# Tag Info

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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 ...

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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 ...

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 ...

4

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 ...

3

Using Lagrange's equation $$\frac{d^2}{dt^2}\left(\frac{1}{r}\right)+\frac{1}{r}=\frac{-m r^2}{l^2}F\left(r\right)$$ And plugging in $$F\left(r\right)=\frac{-k}{r^3}$$ We get $$\frac{d^2}{dt^2}\left(\frac{1}{r}\right)+\frac{1}{r}=\frac{k m}{l^2 r}$$ Which simplifies to $$\frac{d^2}{dt^2}\left(\frac{1}{r}\right)+\frac{1}{r}\left(1-\frac{k ... 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 ... 2 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 ... 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 ... 2 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) ... 1 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 ... 1 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 ... 1 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): 1 The statement is wrong, though sort of true. Gravitational waves are exceedingly hard to create and significant energy is radiated as gravitational waves only for massive stars rotating rapidly at a short distance. In principle the Earth-Moon system radiates gravitational waves, but at such a ridiculously low intensity that it's fair to say it doesn't ... 1 Just solve the second order differential equation obtained from using Newton's Laws i.e.$$F=-\frac{k}{r^3}$$or$$m\frac{d^2r}{dt^2}=-\frac{k}{r^3}$$If you solve this differential equation, then your equation for the path will be of the radius as a function of time. The equation will be a non-central conic. HINT (TO SOLVE THE DIFFERENTIAL EQUATION): ... 1 To get a spacecraft to the Moon we normally use a Hohmann transfer orbit. The fuel is used in two steps: increase the velocity of the scapecraft to put it into an elliptical orbit with its apogee at the Moon. when the spacecraft reaches the Moon increase its velocity again to match the velocity of the Moon. The amount of fuel required is described by the ... 1 The acceleration in a gravitational system is easily calculated using Newton's Law of Gravitation:$$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.

1

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 ...

1

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 ...

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 ...

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