# Tag Info

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I was trying to solve a similar problem, and did, with help on StackOverflow. My question was here: http://stackoverflow.com/questions/16501182/find-first-root-of-a-black-box-function-or-any-negative-value-of-same-function I asked it more abstractly. The way I saw it, you have a ship and moon (for example), and for different values of time they have a ...

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The situation in your question is not clear. Without a sketch, all I can offer is the following. The governing equations of a sliding, slanted rod driven by a force $F$ at the bottom are: $$\ddot{\theta} = \frac{ \frac{\ell}{2} \left( m \cos\theta \left( g - \frac{\ell}{2} \cos\theta\,\dot{\theta}^2\right) + F\,\sin\theta\right)}{I_C + m \frac{\ell^2}{4} ... 0 Since this old question got bumped I might as well add my own answer. In classical mechanics, two masses that interact gravitationally define a two-body problem, which follows Kepler's laws: they will orbit each other in an ellipse, and Kepler's Third Law states that$$ n^2a^3 =\mu, $$where a is the semi-major axis, \mu=G(M+m) and n=2\pi/T, with T ... 0 My solution has been challenged, and this comes from a separate source so I'm posting it as community wiki. The reference is here: http://web.ist.utl.pt/~berberan/data/43.pdf The proposed equation is:$$ P(z) = P(0) \exp{ \left( - \frac{ m g_0 R_0 }{ kT} \frac{ z }{ z+R_0 } \right) }$$You could put this in terms of characteristic height if you wanted. ... 1 Both definitions are fine as long as you're careful with signs. Here's a derivation of your gravitational potential energy using the idea of the negative work done "by the field." (Note the initial negative sign.)$$U(r)=-W_\text{by field}=-\int\vec{F}_\text{field}\cdot d\vec{s}=-\int_{r=\infty}^{r=x}\underbrace{\frac{-GMm\,\hat{r}}{r^2}}_\text{Toward ...

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You don't do any negative work. All that gravitational potential says is that an object at a higher altitude (say, a ball in your hand) is at a relatively higher potential than the same object when it's at the Earth's surface. Mind you, I meant "relatively" higher potential, because it's still negative - gravitational potential is negative everywhere, ...

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Black holes this small will have very high Hawking temperature: $$T_H = \frac{\hbar c^3}{8 \pi G M k_B} \approx 10^{20}\,\text{K},$$ So, before this black hole can fall down even the diameter of an atom it will evaporate through Hawking radiation. As a result, the 1 tonne of black hole mass would be converted into the energy of very high energy particles ...

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As dmckee said in his comment, the black hole would fall towards the center of the Earth. To specifically answer this portion of your question: How dense would rock have to be to form a barrier? There is absolutely no density of rock or anything else that would stop or even slow it down. Even if you created this microscopic black hole on the surface ...

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You cannot use the second kinematical equation because it is valid only when the acceleration due to gravity, $g$ , is constant. This is incorrect for distances comparable to the radius of the earth, and velocities comparable to the escape velocity. The first correctly assumes a $\frac{1}{R^2}$ fall-off of the gravitational attraction on the body due to ...

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If you only consider viscous dissipation within the droplet, this should indeed go to zero in the vanishing velocity limit: the (local) dissipation rate is quadratic in the velocity, so that decreasing the velocity by a factor of $\lambda$ reduces the (local and global) dissipation rate by $\lambda^2$. Of course, the process takes $\lambda$ times longer, ...

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Any reshaping of the droplet will require flow of water inside the droplet and there will be viscous losses. Presumably the energy would come from an increased torque on whatever motor was moving the droplet and substrate.

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A space traveler on an artificial satellite will be in freefall around the planet it is orbiting. So the ink will not experience any acceleration relative to the pen due to the planets gravity. On earth the ink gets sucked up by the pen due to capillary action, but is counteracted by gravity. While in orbit the full "force" of capillary action can be used ...

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One could start from the more general Poisson equation for a mass distribution $\rho$, namely $\nabla^2 E_{pot} = 4 \pi G \rho$ and infer that the solution to this equation for a point mass $\rho = m_1 \delta$ is the familiar $1/r$ potential. See here. Then one would derive the force from the usual prescription $F = -\nabla (m_2 E_{pot})$. As for a more ...

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The only thoughts I have ever had is that you may/should be able to reproduce the GR explained visibility of stars positioned behind the Sun using some sort of lensing effect due to light's intetaction with gravitons. If you can do anything with lasers and horseshoe shaped lenses. Imagine what you could do with gravitons;-)

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Wait, don't bother. I found the embarrassing error. It's the H' value. It should have been: $$H' \equiv \frac{ M_0 G M }{R T } \approx 4,739,000 \text{ km}$$ I just read that output wrong. The plots now seem to fit. Sorry to answer my own question so fast. I suppose whether any of this is right remains an open question.

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"how does it distinguish case a and b?" Well,let's first clear our conceptions about the archimedes' principle. Then I think you will get your answer for yourself. Archimedes' principle states that, a body immersed in a fluid is buoyed up by a force equal to the weight of the displaced fluid. Focus on this."... a force equal to the weight of the displaced ...

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The angular momentum of a massive sun may cause the freely falling spaceship to start spinning in the direction of the sun's angular momentum for an effect of frame dragging. You can take a look at the Kerr metric which describes the behaviour of the spacetime near a massive spinning object. If you're not familiar with general relativity it could be ...

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The only way for a falling object to be made to rotate and translate is if there was a separate force causing this rotation. In an atmosphere this is a net force on one side of the craft whose surface area (and therefore drag) is the highest, causing this part of the craft to rotate away from the direction the entire craft is translating. Essentially ...

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In the situation at hand, you'll never be able to achieve uniform circular motion. $\frac{d\vec{v}}{dt} = \frac{1}{m}\sum \vec{F}_{\text{ext}}$ This is a vectorial equation. If you look at the picture you've drawn, you have forces on the radial as well as the tangential direction. On the radial direction, there is the tension force, and $mg\cos\theta$. ...

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The answer to your question is "Yes, if you want it badly enough. Uniform vertical circular motion implies that an object is moving in a circle, that the plane of the circle is vertical, and that the speed of the object does not change as it moves around the circle. An example of such motion is that of a point on the end of the hour or minute hand of a ...

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I suspect that you may be under the mistaken impression that there is no gravity in space. This is a common belief since we all can see the astronauts floating in "zero g" when on the ISS are some other spacecraft. However, we can easily dispense with this misconception by asking "what keeps the ISS in orbit around the Earth if there is no gravity?". Of ...

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The escape velocity from earth (the speed required for an object to leave earth completely, i.e. travel infinitely far away) is 11.2 km/s. If the object has a smaller velocity it will return eventually. Unless an object is launched straight up, it also has a sideways velocity. This means that it will not fall back directly on top of the launcher. If it ...

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That's exactly the case. If you look at the trajectory of any given spacecraft, you will see that it has a few burns of the rocket engines punctuating very long periods just coasting along in orbit around some other body. For example, the flight path of Apollo 8 has something like eight different rocket burns: launch, translunar and transearth injection (to ...

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In one sentence: More mass means stronger attraction and less buoyancy (they fall faster), but the effect is negligible in most cases.

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Suppose you pick two people at random. From one, you pluck a single hair from their head. Is it possible to tell who had the hair plucked by weighing the people? Technically, plucking a hair makes a person very slightly lighter, so you get a tiny bit of information about who had the hair plucked by weighing the people. But the information is very slight ...

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Newton's original proof was in fact based on geometry (he hadn't invented calculus yet). Richard Feynman devised his own, simpler geometric proof for one of his famous lectures. You can find it in Feynman's Lost Lecture, by Goodstein & Goodstein, and in this article: Paths of the Planets from Hall & Higson. But since it's so much fun, I'll describe ...

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You do no say what information you know and do not know. For example if the cube sinks and $h_1$ is big enough, it is possible that $s_2=0$. But if you know $s_1$ and $s_2$ then it is easy. The volume of liquid displaced is $(s_1+s_2)n^2$ so the extra height (ignoring overflows) is $\dfrac{(s_1+s_2)n^2}{\pi R^2 }.$ So the final overall height of the ...

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1b) fibontic correctly pointed out that your expression for Newton's 2nd law is not correct. It should be $$ma=F_\text{net}.$$ You have an $x$ instead of an $a,$ which is causing one of your problems in part b. By writing $ma=\rho Vg-mg,$ you should be aware that you've already implicitly imposed a coordinate system where up is positive. This is probably ...

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I don't think that you really understand integration. Let me clear this up for you. In that question there is a rod of length l. You know how to calculate gravitational force between two point masses but not in continuous mass bodies. If you apply the formula to find the gravitational force you don't know what to take the distance as because it is ...

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The Earth orbital speed could be increased by bringing the Earth closer to the Sun. That would require orbital maneuvering of the whole planet (planetary retrograde burn to go into a transfer orbit and another retrograde burn to circularize the orbit) which would be felt by the Earth population as acceleration. After the final orbit would be achieved, the ...

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The faster the earths orbit, the further it would go form the sun until it reached "escape velocity". The speed itself would make no difference to you anything you can feel. However, the distance from the sun, you would feel. The sun rotates around our galaxy core at one tremendous speed (and us with it), and you don't feel that.

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Most certainly it does: the variation can be measured by a sensitive acceleration called a Gravimeter (see Wikipedia page with this name) and is the basis for gathering data important for minerals exploration. Bodies of mineral ore distort the Earth's gravity and thus can be found by measuring the variation of the local gravitation as a function of position. ...

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The acceleration due to gravity changes not only on the surface of the Earth (depending on where you are) but also how high up you are (which varies by $\frac{1}{r^{2}}$, where $r$ is the distance from the center of the Earth to you). For more information on how it varies depending on location, maybe this will be of use? GOCE Delivers Best Gravity Map of ...

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why we always choose the center of gravity of the bicycle be the rotational center. We do not do that always, sometimes it is better to use the point in contact with the ground or some other point. We use center of mass when it leads to simpler equations than the other points. In problems dealing with torques or rotations we use the theorem T: the sum ...

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"Focus" is an inconvenient word if you're thinking of changing the potential, because if you do then the orbits are no longer conics and the word kind of loses its meaning. That aside, let me see if I understood your question correctly: Given a gravitational potential that's spherically symmetric around a central point $\mathbf{r}_0$, and which has a ...

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The velocity of an orbit around some central object can be easily calculated for a circular orbit. Let us assume that there is some central Force $F=c\cdot r^\alpha$, where $c$ and $\alpha$ are some constants (for gravity $c=Gm_1m_2$ and $\alpha=-2$). For a stable orbit, this central force must be equal to the necessary centripetal force (not balance the ...

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The best way to explain it (and even the way Kepler's second law can be derived) is by conservation of angular momentum. The latter is given by $$\mathbf{L}=\mathbf{r}\times m\mathbf{v},$$ where $\mathbf{r}$ is the position vector and $\mathbf{v}$ is velocity. Since this quantity has to be conserved for the motion of the object at all times, assuming an ...

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