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If the planet was at a much greater distance from the binary than the distance between the stars themselves (say 10 times) its motion would would be similar to a Keplerian orbit around a star equal to the combined mass of the stars and situated at their centre of mass. It has been shown that such an orbit is stable indefinitely. There would be very small ...


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In figure A the satellite is orbiting in a circular orbit. By the satellite increasing speed it would turn the orbit into a more egg shaped orbit as shown in B. This would rather increase the time for orbit revolution rather than decreasing it When the satellite is approaching B.1 it's velocity will begin to slow down as earths gravity begins to pull it. ...


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Thanks to the hint given by knzhou I figured out that if one wants to give the particle a proper initial velocity of $v_0$, the initial velocity in terms of Schwarzschild coordinates $v_i$ would then be $$\dot{\theta}(0)\cdot r(0) = \frac{{v\perp}_0}{ \color{green}{\sqrt{ 1-{v\perp}_0^2/c^2}}\cdot \color{blue}{\sqrt{1-r_s/r_0}}}$$ for the transversal ...


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There seem to be several confusions here. Massive and massless particles behave qualitatively differently, even if the massive particle is traveling very fast. The minimum radius for a stable orbit for a massive particle is $3 r_s$. Circular orbits above this radius are all stable. Massless particles only have circular orbits at the photon sphere, $(3/2) ...


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As far as I understand your question, you can consider the Moon as continuously falling down to the Earth (or continuously following the moving Earth, if you want) due to the gravity. Nothing compensates gravity. However the problem is that the Moon aims not precisely into the Earth, but a bit away - that was original "error" from the Moon's first day. The ...


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Lets distinguish 'rotation' into 'spin' (rotating around itself), and 'orbit' (orbiting around another object). The orbital motion is due to gravity. The 'centrifugal force' (instead of 'kinetic energy' of rotation, per se) is balanced by the 'gravitational force' [1]. The spin of an object doesn't need an external force (gravity from another object) to ...


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Just to more fully answer your question, here is an example of what differences in distances can mean mean as far as they affect gravitational forces. The planet Jupiter is extremely massive, and one side of one of it's moons, Io, feels a slightly larger gravitational pull than the opposite side. This difference in distance results in a gravitational force ...


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Yes. All objects are gravitationally attracted to one another. Even people. Therefore, the earth will draw you to itself no matter how far out you are.


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Gravity is an attractive force, and classical Newtonian theory is adequate to answer in the affirmative: starting from two objects in space at rest, the attractive potential is 1/r , r their distance, and they will fall on each other. The reason planets and satellites have a stable orbit is due to the solutions of the kinematic equations when there exists ...


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@DavidZ comment in the suggestions is correct: there is no unique, general solution to this problem. Depending on the context, dozens of different heuristic approaches are used in simulations. This is part of the general 'N-body' problem. The most literature is in the context of large N-body, cosmological simulations where the problem is called often ...


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The proper definition of the inclination is the angle between the angular momentum vector and the pole (Z axis) of the reference frame. Defined this way i must be be between 0 and 180°. Other definitions (where i is the angle between two planes) are not rigorous. When 0 ≤ i < 90° the orbit is called prograde or direct. For 90° < i ≤ 180° the orbit is ...


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The solution is the same as the elliptic case, except: $$ t = \sqrt\frac{a^3}{GM} (\theta - e\sin(\theta))$$ becomes: $$ t = \sqrt\frac{a^3}{GM} (e\sinh(\theta)-\theta)$$ and $$ \nu = 2 \arctan\left( \sqrt{ \frac{1-e}{1+e} } \tan\left( \frac \theta 2 \right) \right) $$ becomes: $$ \nu = 2 \arctan\left( \sqrt{ \frac{e+1}{e-1} } \tanh\left( \frac \theta 2 ...


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I assume you are asking about a light sphere falling from infinity towards a heavy one. Well, the potential energy at $r$ is $$P = - \frac{GMm}{r}$$ And kinetic energy is $$K = \frac{mv^2}{2}$$ Total energy is zero, so $P=-K$ or $$v^2 = \frac{2GM}{r}$$


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Found the answer on this site: http://hopsblog-hop.blogspot.com/2013/01/deboning-porkchop-plot.html It's because of the plane change required to get from the first body's orbit to the other's. The gap in the middle represents the necessary delta-v to do a 180 degree plane change. You'll notice that they only show up when doing ballistic transfers directly ...


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First, calculus isn't just really small steps: I can show you limit processes that disagree with any really small step based solution, like making a staircase with smaller and smaller treads. The total "tread plus rise" size remains 2k, while the limit is a line with length sqrt(2)k. However, almost all of the parts of calculus that work in predicting ...


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Since you still seem puzzled I'll try a different tactic here: You're showing the tangential velocity (A) and the radial acceleration (B) and adding them to get the green arrow. What you're missing is that this occurs in a gravity field. As the initial path climbs away from the object it's orbiting it loses velocity. This shows up as a third arrow ...


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After reading the answers from other people, I see that you are still confused, so I thougth i coult take my chances on clearing up your confusion. It seems that your confusion is why does the orbiting body not accelerate to infinite velocities (or crashes to body A) eventually, and the origin of your confusion is in your assumption that the resulting ...


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Your vector sum drawing is incorrect. Considering the simple case of a perfectly circular orbit, there is a tangential velocity vector and a perpendicular (inward) acceleration vector. Strictly speaking, you can't add them because they are different quantities (acceleration vs velocity). At some point in the orbit, the object has an initial velocity V in ...


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The system in its final state is created in periastron ($r$), so that's half the problem right there. To find apastron, use angular momentum conservation: $r'v' = rv$ and solve the energy balance equation for the maximum value of r.


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Because the direction of the velocity changes. The velocity will start to point less and less 'towards' point A and when the distance between A and B is the smallest, the velocity will make a right angle with the radius, which means acceleration vector also makes a right angle with the velocity. At this point the radial component of the speed is zero and the ...


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This question points out the importance of symplecticity in physics. In an orbital simulation, suppose one simply advances state via $$\begin{align} \boldsymbol x(t+\!\Delta t) &= \boldsymbol x(t) + \boldsymbol v(t)\, \Delta t \tag 1 \\ \boldsymbol v(t+\Delta t) &= \boldsymbol v(t) + \boldsymbol a(t)\, \Delta t \end{align}$$ where $\Delta t$ is a ...


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You have to consider the limit of infinitesimally short time, in which the (vertical on the paper) component of velocity is infinitely short, and thus also the angle changes for an infinitesimal amount. In this limit, the correction to the length is quadratic in the time step and vanishes exactly in the physical limit of continuous time. Pythagoras: ...


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You sound disappointed that the earth hasn't yet collided with the sun. Applying random formulas to a physical situation is never a good idea. I suspect that the formula you are having issues with is for two bodies in space which are not in orbit around one another, as unfortunately the earth is with the sun. When one body orbits another, the motion of ...


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The answer you refer to is to a question where both bodies are initially motionless. That's not the case for the Earth/Sun system. The Earth orbits the Sun. The gravitational force the Sun exerts on the Earth provides the centripetal force needed to keep the Earth in a stable orbit. In a sense the Earth is constantly free-falling towards the Sun but ...


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As stated in the answer you link, that formula is for 2 bodies starting from rest. The Earth and Sun are not at rest relative to each other, they are in orbit at a relative tangential speed of nearly $30\,{\rm km}\,{\rm s}^{-1}$.


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Any body travelling with increasing velocity increases its kinetic energy $KE$. Since in your system: $$KE+PE=\text{constant}$$ where $PE$ is potential energy. Therefore increase in $KE$ results into decrease in distance between the objects (so as increase $PE$). Note: $KE$ is always positive. $PE$ can be positive or negative. $PE$ is negative for bound ...


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Interesting question. Taking a stab at it - not absolutely sure this is correct, but let the comments begin. In the frame of reference of Earth, the light travels straight out to the reflector, and straight back. You are asking about the case where an observer is in a reference frame that is moving with respect to Earth/moon, and the picture would have to ...


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If you're thinking about stable orbiting systems the big difference between gravity and the magnetic force is that magnetic monopoles do not exist. The simplest source of a magnetic field is the magnetic dipole. By contrast gravitational monopoles exist but gravitational dipoles do not. The Sun and the Earth are both (approximately) gravitational monopoles, ...


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John Rennie already gave the practical answer considering the atmosphere, noting that without doing anything objects near the ISS will deorbit quickly from drag. But that's letting reality get in the way of a good physics problem. I'll show that while a human can't send a ball crashing into the surface in one orbit, they can come close. The ISS is listed as ...


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On a practical note, there was a space walk a few years back when they replaced a failed ammonia pump that was the size of a refrigerator. The astronaut simply gave it a swift push away from the station knowing that it would soon deorbit fast enough that it wouldn't be a collision hazard. PS- If you have Kerbal Space Program, this would be a fun thing to ...


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The relation between velocity v and distance r at which a small body orbits a much larger one of mass M is given by $v^2 = GM(\frac{2}{r} - \frac{1}{a})$ where a is semi-major axis. The perihelion is $p = a(1-e)$ where e is the eccentricity, given by $e^2 = 1 - (\frac{b}{a})^2 $ and b is the semi-minor axis. If you don't have values for G and M you can ...


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You don't need to throw the ball! At the altitude of the ISS the atmosphere is thick enough that it loses 50-100m of altitude every day due to the drag. At that rate over your ten year timescale the ISS would lose 180 to 360km. When you take into account the increased drag at lower altitudes ten years is enough to bring the ISS crashing to a fiery end. So ...


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It is not clear at which level you want to simulate this. Fine grained If you want to do this on a very fine grained level, you just need so simulate Newtonian mechanics and gravity and it should emerge from itself when you have a trajectory set up. This means that you compute the gravitational force to be $$ \vec F = - G \frac{Mm}{r^3} \, \vec r \,, $$ ...


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P.S.2 - Done on Earth. One cannot watch the skies! If you insist on this, it cannot be done. Your condition is equivalent to us living on a permanently clouded world. In that case, we would experience alternating light and dark periods and would have no way of hypothesizing what was happening. Understanding the universe would have to wait for the ...


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There is! It corresponds just to the tides due to the Sun. Let us suppose that the Earth is not rotating around the Sun, that is, we are not in free fall towards the Sun. In this case the liquid from oceans would accumulate nearest the Sun. The effect would be only one daily tide. On the other hand when we fall towards the Sun there are accumulation of ...


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The most straightforward observation to show that the Earth moves is stellar parallax. If you take photographs of a groups of stars over a period of six months (half an orbit), some of the stars will seem to shift in position compared to the others. These stars are much closer to Earth and so seem to move more. This is similar to how, when you are riding in ...



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