# Tag Info

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Given the page, I am assuming you're asking about this image. It shows the orbit of the station around the earth as a red line. From this view, the position of the line is approximately fixed around the center of the earth, with the angle almost fixed with respect to the stars (inertial frames). In this view, the earth turns to the right (west to east) ...

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No. The ISS is in low-earth orbit, so it won't maintain a specific ground-track along the earth. Wikipedia has a picture of the orbit at different times that I'll attach below (open source image): The orbital period of the space station is ~90 minutes or so, so the earth will rotate by that much underneath it when it is in its orbit. From LEO ...

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One has to distinguish between, on one hand, an orbit and an orbital motion, which are classical notions; and on the other hand, an orbital, which is a quantum mechanical notion, cf. above comment by dmckee. If the question is really Why quantum mechanics?, then have a look at e.g. this Phys.SE post and links therein. Here we will assume that OP accepts ...

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Electron in a ground state hydrogen atom has zero angular momentum $L^2$, l=0. Moon has a huge angular momentum. Therefore it is a poor comparison. If moon would have zero angular momentum, in classical physics, it would fall down and hit earth. Electron in an hydrogen atom, in l=0 state gets constantly pulled to the center, but this is countered by the ...

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Can we transfer burn to another planet at any time? Yes... if we have a big enough rocket. The problem of figuring an orbit that gets you from point A to point B in a certain time is called Lambert's problem, and it turns out that there is a solution that is relatively easy to calculate—by which I mean that it does require solving equations ...

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Space is big. Most likely (with many nines), it will miss the earth since they are not on the same orbit. It is possible that it could interact with the earth at some point in the future. However, the spacecraft is only a little more than 1000kg. As a single object in a heliocentric orbit, the risk from it is nearly zero. Tiny metal fragments in earth ...

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To determine transit UTC transit time of a celestial object; (360 plus your Longitude) minus star SHA= GHA Aries Get GHA Aries from The Nautical Almanac here; www.TheNauticalAlmanac.com To determine transit time of the Sun on your meridian (Longitude); Sun GHA = your Longitude. Let's say you're located at W 078 degrees 42 minutes. Date- November 18, ...

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The shape often doesn't matter. Most orbits are very far away. In the solar system, the sun and planets are modeled as points. Even ones like Earth, which has a large moon. This gave extremely good results. For example, the orbit of Mercury is close to an ellipse. It is perturbed by the attraction of other planets, primarily Jupiter. Because of this, the ...

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Depending on the dimensions and mass of the linear object you mention, rather than a normal spherical one, at distances relatively far away from the object, space debris would orbit it as though it were spherical. This is because the centre of gravitational attraction would, in effect be a point and a spherically symmetrical attractive force field would ...

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63,360 x 238,900 = 15,136,704,000(inch now) (1.5inch x 4,000,500,000 = 6,000,750,000) 15,136,704,000 - 6,000,750,000 (inch then ) = 9,135,954,000 inches. 9,135,954,000/63360=144,191.19318181 (converting it back to miles) Roughly 144,000 miles 4.5 billion years ago But due to the tidal friction, caused by the tidal bulge, the earth loses energy and is given ...

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You're on the right track. "How do we know that the space shuttle passes throught the point P after losing speed ?" The assumption (in these types of problems) is that the thruster is applied for a very short time compared to the duration of the orbit. In that way, we can assume that applying the thruster is effectively instantaneous. So if the ...

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It is assumed that the spacecraft fires changes its velocity in an instant, not over a period of time. Its velocity is decreased exactly at the point in time it passes through P. It is true that spaceships in lower circular orbits have greater orbital velocities, but in elliptical orbits the velocity changes with the distance between the two masses (since ...

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The answer to your question can be found in the description of the engineering of the monument You can read the whole story at that link; I will just quote the most pertinent statement: Using the statistical mean of the 100-year data, the altitude and azimuth angles for the structure were adjusted to provide time/error fluctuation of plus or minus 12 ...

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I think that instead of considering the amount of time spent in the shadow of a planet with a significant radius $R_{earth}$ that it may make the analysis clearer and simpler if you instead consider the limit of some very small object with an infinitesimal radius casting a shadow. Let's call the radius $\text{$\Delta $R}$. What you're then asking is if it ...

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This is the elliptic case of the radial Kepler problem, the equation for time as a function of position is $$t(r) = \sqrt{ \frac{d^3}{2 g} } \left( \arccos\left( \sqrt{ \frac{r}{d} } \right) + \sqrt{ \frac{r}{d} \left(1 - \frac{r}{d} \right) } \right)$$ where t is the time, r is the position, d is the initial (maximum) separation, and g=G(m1+m2). In ...

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Actually it is similar to the solution of a body going through the center of the earth through a drilled hole . Neutrinos could go through such an orbit with zero angular momentum as the probability of interacting and disappearing is very small. Planets cannot. If Goldstein is in a chapter for planetary orbits (hint Keplerian) it is obvious why the one ...

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The two equations of motion reduces down to one equation of motion by considering the separation $x=x_2-x_1$ and the separating acceleration $\ddot{x} = \ddot{x}_2 -\ddot{x}_1$ $$\ddot{x} = -\frac{G (m_1+m_2)}{x^2}$$ or $\ddot{x} = -K/x^2$ with $K=G (m_1+m_2)$ This can be re-written as $\frac{{\rm d} \dot{x}}{{\rm d} t} =\frac{{\rm d} \dot{x}}{{\rm d} ... 0 What you have is a system of coupled differential equations. Say the position of the masses are$m_1$and$m_2$. The positions are$x_1(t)$and$x_2(t)$. We'll assume that$x_1<x_2$. Note that they will stay on a line, so it suffices to consider one dimension. Now, we use$F=mx''$to construct our ODEs: $$G \frac{m_1 m_2}{(x_2(t)-x_1(t))^2} = m_1 ... 1 I assume that friction is an external velocity dependent force in your simulation code. Since you have such external forces, your total energy, total angular momentum, total momentum are likely not to be conserved. In your case, the friction is a phenomenological external force, but similar behavior could also be simulated with a large particle, moving in a ... 3 According to conservation of momentum, the center of mass of a system cannot accelerate without external forces. In other words, if the center of mass starts out at rest (which is generally a good procedure in simulations), then it should always stay at rest. It is normal for numerical errors to introduce deviations, but the motion you are seeing looks ... 0 I'm trying to answer my own question. Please feel free to comment and complement me! The earth goes around the sun and it is fixed with respect to it because: ... I have three reasons to give to the man and all things he can check in first person: Mercury and Venus, the other inner planets have phases if you look at them through a telescope. If they were ... 1 The acceleration in a gravitational system is easily calculated using Newton's Law of Gravitation:$$F = \frac{Gm_1 m_2}{r^2} \hspace{0.2in}\&\hspace{0.2in} F = ma$$For two bodies (like you describe), it is just this simple. If there were more bodies (a so-called 'n-body system'), you can just add the forces between each pair of bodies. 0 Also, there is parallax for the closer stars so that over the course of a year, the nearby stars will move back and forth with respect to more distant stars. The measurements are good to about 100 parsecs which equates to the limit of measurement of 0.01 arcsecond. The only way to explain the back and forth motion of the parallax is that the earth is moving ... 0 Quite simply, the Sun moves relative to the stars. If you could see stars during the daytime then this would be somewhat easier to measure, but as it is you need some to look at night. Choose a time (say, 9pm) after dark, and go outside and look at the stars everyday for about one or two months. Observing at the same time will keep the Sun in the same ... 0 Another proof: relative to stars, the day is 23h56, while relative to Sun, the day is 24h. https://en.wikipedia.org/wiki/Earth%27s_rotation So Earth rotation couldn't expain both stars and Sun rotation, there must be a differential rotation between these two. 0 The historical "proof" was about the motion of planets along the year, but it needs some patience and observation. Earth-centric model yields pretty complicated cycloidal motion for planets, while Sun-centric yields simple ellipses (with moderate excentricity). Also the seasonal tilting of the axis is alot simpler in the second case, since the axis keeps a ... 2 The elements you give describe an idealised orbit that does not exist in reality. Those numbers are parameters to an approximate model. Earth's closest distance to the sun is different each and every year, by a lot (about 20,000 km in fact). Are there any exact data about Earth's orbit? There are certainly far better models than the 6-parameter ... 0 As David Hammen points out, there is nothing special about the points B and C. But if you insist, those are the points where the eccentric anomaly E is \pi/2 and 3\pi/2, respectively. For an idealized orbit, Kepler's equation gives$$ \frac{2\pi}{T}(t-\tau) = E - e\sin E, $$where \tau is the moment of periapsis (January 4) and T=365.2596 days is ... 0 The position of equinoxes is far more complicated than I thought, can someone explain how you determine with a certain accuracy the middle points of the ellipse B and C? Your points B and C will not help in your understanding. Point C was before the end of September, point B in March, how can we determine the position with a certain accuracy? From ... 0 If you just want to find the dates for various events, the the following link provides a list of sources for astronomical calculators of various types. http://www.midnightkite.com/index.aspx?URL=Software The US Naval Office also is a good source of this sort of astronomical information. http://aa.usno.navy.mil/index.php From the data you present in your ... 5 Tyco Brahe Observed Mars. And as the Mars is out side us, and rotates slower, it has an particular character that it even moves to "wrong direction" in the sky for a while. It must have been partially luck, that 5 of these observations is measures with enough accuracy this important point in orbit. (see link) Or maybe this was exactly the interesting ... 19 Are there any exact data about Earth's orbit? No. There are always measurement errors. There are however very good estimates. The best estimates come from three competing organizations, the Jet Propulsion Laboratory (the Development Ephemeris models), the Russian Institute for Applied Astronomy (the Ephemerides of the Planets and Moon), and the IMCCE ... 28 I generally regard NASA as authoritative, and they report the orbital parameters on their Earth Fact Sheet. I note that they disagree with Wikipedia about the aphelion though they agree on the perhelion, semi-major axis and eccentricity: NASA Wikipedia Aphelion 152.10 151.93 Perhelion 147.09 147.095 Semi-major 149.60 ... 1 To get a spacecraft to the Moon we normally use a Hohmann transfer orbit. The fuel is used in two steps: increase the velocity of the scapecraft to put it into an elliptical orbit with its apogee at the Moon. when the spacecraft reaches the Moon increase its velocity again to match the velocity of the Moon. The amount of fuel required is described by the ... 0 The answer is (d) and there's nothing ambiguous about it. a=F/m, true. But Newton's law of gravitation has F = GMm/r^2 \propto m, so that a = GM/r^2. Since r is the same for both, their acceleration (centripetal) is the same. Inversely proportional to their respective masses, yes, but still the same. The inverse relationship between acceleration ... 1 None of the given answers are the correct explanation. The reason that the astronaut doesn't float away is because the acceleration due to gravity is the same for the astronaut and the spacecraft. It's what is keeping the spacecraft in orbit and it doesn't change just because the astronaut steps outside. The force of gravity on the astronaut and the ... 4 Let's look at each answer in turn: a) The force of gravity acting on the astronaut and spacecraft is negligible Wrong. To be in orbit, gravity needs to be acting. If it were negligible, they would just head off in a straight line in space. b) The spacecraft and the astronaut are in orbit around the Sun with the Earth This is true, but irrelevant. ... 5 As a general rule (regardless of the definition of V) we have:$$\frac{\mathrm{d}}{\mathrm{d}t}(\bf{V}\cdot\bf{V})=\frac{\mathrm{d}\bf{V}}{\mathrm{d}t}\cdot \bf{V}+\bf{V} \cdot \frac{\mathrm{d}\bf{V}}{\mathrm{d}t}\frac{\mathrm{d}}{\mathrm{d}t}V^2=2\left(\bf{V}\cdot \frac{\mathrm{d}\bf{V}}{\mathrm{d}t}\right)$$... 2 Consider some small object orbiting the Earth. By small I mean that the mass of the object is so much smaller than the mass of the Earth that we can take the Earth to be fixed i.e. the object can't move the Earth by any measurable amount. If the mass of the object is m, the mass of the Earth is M and the distance to the object is r then the ... 2 No it's not a coincidence. The linear eccentricity, c, is the distance from the centre of the ellipse to either of the foci. This diagram shows an orbit with this marked - for clarity I've made the orbit very eccentric: The eccentricity that you quote is defined as:$$ e = \frac{c}{a} \tag{1} $$where a is the semi-major axis. The lower diagram ... 1 Do you remember the formula for centripetal force in circular motion, which in this case is due to gravity. As a refresher:$$F= \frac {m_2v^2}{r} $$Where F is the centripetal force which is due to gravity and is given by:$$F = G \frac{m_1m_2}{r^2}$G$is the universal gravitational constant$6.67408 × 10^{-11} m^3 kg^{-1 }s^{-2}m_1\$ is the ...

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