# Tag Info

40

Let's assume mass of the person plus spacesuit to be $m_1$=100kg Asteroid density: $\rho=$2g/cm$^3$ (source) that is 2 000kg/m$^3$ 15km/hour is a good common run. That's roughly v=4m/s The orbital height is negligible comparing to the radius, assume 0 over surface. Linear to angular velocity (1): $$\omega = {v \over r }$$ Centripetal force (2): $$F = ... 37 One thing to keep in mind is that objects that are bound gravitationally actually revolve around each other around a point called a barycenter. The fact that the earth looks like its revolving around the sun is because the sun is much more massive and its radius is large enough that it encompasses the barycenter. This is a similar situation with the Earth ... 31 Anything the mass of a star is going to get hot like a star and fuse hydrogen like a star. In other words it will be a star not a planet! While it's technically possible to have a rocky planet the mass of a star, in practice when stellar systems form there aren't enough metals available to build such a large object. Large objects are invariably built from ... 31 Your understanding is correct. There cannot be a geostationary satellite at the poles, basically because it would have to be at rest, which cannot happen as it would get pulled by the earth's gravity and eventually crash to the surface. In fact, there cannot be a geostationary satellite anywhere else, except above the equator(in an equatorial orbit). ... 20 The accepted answer by udiboy is completely right; however theoretically it is not only gravity that acts on a satellite, but also light pressure from the Sun and Earth. Given a sufficiently light and large solar sail (implausible at the current technological level), it is possible to counteract acceleration due to gravity and enter a totally non-Keplerian ... 17 No, not by jumping. Jumping gives you an acceleration only from the location on the surface. As soon as you leave the surface, you have no way of adjusting your orbit. Either you reach escape velocity, or you will return to your initial location after exactly one orbit. The only way to prevent this would be to have an additional acceleration once you ... 13 Specifically what that is referring to is the 'inverse-square law', nature of the gravitational force, i.e. the force of gravity is inversely proportional to the square of the distance: F_g \propto \frac{1}{d^2}. If you expand this concept to that of general power-law forces (e.g. when you're thinking about the virial theorem), you can write: F ... 12 OK, I tried to do the math here. Something remotely resembling maths, at least. Assumptions: It is possible to reach an orbital/horizontal speed of v_O = 5\textrm{ ms}^{-1}, for example by running. The density of the object to orbit is similar to Earth's density, i.e. \rho = 5500\textrm{ kgm}^{-3}. We want to orbit at a height of 2\textrm{ m} above ... 11 No. Any circular orbital velocity is about 70% (1/\sqrt{2}) of the required escape velocity. To find circular orbital velocity, equate the centripetal force to the force of gravity:$$ \frac{m v^2}{r} = G \frac{ M m}{r^2} \rightarrow \boxed{ v_\textrm{circ} = \sqrt{\frac{GM}{r}}} To find escape velocity, equate the magnitude of the potential energy to ...

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In 1619, almost a century before Newton published his groundbreaking Principia Mathematica, Johannes Kepler made a revolutionary contribution to observational astronomy. He noticed that the square of the orbital period $(P^2)$ (the time that it takes for a planet to go around the sun) of a planet's orbit is directly proportional to the cube of the length of ...

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Typically, a star (or stellar remnant, such as a neutron star, white/black dwarf, or black hole) will be the most massive thing in the area, by far. Planets, even gas giants, are a small fraction of the mass of a typical main sequence star. Now, as in Hal's answer, the relative mass of the planet and its star does make the center of mass, the barycenter, of ...

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Something like this, along with the associated article on Wikipedia, might help: And if you "learn by doing" and are willing to have a bit of fun while you develop a sense of the "map" there's a boardgame (of all things) that treats this topic fairly accurately (at least if what you're looking for is some intuition about how the $\Delta\text{V}$ map feels ...

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Actually, both are quite possible. In the general case, for an arbitrary elliptical orbit, what you'll tend to find is that B is true (granted there will be some precession, but not usually in line with the Sun). However, it is possible to set up an orbit (such as a Sun-synchronous one) in which A is true. But orbits such as this require planning and precise ...

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For an object in low earth orbit (at 100+ miles above the earth's surface) the speed needed is about 17,000 miles per hour. Even if a trebuchet could achieve that speed on the earth's surface, you would have at least three problems: The object would IMMEDIATELY burn up in our dense atmosphere. Think about the space shuttle which is going at orbital speed ...

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From the Sun's center, always. When you deduce the equations of motion of planets, you're always calculating from the center. Plus, the results don't change when the Sun blows up as a red giant, or collapses as a dwarf. But even if you measure from the surface, in most cases it won't make a huge difference. In Earth's case, it's a 0.5% error. It would be a ...

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Hints: Prove that the angular momentum $L^{ij}:=x^ip^j-x^jp^i$ is conserved for a central force law in $d$ spatial dimensions, $i,j\in\{1,2,\ldots ,d\}.$ Choose a 2D plane $\pi$ through the origin that is parallel to the initial position and momentum vectors. Deduce (from the equations of motion $\dot{\bf x} \parallel {\bf p}$ and $\dot{\bf p} \parallel ... 7 Since the calculations are already in others' answers, I'll just refer to this great, classic xkcd. Deimos and Phobos, the two small moons of Mars, match (or almost match) the criteria SF and Claudius derive. As Munroe points out, (The diagram is a representation of the gravity wells of both moons, represented by their height at constant Earth surface ... 7 I was so surprised by the types of graphs I saw for this that I felt compelled to add an answer. As mentioned in the Wikipedia article I linked to, there are two radii that are of interest in addition to the Schwarzschild radius$r_s$. Those radii are the "Innermost Stable Circular Orbit" (ISCO)$ r_{outer} \approx2a^{2}/r_{s}$and the "Innermost ... 7 As Brandon mentioned, two small objects couldn't orbit each other near a significant gravitational field. The Hill Sphere "approximates the gravitational sphere of influence of a smaller body in the face of perturbations from a more massive body." Therefore, your pebble's Hill Sphere would be too small to permit orbits near Earth. The Wiki article has a ... 7 "The resulting orbit resembled the Treyarch logo, which I now suspect was inspired by physics demos in the company's early history." Yes, indeed they are elliptical, but there are also extremely tiny deviations from this general case (which are extremely difficult to observe, in general, even after some time). It's important to note that computers have a ... 6 Orbits are funny things really The key thing is the distance from the object that they are orbiting, and the speed at which they are going. Spacecraft routinely speed up/ slow down their orbital speed, to affect where they are orbiting around a particular spacecraft. So, the real answer is, Neptune pulls Uranus closer when it is ahead of it in orbit, and ... 6 I'll try to answer it by considering radial deviations from a circular orbit. First we have to assume two things about our n-dimensional universe: Newton's second law still holds, that is, for a particle's position vector in n-dimensions$\vec{x} = (x_1, x_2, \cdots x_n), \begin{align} m \ddot{\vec{x}} = \vec{F}, \end{align} where\vec{F}$is some ... 6 The orbit of a satellite is determined by 6 parameters, the orbital elements: The eccentricity$e$The semimajor axis ($a$) The inclination ($i$) The longitude of the ascending node ($\Omega$) The argument of periapsis ($\omega$) The mean anomaly at a specific time ($M_0$) The plane of reference is the equator, and the reference direction is the Vernal ... 6 As noted you can't be stationary without a really big solar sail or magic rockets. The usual solution to this problem are orbits with long hang times in view of a pole, specifically the Molniya orbit, commonly used by the Russians with a lot of high-latitude territory. Multiple Molniya satellites can provide very good coverage. 6 As Guillermo Angeris correctly pointed out, this is essentially a numerical roundoff problem, not a physical situation. As a physical example, there are sungrazing comets that get very close to to Sun, yet they maintain their original elliptical (or hyperbolic) orbit, without the orbit precessing a full third of a circle as you seem to be seeing. ... 6 In Newtonian mechanics, with nothing to slow the orbiting object (or otherwise obstruct it), energy is conserved, so the orbit is maintained indefinitely, and can continue so at any altitude. As soon as you introduce an atmosphere, there is a mechanism to remove energy from the orbiting object, so the orbit decays until it makes contact with the body it was ... 5 An orbit is a closed trajectory of a classical dynamical system. Properties of orbits or related to orbits are referred to as orbital properties. http://en.wikipedia.org/wiki/Orbit_(dynamics) An orbital is a single-electron wave function for an atom or molecule in the Hartree-Fock approximation. (There are also hybrid orbitals for electron pairs in a ... 5 If the catapult accelerates the object from rest to$v_{e}$over a length$L$at a constant acceleration$a_{e}$, then the acceleration will be $$a_{e}=\frac{v_{e}^{2}}{2L}$$ so the force during acceleration of an object of mass$m$will be $$F=m\frac{v_{e}^{2}}{2L}$$ For$v_{e}$of Earth, when$L$is ... 5 What does this mean? It means that there won't be any (periodic) orbit anymore; the answer to your title question is therefore that it will cease to exist. The value of$r$will just monotonically decrease. Obviously, when it falls below the event horizon, there's no way for the particle to return outside the black hole i.e. to values of$r\$ greater ...

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This is really just a footnote to Kitchi's answer and Hermann's comment to it: Dynamical friction won't cause all the stars in the galaxy to swirl down into the black hole like water down a plug hole. Apart from anything else conservation of momentum forbids this. Dynamical friction causes a type of sorting. In interactions between stars on average the ...

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