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71

The trouble with orbital mechanics is that it rapidly gets exceedingly complicated and hard to make intuitive sense of. However I think there is a reasonably straightforward way to show how little effect GR has on an Earth-Moon transfer orbit. But this takes a little preparation so bear with me while I give a short introduction. I hope everyone who reads ...


42

The Jet Propulsion Laboratory has incorporated general relativistic effects in its numerical integration of the planets since the mid to late 1960s. For example, the JPL DE19 ephemeris, released in 1967, incorporated relativistic effects in its modeling of the solar system. This didn't help much. Had they ignored relativistic effects there would have been ...


24

I'll start the ball rolling on this one. My GR knowledge is probably not good enough to make this a truly satisfying answer... The gravitational acceleration for an object moving radially at non-relativistic velocities in the Schwarzschild metric is modified by a factor $(1 - r_s/r)(3[1-r_s/r] -2)$, where $r_s = 2GM/c^2 = 0.00885 m$ for the Earth. If we ...


22

A few sanity checks without actually computing anything: First, the error due to neglecting general relativity is so small that it didn't affect prediction of lunar eclipses and wasn't actually noticed anywhere except in Mercury's orbit (at least not until they purpose-built experiments to detect minor discrepancies). I know this doesn't give a completely ...


22

But can someone please explain why this is without using pure algebra? I will try without a single formula. In Newtonian gravity, the gravitational force on a particle is proportional to the particle's gravitational mass; the more gravitational mass, the more the gravitational force. In Newtonian mechanics, the acceleration of a particle, for a given ...


11

You certainly know that all things fall at the same rate regardless of their mass (neglecting friction). An orbiting body is not different from a falling body in that the only force acting on it is the gravity of the thing it orbits, so there is no reason its mass should influence its orbit.


7

Why does the mass of the orbitting object have no effect on its revolution at all? It does have an effect! However, the effect is immeasurably small if the orbiting object itself has a very small mass compared to the object it is orbiting. The most massive object we humans have put into orbit is the International Space Station, with a mass of 419.5 ...


5

Orbit is free-fall around a body, so it is for the same reason a feather falls as fast as a bowling ball (in a vacuum). In free-fall, $F = mg$ And we know that, $F = ma$ So we can substitute, $ma = mg$ And divide by $m$, $a = g$ Thus, no matter what mass is, acceleration equals $g$.


5

Your question highlights a common misconception. A satellite in orbit around the Earth is accelerating towards the Earth right now. Any object moving in a circular path has an acceleration towards the center of the circle because the direction, and therefore the velocity, of the object is constantly changing. This acceleration, called centripetal ...


4

You are no doubt familiar with the apocryphal experiment of Galileo showing that falling bodies fall at a rate independent of their "weight". [ We should really say mass.] An orbiting body is just a special kind of falling body, albeit one that manages to miss the ground due its sideways motion. Hence the term "free fall" as used regarding astronauts or ...


4

You raise an interesting point about the role of experiment and falsifiability in science. Despite a long-standing anomaly in Mercury's perihelion, Newton's theory of gravity itself wasn't heavily questioned, let alone rejected or falsified. Rather, auxiliary assumptions were concocted that saved Newton's theory, such as an erroneous mass of Venus, a planet ...


3

Since he is falling at terminal, hence constant, velocity, he is experiencing NO net (total) force. There are (at least) 2 forces acting on him though, which are his weight (900N) and the air resistance. Because he has constant velocity, i.e. he is not accelerating, these forces must balance, i.e. the air resistance is 900N upwards whilst his weight is 900N ...


3

The best paper on I've found on tidal tails is Reshetnikov & Sotnikova (2000). Their simple description of tidal tail formation is: To understand the development of tidal tails, one must recall how the water surface of the oceans get stretched radially by differential gravitational attraction exerted on it by our Moon. The differential forces ...


3

Consider that it would not be particularly difficult to do an Apollo-type landing if each of your relative velocity, range, and angular measurements were off by +/- 5%. You could simply make small iterative corrections along the way, until the absolute values were small enough to make the relative errors inconsequential. At worst you'd need to carry ...


3

Dark matter does not readily "accumulate". If(?) it exists then it interacts very weakly with normal matter and is primarily influenced by gravity. The Earth's gravity is far too small to make a local concentration of dark matter. The local dark matter would be moving in the Galactic potential at speeds similar to that of the Sun around the Galaxy ($\sim ...


3

In binary systems, each object is so affected by the others gravity that they have significant orbit. The sun has so much inertia that the earth's pull barely moves it, but the earth certainly revolves around the sun. In the reference frame of the Earth however, the Sun does revolve around the Earth.


2

Both the the Earth and the Sun orbit around the solar system barycentre. This is defined as the centre of mass of all the bodies in the solar system. Because the Sun contains the vast majority of the mass of the solar system then the barycentre is very close to the Sun. The picture below, from the wikipedia entry on the solar system barycentre, has the ...


2

On a planet with gravitation ten times stronger than that of Earth's, will an object fall faster? Yes. It will have ten times greater acceleration. If it starts out at the same height as on earth it will be going faster when it hits the ground and it will hit the ground in less time. If I drop two apples about 8000 meters away from the surface ...


2

Newton actually calculated, using a lot of obscure geometry and limiting concepts, the orbits that various force forms would generate. One of those forms was the inverse square force. If you want to know what the others were, find a copy of the Principia and wade through it. The result wasn't published until Edmund Halley asked Newton if he knew the nature ...


2

This is a classic misconception that most people share at some point in their lives. For centuries, we struggled to understand this point. For example, the famous Aristotle expresses your misconception that: continuation of motion depends on continued action of a force i.e. you see a ball moving upwards, and think that there must always be a force ...


2

You ask why do heavier objects decelerate faster when moving away from the earth That question can't really be answered, because they don't. The force of gravity is a function of the distance - this is the famous inverse square law. But the force is proportional to the mass of the object - so objects with twice the mass experience twice the force. ...


2

It would be more convenient to use the fact that $F\Delta t = m\Delta v$. There is a constant force of gravity pulling you down, and an occasional force of the foot pushing you back up. Averaged over time, the two must be equal as there is no net change in vertical velocity. This means that $F_{strike} \approx \frac{W}{\Delta T}$ where $W$ is the weight and ...


2

There isn't a simple closed form expression for the distance as a function of time in general relativity. However if you're just interested in how big the difference is I think there is a nice way to see this. For simplicity let's take a falling object with zero total energy. In the Newtonian case this means the kinetic energy is equal to the potential ...


2

The mistake in your reasoning is that the mass elements that are closer to you (on the closer side from the center) exert more gravitational force than those that are far away. So it is unjustified to treat the mass distribution as a mass sitting at the center. Imagine for instance a linear mass of infinite length, then according to your intuition, the ...


2

Suppose you start with a proton and an electron separated by a large distance. The mass of this system is just $m_p + m_e$. Now let the proton and electron fall towards each other under their mututal electrostatic attraction. As they fall they will speed up, so by the time the proton and electron are about one hydrogen atom radius apart they're moving with ...


2

You are writing (apart from a minus sign) the tidal tensor, which is the traceless part of the Hessian of gravitational potential, for a point mass in the origin. This tensor contains information about the piece of gravitational force which can't be removed by choosing a free falling reference frame. I'm not sure to understand your problem, however the tidal ...


2

Let's consider a hover craft or a rocket that will accelerate up to a certain elevation and keep hovering at it, as a better vehicle for our purposes than a plane. Let's forget about air momentarily. What will happen is that the moment our air craft leaves contact with the ground, its initial tangential velocity will be the same tangential velocity of ...


2

First of all, it appears that you assume the centre of mass of the two-body system lies in the middle of the path between these bodies. This, however, isn't true. The centre of mass will lie closer to the star because that is where the majority of the mass is situated. If we define $r_M$ as the distance between the star and the c.o.m., and $r_m$ for the ...


1

It seems that you are mixing up the distance between the planet and the star and the distance between the star and the centre of mass (both are called $r$ in your derivation). Try to use $r = r_M + r_m$, where $r$ is the total distance between planet and star, $r_M$ is the distance between the star and the combined centre of mass,and $r_m$ is the distance ...


1

There has indeed been such research about the Pioneer anomaly: Two spacecrafts launched in the 1970s into the outer solar system did not move quite as expected (as calculated due to gravity and solar wind) after ca. 1980. Only in/after the 2000-2010 decade the source of the discrepancy, accidental thrust by thermal radiation, became generally accepted ...



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