# Tag Info

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In electromagnetism there are both positive and negative charges. Hence the force due to electric charges can be attractive or repulsive. Gravity, when treated as a classical force field, can only be attractive, there are not two types of "gravitational charge". What this means is that in electromagnetism, a given medium, may contain both positive and ...

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First remember that $k = \dfrac {1}{4 \pi \epsilon_r\epsilon_o}$ where $\epsilon_r \ge 1$ It is because a medium can be polarised by an external E-field. The dipoles so set up produce the external E-field produce an E-field in the opposite direction so the net E-field (the sum of the external and dipole produced E-fields) is smaller. Thus the force a given ...

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The mentos thrown faster than terminal velocity will experience greater drag until it once again has terminal velocity (note - terminal velocity is only ever approached asymptotically, but we can decide to call "close enough"="equal") If one thing is going faster than another for a while, after which they are traveling at the same speed, then the faster ...

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The terminal velocity is by definition the velocity for which the gravitational force is equal to the friction force of a falling body. As the friction force grow with speed (linearly for laminar flow, quadratically for turbulent flow) an object going faster that the terminal velocity will have his friction force bigger than the gravitational force and thus ...

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There's a lot of text here, hopefully it's not too much and makes at least a little sense. $\ddot\smile$ There are several units for gravity, depending on how you model gravity and what you're looking to calculate. Energy, in Joules ($J$); or spacetime distortion, but I don't know the units, both measure the "absolute" amount of gravity near a point, and ...

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using G ,u can find magnitude of acceleration of any Celestial body.Earth has different radii of curvature at poles and equators. so magnitude of gravity is different at different earth positions

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You need to use energy because the value of $g$ varies with height. So something like $\frac 12 m v^2 - \frac{GMm}{R} = 0 - \frac{GMm}{R+h}$

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Gravity (g), is the acceleration that the Earth imparts to objects on or near its surface. It is measured in metres per second squared (m/s2). The gravitational force on Earth is different from that of other planets as the force is determined by the size of the planet, thus why when astronauts walked on the moon they were almost floating. Gravity also ...

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Here's the way I understand it. Someone will correct me if I'm wrong I'm sure. Using the earth as an example, and assuming a constant mass density, for an object outside of the earth’s surface, it appears as if all the mass of the earth is concentrated at its center. This is a form of Gauss’s law for gravitational objects. If you could go to the exact center ...

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If at any time the speed of the planet in the reference frame of the star exceeds the escape velocity $\sqrt{2GM_\star/r}$, where $M_\star$ is the mass of the star and $r$ is the distance from the star to the planet, it will escape in a hyperbolic trajectory (or straight line if $M_\star\rightarrow0$). As noted in the other answers, the result of the ...

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The scenario you suggest is of course hypothetical, but in all cases you must conserve angular momentum and mass/energy. So for example: If you have a way of removing mass from a star in such a way that the mass disappears outside the orbit of the planets (in astrophysics this is accomplished simply by mass loss - either the star has a wind that expels mass ...

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Your final question very much correlates with a famous thought experiment.If the Sun was suddenly removed the planet s will still continue to stay in orbit. For 8 minutes and 20 seconds. This is because the speed of the space time fabric or simply putting gravity travels at the speed of light. That is, the earth will be devoid of sunlight and will move ...

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A coarse correction would have to be made because of the difference in the curvature of space in that section of space when a planet is deleted. The sun holds the planets in orbit and they would fly away as if the string to a tether ball was cut to all the planets.

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Unlike the centre of masss the centre of gravity is not an intrinsic property of a mass distribution. The centre of gravity only becomes distinct from the centre of mass when there is an external gravitational field that is non-uniform i.e. varies significantly across the distribution of masses. If the external gravitational field is higher on one side of ...

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The multipole expansion in its most general form reads $$V(\mathbf r)=\sum_{l=0}^\infty \sum_{m=-l}^lQ_{lm}\frac{Y_{lm}(\theta,\phi)}{r^{l+1}},$$ where the $Q_{lm}$ are the multipole moments of the system and the $Y_{lm}$ are spherical harmonics, and in this form it is applicable to bodies of any shape, and located at any point in space, so long as the ...

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I think you are puzzled because you are not distinguishing the gravitational potential produced by the system and the one acting on the system. The multipole expansion you link to, is performed on the expression for the potential produced by the system on a distant point. In this case the system has a mass center, but no gravity center can be defined for it ...

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Because it's an excellent approximation. The gravitational potential from a spherical mass $M$ of radius $R$ is, to second order \begin{align} U(r) & = -\frac{GM}{r} \\ & = U(R)+\frac{dU}{dr}(r-R)+\frac12\frac{d^2U}{dr^2}(r-R)^2 +O\left(\left(\frac{r-R}{R}\right)^3\right) \\ & = -\frac{GM}{R}+\frac{GM}{R^2}(r-R)-\frac{GM}{R^3}(r-R)^2 ...

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