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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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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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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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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) ... 14 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 ... 12 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$is a ... 7 Thanks to the hint given by knzhou I figured out that if one wants to give the particle a proper initial velocity of$v_0$, the initial velocity in terms of Schwarzschild coordinates$v_i$would then be $$v_i = \frac{v_0}{ \color{green}{\sqrt{ 1-v_0^2/c^2}}\cdot \color{blue}{\sqrt{1-r_s/r_0}}}$$ for the transversal component, and the same without the blue ... 5 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 ... 5 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 ... 5 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, ... 4 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 ... 3 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 ... 3 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 ... 2 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 ... 2 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 ... 2 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. 2 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 $$v^2 = \frac{2GM}{r}$$ 2 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 ... 1 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$30\,{\rm km}\,{\rm s}^{-1}$. 1 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 ... 1 @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 ... 1 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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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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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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