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3

Now imagine that the tower is a little bit higher, and an object is released from the top of the tower. The object will float away and "climb" higher. Does the object move slower or faster around the world than the tower? Slower. Provided that the space elevator is geo-stationary, the higher you go above geosynchronous orbit, the greater your velocity ...


2

If you are asking the question "How high does a person have to be in order to be "weightless" because gravity is canceled from the rotation of the Earth?" as paraphrased by CoilKid, then what you are talking about is a Geostationary Orbit. You can only achieve this orbit in a equatorial plane for the orbitting object to seem truly stationary with respect to ...


2

Think about this. If gravity is of equal magnitude as the centrifugal force, a satellite there will have an angular velocity same as the earth rotating angular velocity. However, this is just a geosynchronous satellite! The equation is $$\frac{GMm}{r^2}=mr\frac{4\pi^2}{T^2}$$ where $T=24h$. Hope it helps!


1

It is incorrect to think that Newtonian gravity seems to dictate that we can treat all masses as point masses. Indeed, you can't do that, and one of the consequences of that fact is the orbital decay that you refer to. The theorem you're thinking of holds only for bodies with perfect spherical symmetry, and states that for such a body, the ...


0

Newtonian gravity seems to dictate that we can treat all masses as point masses. This is the source of your confusion. This is not true. One can treat an object with a spherically symmetric mass distribution (i.e., density is a function of distance from the center of the object) as if were a point mass for points outside of the object in question. ...


0

Point masses are an approximation that describe the motion of the center of mass of objects which do not have their rotational and internal vibrational degrees of freedom excited. If that assumption is incorrect, then so is the treatment of Newtonian problems by using coordinates of centers of mass alone. Orbital decay can only occur if both energy is lost ...


0

A very late answer, one that I hope adds to the excellent answers by Mark and Luboš. From the perspective of Newtonian mechanics, there's nothing wrong per se with using a geocentric point of view. Such a point of view does require adding fictitious forces and torques that would otherwise be absent in an inertial perspective, but if makes sense to do that, ...


4

How far ahead can we predict solar and lunar eclipses? NASA has uncertainty calculations that show how certain we are about when eclipses happen. From a back of the envelope, the eclipses will likely vary by a full day 35 thousand years from now. That said, we have eclipse seasons, so we know eclipses will continue to happen, and at roughly which time of ...


42

On predicting planetary orbits A number of studies have shown that the inner solar system is chaotic, with a Lyapunov time scale of about 5 million years. This 5 million year time scale means that while one can somewhat reasonably create a planetary ephemeris (a time-based catalog of where the planets were / will be) that spans from 10 million years into ...


0

Look I know link only answers are terrible, but I don't want to ruin the surprise. Check out this link. That's on Wikipedia! It's safe to say that if wikipedia knows something, the experts know quite a bit more. We know the dynamics of the Sun-Earth-Moon system really well including the perturbations and the most likely sources of future upsets, so barring ...


-2

Given that we exist at all, we can confidently back-track the positions of the planets (and our moon) for a couple billion years. I see no reason, barring rather massive exo-system-sourced objects showing up unexpectedly, that the positions will go chaotic enough to be unpredictable any time in the next couple billion years. I suppose it might depend a bit ...


0

I would agree that a circular orbit with full tidal lock is more "in equilibrium" (everything is time independent in the rotating frame) than one in which the orbiting object had zero rotation in an absolute Cartesian reference. There is however an important notion for determining whether something is rotating, and that is its angular momentum. An object ...


0

You want to know the ratio of the speeds, i.e. $v_{neptune}/v_{earth}$. You can just divide the two equations you already have. Now, what is the ratio of the radii of the orbits? Your equation should tell you how to relate this to the orbital speeds.


2

Are retrograde capture orbits “easier” than prograde capture orbits? The answer is not just yes, but a rather emphatic yes. This is why the irregular moons of Jupiter predominantly have retrograde orbits, and why all of the outer moons of Jupiter have retrograde orbits. This is also why NASA has been interested in capturing an asteroid and putting into ...


2

Well, you have a spherically symmetric potential, so you are going to have some integrals of motion. Since you have a potential force, your total energy is going to be conserved. The other integral is angular momentum $\vec{L}$. You can check it by differentiating it's definition in time and using things you already know. Since angular momentum is not ...


4

Suppose the Earth was stationary, and the Moon revolved around the centre of the Earth: If $v$ is the orbital velocity of the Moon then at point $A$ the linear momentum of the Moon is $mv$. Half an orbit later, at point $B$, the velocity of the Moon is $-v$, because it's in the opposite direction, so the momentum of the Moon is $-mv$. If the Earth is ...


0

Maybe it's best to forget space curvature as described in general relativity; while that theory implies major changes to the foundations of celestial mechanics, it is set up in a way that reproduces Newtonian mechanics when gravitational fields are of moderate strength, and leads to only very tiny adjustments of the actual orbit calculations. Movement of an ...


5

If the radius of the torus is $r$ and the tangential velocity is $v$ then the acceleration towards the centre is: $$ a = \frac{v^2}{r} \tag{1} $$ So just set $a = 9.81$ m/sec$^2$ and you have your answer. If you prefer to have the answer in terms of the rotation rate then use the equation $v = r\omega$ to rewrite equation (1) as: $$ a = \omega^2r $$ ...


5

Good question! Many astronomers think that the moons of Mars, Phobos and Deimos, are captured asteroids. Others object precisely because of the issues that you raised. Capture is not easy. Sans a collision, capture is impossible in the Newtonian two body problem. A hyperbolic trajectory stays hyperbolic. On the other hand capture in the multi body problem ...


1

Physically, how can it be that tidal friction on Earth makes the Moon do something? I know it is because conservation of angular momentum, No, conservation of angular momentum alone can't predict that one object will lose angular momentum and another will gain. It would be equally consistent with conservation of angular momentum if both stayed the same. ...


4

The only way to measure "motion" of the moon is relative to objects in its (close) vicinity. In your case, the only objects might be stars, or clouds. Clouds are a terrible reference object; and if there were any optical effects in the sky, the stars would move in concert with the moon. Our ability to look at an object in the distance without any reference, ...


1

Let's look at what happens if we put numbers into your equations to see if there actually is a violation of the conservation of energy. Using numbers for Earth from wikipedia we have a velocity of $29.78*10^3\frac{m}{s}$ for orbital velocity and a mass of $5.972*10^{24}kg$. For say the Voyager 2 we have mass of about 730 kg and a speed of ...


1

I read that the KE a free-falling ball acquires is not originated by the attracting body but that energy was actually stored in the ball when it had been lifted to the height it dropped from. In this way, it was said, gravity is subject to the conservation of energy principle and cannot change the total energy of an object. Consider now the ...


33

Energy is in fact conserved, even in gravitational slingshots. After the slingshot, the velocity of the spacecraft may indeed change, which means its kinetic energy will also change. If this happens, the energy increase (or decrease) will be made up by a commensurate decrease (or increase) in the kinetic energy of the planet. In plain English: The planet ...


84

Cory, here's a different way of thinking about gravity assists that may help: First is my short answer for readers in a hurry: What is really going on is a giant game of pool, with fast-moving planets acting as massive cue balls that impart some of their energy when they whack into tiny spacecraft. Since you can't bounce a spacecraft directly off the ...


7

Gravity assists don't change speed in the two body problem. An object approaching a lone gravitating body will enter and leave the vicinity of that body with exactly the same speed. All that a lone gravitating body can do is change the direction in which the object is heading. The body that provides the assist needs to be moving with respect to the target ...


6

Wouldn't the centrifugal/-pedal forces between the planet and Sun make the planet revolve around the Sun in a perfect circle? Is there a 'third force' that attracts planets at a specific point (a hypothetical reasoning for elliptical orbits)? I need to rant a bit, not at you, Hassaan Qazi, but at whoever taught you (and who teach so many others) that ...


2

When looking at the mathematics of Kepler orbits (a two body problem), elliptical orbits are just as stable and periodic as circular orbits. And an elliptical orbits will not be pulled into an circular orbit. When only considering Newtonian gravity, and there are no perturbations by other bodies, an elliptical orbit will not change its trajectory over time. ...


1

why don't planets revolve in the special ellipse known as a circle? You've answered your own question: the circle is special and thus, highly improbable. In fact, a perfectly circular orbit is essentially impossible and is, in fact, an idealization that can only be approximated. Think about it - a genuine circular orbit is like a genuine sinusoidal ...


3

Consider a planet near a star at one moment in time. They are separated by a certain distance $r$, and so the planet feels the corresponding $GM_\mathrm{star}/r^2$ acceleration toward the star. That's due to the only force on the system. But of course velocity is not necessarily in the direction of acceleration; acceleration merely acts to change the ...


2

Generally, if they start off as circular, and the planet is disturbed in its orbit (by passingt planets or commets), they come to be elliptic. One should note that stable elliptic orbits are a feature of three dimensions only. They don't work in four or higher dimensions.


0

So after a bit of research, i came up with this answer. As i had said before we know that in orbit, the mass of the satellites does not matter to find the velocity. so if we use: $$v = \sqrt{GM/R}$$ as M being the mass of earth i can find the final velocity of both satellites after collation. $$Vf = (m2v2 - m1v1)/ (m1 + m2) $$ I came up with 4407.87 m/s If ...


0

Luboš Motl addressed the reasoning very well, but for the mathematics: Richard Fitzpatrick's free e-book A Modern Almagest: An Updated Version of Ptolemy’s Model of the Solar System presents a modernized version of Ptolemy's Almagest. The following classic paper shows how epicyclic astronomy can be re-expressed in the modern mathematical idiom of complex ...


1

We have to make a few assumptions here, because your question as posed is a bit incomplete. 1) collision is inelastic 2) satellites don't disintegrate on collision (see user58220's interesting observation) 3) there is no atmosphere (so drag is not an issue - let's see if the satellites end up in an orbit above the planet's surface) 4) you said "crash into ...


1

Introducing a side issue to this question: Using $v$ for the orbital velocity at $1000$ km altitude, the total kinetic energy of the two satellites just before the collision is $$KE_{\text{Before}}=\frac12400v^2+\frac12100v^2=250v^2$$and using the final velocity of John Rennie above, the total kinetic energy of the combined satellites just after the ...


5

Unless I'm missing an easy way to do this problem it seems a surprisingly hard one. This diagram shows the problem (I've exaggerated the altitude of the satellite to make the diagram clearer): The satellites are in circular orbits (dotted line) at a distance $r$ from the centre of the Earth, so their orbital velocity as (as you say): $$ v = ...


0

I can only assume that the satellites are assumed to meet head-on (rather than at an angle), and that the collision is modeled as inelastic (rather than an explosive collision where pieces fly everywhere). If so, you have the velocity of each and the mass of each. You can use the conservation of momentum to find the new velocity of the combined-mass object ...



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