipse ...


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This doesn't violate Newton's second law. Lifting the wheel over his head by holding it on the end of a long handle is hard because of torque (imagine lifting a broom holding it in the middle of the handle compared to just at the end of the handle). If he had been holding the wheel-handle arrangement at its center of mass, he would have just as easy a time ...


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The flywheel becomes lighter because the force that is usually directed downwards (because of gravity) is redirected to its side, so instead of the weight being used to pull the flywheel down, it is busy spinning it, making it lighter. Basically, because the direction of force is not towards the ground in a direct way, it makes it easier to lift. I think ...


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Usually, in textbooks we think of two masses on a frictionless surface, they collide, and we analyze momentum, energy, and velocities before and after the collision. There is no external force except the forces during the collision interaction. In physics, collision happens when the momentum of a body is changed due to other body,this does not mean that ...


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I will try to explain it mathematically. Our orbital velocity is given by formula: $$v_{\text{orbital}}=\sqrt{\frac{GM_{\text{star}}}{R_{\text{orbit}}}}$$ It’s obvious that numerator terms are not going to change. So, if the orbital velocity changes, the radius of the orbit changes. Since they are inversely proportional, if one increases, the other decreases ...


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Assume a high circular orbit above a planet. If you want to drop to a lower orbit, you have to do a small retro "burn" (fire your rocket engine to provide thrust opposite the direction that you are traveling) to reduce your tangential velocity a bit. If you don't slow down too much, you will go into an elliptical orbit, gain kinetic energy as you ...


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You need to know the (conserved) value of the Runge-Lenz vector $\textbf p\times\textbf L-mK\hat{\textbf r}$. This points in the direction from the centre of force towards the periapsis, which is your $\theta_0$. The vector lies in the plane of the orbit, as you can see from its definition, and its magnitude gives you an alternative route to the constant $C$....


3

Clearly, $L$ varies with $r$ somehow. So, let's work from first principles; in fact, let's not restrict ourselves to circular orbits. The Binet equation proves closed orbits are elliptical with semi-latus rectum $\ell=\frac{h^2}{G\mu}$, with $h$ the orbiting body's specific angular momentum and $\mu$ the orbiting and central bodies' reduced mass. This ...


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Even though the argument is completely general, let me confine to the simple case of circular motion. The first derivation mentioned in the question is valid. The starting formula, equating the expression of the modulus of acceleration in a uniform circular motion to the acceleration due to Newton's force, is valid for all the circular motions, each ...


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From that wiki page: "The Hill sphere for Earth thus extends out to about 1.5 million km (0.01 AU). The Moon's orbit, at a distance of 0.384 million km from Earth, is comfortably within the gravitational sphere of influence of Earth and it is therefore not at risk of being pulled into an independent orbit around the Sun."


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In a typical Cavendish experiment, the timescale for the masses and the torsion fiber to equilibrate after an adjustment to the setup is hours. In low-Earth orbit, your laboratory orbits the Earth every 90 minutes. That’s a lot of background rotation to cancel out. I’m pretty confident it’s never been done. The closest thing to a Cavendish-type experiment ...


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I do not know if the Cavendish experiment has been duplicated in free-fall, but since the rotational plane of the weights originally used by Cavendish was horizontal, and the weights themselves were vanishingly small compared to the earth, you'd get the correct result even in the presence of gravity.


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I am assuming "notably harder to accelerate forwards" means that the car loses traction sooner at low gravity vs normal gravity. The reason that is the case is traction is lost when the static friction force, which is responsible for acceleration, reaches the maximum possible static friction force sooner, which is $$F_{f-max}=\mu_{s}mg$$ Where $mg$ ...


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Assuming that the car is a wheel-driven vehicle, then the horizontal force that accelerates the car is actually the friction between the car's tyres and the road surface. The maximum friction force is proportional to the vertical normal force exerted by the road on the car, which is in turn (on a level road) equal to the car's weight. In lower gravity the ...


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Given parameters Symbol radial distance vector $\vec{r}$ veloicity vector $\vec{v}$ standard gravitational parameter $\mu$ semi-major axis a 1. Calculate the specific angular momentum vector, $\vec{h}$ Specific angular momentum is the cross product of the radial distance and velocity vectors, and is useful for finding many of the other Keplerian ...


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This is because the typical car accelerates by exerting a force on the ground. The reaction force the ground exerts on the car then propels the car forward. In a low gravity situation, the car is not in solid contact with the ground, there is no reaction force, and hence it cannot accelerate. Note this doesn't apply if the car uses a jet engine, which works ...


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There is a really nice answer by @Farcher, I am submitting my answer because I believe in certain cases a picture is more then words, and to add some interesting facts. In your case, the raindrop does the same thing, it skydives, first, it accelerates, until increasing air resistance cancels gravity (acceleration), after which point the raindrop falls at a ...


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Because the box remains at rest. (I know that's a ball but pretend is a box). The normal force (Fn) and the force of friction (Ff) must balance for the box to remain at rest. That's why Ff increases as Fn decreases. So in that context what does this mean? Ff = µ Fn This tells you the max Ff that friction can provide to keep the box from sliding. To make ...


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You are probably thinking the static friction force, $F_f$, acting up the incline always equals $\mu_{s}F_n$. It doesn't. Static friction $F_f$ is a variable force that equals the force acting on the object down the plane, $mg\sin\theta$, up until the static friction force equals the maximum possible static friction force, or $F_{f}=\mu_{s}F_n$, is reached ...


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Let's pretend that gravitational acceleration does depend on the mass of an object. Let's then say that we have an arbitrary number of objects of the same mass, and that each have the same gravitational acceleration. The system has a different mass than each of the smaller objects it's made of, so if the gravitational acceleration of the system depends on ...


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If you have two identical planets and drop a 10 kg ball bearing (for example) from 10 meters on one and repeat the experiment with a 100 kg ball bearing on the second planet, the 100 kg object will reach the planet surface quicker, though the difference in time would be basically imperceptible. This is because from the reference frame of the center of ...


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Acceleration due to gravity doesn't depend on mass of falling object can be proved mathematically by using Newton's law of gravitation. $$ F= \frac{GMm}{r^2} $$ $$ F= mg $$ (Symbols having usual meaning here) Equating these two, $$ mg=\frac{ GMm}{r^2} $$ $$ g= \frac{GM}{r^2} $$ From derivation you'll see that the final equation doesn't contain 'mass of the ...


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