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You must assume circular orbits, as stated in another answer. The ground speed will also be variable due to inclination, unless it's equatorial. An extreme example is an east-west orbit versus a west-east orbit. In one case, the ground velocity is added, in the other case, it's subtracted. But with the query limited to equatorial orbits, we can continue. ...


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Geostationary satellites travel in circular orbits about the equator. If you express the velocity as angular velocity, it will equal the angular velocity of Earth's rotation, about $7.3 \times 10^{-5} radians/second$. If you had another satellite in circular orbit about the equator, but not geostationary, you could subtract the angular velocity of the ...


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The mass $M$ in the equation for orbital velocity would be equal to the mass of the Earth (assuming that the mass of the satellite can be neglected) and thus constant. The gravitational constant, $G$, like the name suggests is constant as well. But the radius of the (circular) orbit, $r$, can vary. So by using different altitudes at which the satellite would ...


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As the formula you wrote suggests, it depends on $r$, the distance between the satellite and the centre of mass. $3.1\:\mathrm{km/s}$ is for geostationary orbit, and $8000\:\mathrm{m/s} = 8\:\mathrm{km/s}$ is for low Earth orbit. Please see Wikipedia for more details.


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Let's assume the light from the Sun is parallel, then the shadow of Earth looks like this: The dotted line is the orbit of the satellite at a height $h$ (I've exaggerated the height a bit to make the diagram clearer). All we have to do is calculate the angle $\theta$, because the time the satellite is in the Earth's shadow is simply: $$ t = \tau ...


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Earth has few and relatively tiny sattellites other than the moon specifically because of this large moon. Note the mass ratio of our moon to our planet. It is the highest in the solar system by a large amount. This one large sattellite will over time sweep up and aggregate other smaller sattellites. Put another way, we probably did have other smaller ...


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The moon is by far the largest of earths natural satellites, but is by no means the only one. There are many much smaller bodies which either orbit earth or have an orbit around the sun which is extremely similar to that of earth. The moon is the most significant and largest as it is the result of large amounts of matter clumped together that was thrown off ...


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When the the earth was formed. I guess only the moon was close enough in proxmity to be attracted by the gravity of the earth or rather earth cant hold anymore natural sattelites but the moon.


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As indicated in the comments, this is probably a question for Computational Science. But I will try to answer your question in such a manner that it may be on-topic enough to stay here. I think you can solve this by using Python's atan2(y,x) function, rather than defining $z = y/x$ and using atan(z). Let me address this concern first: ...except where ...


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Planck: 13.82 Gyr; 68.3% dark energy, 26.8% dark matter, 4.9% baryonic matter. http://arxiv.org/abs/1306.5534 There is no dark matter in the solar system. Dark matter inside Saturn's orbit is less than 1.7×10^(-10) M_solar. Dark matter is repeatedly reparatmeterized curve-fitting with no empirical composition. Dark matter phenomenology is wholly ...


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The answer is because dark-matter has relatively constant density, as has been given explicitly in another answer. Then, it logically follows that the impact on the Milky Way due to this low density. To show this step, I will establish a figure of merit. $$ FOM = \frac{M_{dark}}{M_{normal}} $$ That is, the ratio of dark matter within the area of ...


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Dark matter collects in larger quantities (thus a higher proportion relative to matter) in the centre of galaxies compared to in the centre of stellar systems such as the solar system. galaxies are not very dense, as stellar systems are sparsly spaced. So even though on a galactic scale the dark matter is in high ratios, on a stellar scale the ratio is ...


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Dark matter would affect planetary motion, but the influence of dark matter on planets in our solar system is too small to detect even currenlty, due to the low concentration of dark matter compare to ordinary matter in our solar system. See Constraints on Dark Matter in the Solar System. The density of dark matter is very low, $ <~10^{-19} grams/cm^3$, ...


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$\theta$ is the angle made between the positive x-axis and the spacecraft's velocity vector (all at closest approach). Don't you need $\theta$ to be the angle of the spacecraft's velocity relative to Mars? Try subtracting the Mars velocity vector from the spacecraft's velocity vector, calculating $\theta$ for the resultant vector and using that angle ...


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You've used the gravitational constant with only three significant digits. So it's no surprise that your answer isn't accurate to five significant digits. Instead of $G$ and $M_\odot$ separately, you should use the product $GM_\odot$, known as the standard gravitational parameter. Its value is known very accurately: in the link, you'll find $$ GM_\odot = ...


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To expand on Prahar's answer, let me run some numbers to try and convince you this is reasonable. Your answer is correct to within one part in 104: $$ \frac{365.256363004}{365.2075}\approx 1.000133795. $$ The main perturbing influence on Earth's orbit is the gravitational pull of Jupiter, whose mass is about 1000 times smaller than the Sun, and which orbits ...


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Kepler's 3rd law assumes that the Earth travels in a perfect ellipse with the only gravitational force on it being from the Sun. Further, Kepler's laws are derived from Newtonian gravitation. In reality, the orbit of the Earth is affected by the gravitational pull of other planets, and by the effects of General Relativity and is therefore not quite ...


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I think there's a flaw in your reasoning. From a parent-body-fixed reference frame, an object coming in retrograde would actually appear to move faster than a prograde object. In any case, the only way a retrograde capture would be more likely is if the orbital path of the incoming object takes it through the parent body's atmosphere, in which case a ...


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No. The n-body problem is a problem of calculation, not of determinism. There is nothing non-deterministic about Newtonian mechanics, it is just hard to calculate. By contrast, a quantum system is generally accepted to be non-deterministic by its nature; Einstein disagreed with this idea, hence his statement that 'God does not play dice'.


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Orbits are pretty complicated. Most texts on this deal in terms of predicting positions at a specific time rather than just a simple ellipse, because that model while correct is too basic. As others mentioned position estimation makes it more complex because there are more parameters involved. I'll just pull some stuff for you on the basics. For standard ...


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It depend on which "many relationships" you are referring to. A better model would be a Kepler orbit with an eccentricity ($\epsilon$, a measure of how elliptical an orbit is) smaller than one. This model describes a two body problem. But the downside of this model is that it can not explicitly be expressed as a function of time. Namely you would have to ...


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I think your definition of sphere of influence is not correct. You could also be confusing the sphere of influence with the Hill sphere. The sphere of influence has mainly an application in the patched conic approximation. And the word sphere is even another approximation. A related question asks about the derivation of the radius of this sphere. I also ...


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The two relations that you give, are strictly valid only for circulation motion. You give an example of circular motion with variable radius, but, that is not circular motion anymore. Also, how would you define the centre of the motion, and thus angular velocity or cetripetal accelaration? If you have a well-defined case, you can always derive similar ...


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Take for example the equation for the velocity: $$ v = r\omega $$ In circular motion the velocity is always normal to the radius. In non-circular motion, e.g. an elliptical or hyperbolic orbit, the velocity is not normal to the line joining the object to the focus of the orbit. However the velocity can be expressed as a vector sum of a normal, $r_\perp$, ...


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A Lagrange point is a position relative to a system of two-body gravitational objects at which a third negligible small object, if its velocity is correct too, would not move relative to the two other objects. The two gravitational objects will orbit around the center of mass of both objects together (also called the barycenter). I am not sure, but I also ...


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For your first question: Yes gravity is mediated in all directions equally. As for your second question: I assume you are asking about elliptic orbits. If this is the case then, from the law of conservation of angular momentum, when the planet is closer to the sun it has greater velocity, and when its further out it travels slower because $\vec{L}$ = ...


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Yes, the Sun and our whole Solar System are revolving around the centre of our galaxy, the Milky Way. Milky Way is a spiral galaxy and hence has four major spiral arms and a central buldge. The Sun (and, of course, the rest of our solar system) is located near the Orion arm, between two major arms (Perseus and Sagittarius).


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Consider a lift with its rope snapped. The lift would be falling freely. An observer is inside the lift (tough luck for him!) releasing the ball just at the moment of the free fall. Since the ball and lift would be falling freely the ball would appear to float. Thus, to the observer in the lift, it would seem as if no force is acting upon the ball, using ...


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Black holes pull matter in thus creating a pulling of matter or rotation (gravity) force that matter rotates or spirals around thus creating our shape form and rotational direction of our galaxy. If our sun doesn't die first then yes one day we will be pulled to the center of our galaxy. Every galactic year our solar system speeds up and grows closer to the ...


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Would the asteroids count? Or maybe the rings around Uranus or Neptune? I don't think the rings on Saturn would fit in your requirements since they are too broad.


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You should not try to rephrase things in terms of some fuzzy smeared out averaged classical physics, or that the electron "really" is moving in an orbit or something like that. What is angular momentum? Angular momentum is the quantity that is conserved in a rotationally symmetric system. You can work out that in quantum mechanics the (orbital) angular ...


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In an atom, the electron is not just spread out evenly and motionless around the nucleus. The electron is still moving, however it is moving in a very special way such that the wave that it forms around the nucleus keeps the shape of the orbital. In some sense the orbital is constantly rotating. To understand precisely what is happening you need to use ...


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The electron is still translating in space with a certain momentum. This can be represented with an angular momentum operator. Even though the translation of the electron is not "continuous" as in classical mechanics there is still a quantised way of translating about the nucleus, with associated momentum, and position.


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I do not understand the source of the orbital angular momentum. Is it intrinsic like the spin? No. Despite the fact that the picture of the orbiting electron is not quantum-mechanically correct, orbital angular momentum is still written in terms of position and momentum of the electron. More precisely, orbital angular momentum is represented as a ...


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The planets appear to have formed by accretion of dust and gas around initially-small nuclei in orbit around the sun. As orbital radius (distance from the sun) increases, more material is available (in a uniform dust/gas disc) to accrete, so you get bigger planets further out; bigger planets sweep up more of their neighbors, so they tend to be spaced further ...


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As you have said, the general solution for the Kepler problem is an elliptic orbit. The shape of an ellipse is determined by its semi-major axis and the eccentricity. The semi-major axis is determined entirely by the energy, but the eccentricity depends also on the angular momentum. Both energy and angular momentum are conserved in the two-body problem, so ...


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In a two body system the orbit can have any eccentricity you want. The key point is that the eccentricity cannot change. In a three or many body system the bodies perturb each others orbits and the eccentricity can change with time. For example this graph shows the eccentricities of the rocky planets as a function of time. These changes are mainly caused by ...


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It's because of Conservation of Angular Momentum. When all these systems are in the process of creation, all motions not existing in the same horizontal plane cancel each other out which confines them into that plane. The system continues to exist that way due to conservation of Angular Momentum.


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it is because of the way the solar system formed from a cloud of gas. the molecules of gas (mostly hydrogen) were pulled together by gravity which formed a spinning disk of gas around our sun, thus forming the planets all on the same field.


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To the original poster: You appear to be operating under the "hollywood" misconception that a black hole somehow "sucks harder" than the same amount of mass in a non-black-hole form. However, this false "black holes produce an enormous sucking" misconception is one of the many, many concepts of physics that "hollywood" gets totally wrong; a black hole of a ...


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The orbital energy of two bodies is defined as the energy required to separate them to infinite distance: $$ E = E_1 + E_2= \mu \frac{v^2}{2} - \frac{ G m_1 m_2 }{ r } = \mu \left( \frac{v^2}{2} - \frac{ G m_2 }{ r } - \frac{ G m_1 }{ r }\right) = - \frac{ G m_1 m_2 }{ 2 a } = - \frac{ G m_1 m_2 (1+e) }{ 2 d } $$ where: $ \mu $ is the reduced mass, v is ...



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