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With the data you linked to: http://spacemath.gsfc.nasa.gov/earth/6Page58.pdf A linear fit seems fairly reasonable for the given data, but using it to predict the near future is not really possible, since the data deviates from a linear trend a fair amount. I used linear regression to get that the change in the number of days in a year versus the number of ...


1

The basic answer is that there is no such orbit that can survey the entire surface of an irrotational planet in the same manner as satellites do for Earth. Conservation of angular momentum dictates this: the satellite's plane of orbit cannot change without a torque acting on it. Although there is a force acting on the satellite (the gravity from the ...


0

He isn't talking about the planet's motion on the epicycle (Second Anomaly), but about how Mercury's deferent moves, drawing the epicycle toward and away from the earth. To my knowledge, the equant is placed about .05 of the deferent's radius away in the direction of Scorpio, and the deferent center revolves clockwise around a point twice as far in that ...


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Those objects are orbiting closely to SgrA${}^{*}$, certainly, but they are not orbiting closely enough to exhibit significant time dilation effects. In particular, consider the Schwarzschild spacetime. The inner most stable circular orbit around the central obect is at $r = 6M$, three Schwarzschild radii away. This makes the time dilation factor: ...


5

If I'm interpreting your post correctly, you may be misunderstanding time dilation. Time dilation will not cause the stars to seem to move more slowly. The apparent velocity of a star in your frame of reference is the apparent velocity, and relativity will not change it. What time dilation would change is the apparent rate at which a clock moving with the ...


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Apparently no stable orbits in 2+1 dimensions, either: http://arxiv.org/abs/gr-qc/9303005 As for 1+1 dimensions, there is no gravity (the Einstein tensor is a topological invariant, so spacetime is static)


6

Pluto and Charon orbit around their common center of mass. And their common center of mass orbits around the sun, following an elliptical trajectory. Analogously, the Earth and the Moon orbit around their common center of mass, and their common center of mass orbits around the sun. All of this is well described by Newtonian gravitation. The only ...


1

I don't quite know what you mean by "breadcrumb", but this NASA page talks about satellites at Earth's L1 and L2 points, where they follow Earth's orbit around the Sun precisely without actually orbiting the Earth. Unfortunately these points are "unstable equilibria" and are a little like balancing a broom on its end in your hand: you have to actively ...


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Sure. My initial thought was to look in particular at solar observatories, because I know that several have been moved away from the Earth. Wikipedia has a list. The short version seems to be Pioneers 5-9 which are almost exactly what you asked for. Solar and Heliospheric Observatory at the Sun-Earth L1 point. Solar Terrestrial Relations Observatory ...


2

There is no general solution for the two body problem in general relativity. But! There are a few solutions for specific two body problems. These include the Curzon-Chazy metric (Two particles on a cylindrically symmetric axis) $ds^2 = e^{-2\psi} dt^2 - e^{2(\psi - \gamma)} (d\rho^2 + dz^2) - e^{2\psi} \rho^2 d\phi^2$ and the Israel-Khan metric ("two ...


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Under the GR framework there is no known solution. Because the stars and planets are evolving we can say for sure that there is a well-defined 2-body problem and solution. GR was invented to describe gravity at large ant the problem starts with the temptation to use GR in local fields. The contradiction between 'space expands' and 'orbits do not expand' ...


1

You're right that the kinetic energy of the spacecraft is the same both before and after the planetary encounter—in the reference frame of the planet (or, technically, the frame of the planet-spacecraft CM.) But the fact that the kinetic energy is the same before & after in one frame does not mean that the kinetic energy will be the same before & ...


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There is simply no closed-form 2-body problem in GR. The reason is as follows: The governing equations of GR are the Einstein field equations. One obtains the metric tensor as a solution to the field equations which describes local geometry of spacetime, which itself is determined by the local energy-momentum tensor which induces spacetime curvature, etc. ...


2

Define the origin of your coordinate system at the top of the hill with the vehicle travelling in the $+x$ direction with initial velocity $v$ and initial position $s=(0,0)$. The road then "falls away" from the vehicle at a rate of $v \tan(\theta)$, so the altitude $h$ of the vehicle above the downward sloping road can then be written as a function of time ...


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It just oblique projectile motion U= speed of truck h= height of hill If you want to calculate how far from hill the truck land after flight use U multiply by the whole root of 2H/g g is gravitational acceleration =10m/s2


2

First I will make a few assumptions: the hill is flat at the its top (so its surface is perpendicular to gravity); the side of the hill is flat and makes an angle $\theta$ relative to the top; the end of the hill is sufficiently far away such that it will never be reached during the "jump" sideways of the top of the hill; the tradition from the top to the ...


3

The car becomes "weightless" when the curvature of the road is sufficient that the car does not stay connected to the road. Angles don't really matter - what matters is a change in the direction of the road. As you know, an object going around in a circle needs a force $F = \frac{mv^2}{r}$ to stay in the orbit at radius $r$. Normally, when you drive over a ...


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While calculating the orbital parameters works in this case, a simpler method is just to use energy and angular momentum conservation (as mentioned by fibonatic.) First, from energy conservation we have $$ -\frac{GMm}{r} + \frac{1}{2} m v^2 = -\frac{GMm}{r_p} + \frac{1}{2} m v_p^2 $$ Second, we have angular momentum conservation. At perigee, the satellite ...


3

The Schwarzschild radius $r$ is not simply the distance to the centre of the black hole. If you measured that distance by letting down a tape measure you'd find the distance was substantially greater than $r$. See for example my answer to How much extra distance to an event horizon?, where I do this calculation. We actually define $r$ to be the ...


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I would do a hand analysis at first. You have two torques on the arrow. One is the drag torque, which is maximum when the arrow is transverse to the orbital velocity, as in your initial condition. The second is gravity gradient torque, which will be maximum when the arrow is horizontal. Compute each of these for your arrow. If one is much greater than ...



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