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Finding the potential radii is actually quite simple. I already have: $$V_{eff}(r) = \frac{L^2}{2mr^2}-\frac{GMm}{r} - \frac{GMmL^2}{r^3(mc)^2} $$ I was mistakenly graphing $1/r^2 - 1/r - 1/r^3$, when in actuality it makes more sense to take $1/r^2 - 1/r - 0.1/r^3$, since $1/r^3$ is sufficiently small. Since the orbits are circular, the potential radii ...


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Because gravitational force increases with proximity and mass, so areas of higher density and/or elevation exert a greater pull on a satellite, creating what could be called a "speed bump", causing the orbiting body to be slowed/pulled down every time it passes that point, if only a little bit. Also, when you orbit something you are countering its net ...


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Because the moon, like many planetary bodies, is not homogenous in its composition and doesn't express an absolutely even gravitational pull from every side. When you orbit a body, you are countering the gravitational pull of its entiriety. When you come over a spot with higher density and/or proximity, you are pulled more due to the fact that ...


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Your equation has the form $$ V_\text{eff}(r) = \frac{\alpha}{r^2} - \frac{\beta}{r} - \frac{\gamma}{r^3} $$ If you set $\alpha=\beta=\gamma=1$, then you're overestimating the $r^{-3}$ term, which is supposed to be a small correction. You will only find two extrema if the derivative has two roots: $$ V_\text{eff}'(r) = -\frac{2\alpha}{r^3} + ...


2

It's close enough to the earth to be well within the range of solar eclipses. There's always going to be a solar eclipse somewhere in space because the moon will always cast a shadow behind it, well, except for when the moon is eclipsed by the earth. But the moon's shadow passes over the earth just a small percentage of the time. Given that the space ...


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Try to answer these. In equilibrium what balances gravitational force and centrifugal force? What is the expression for kinetic energy? What is the time period of a body in uniform circular motion with given velocity and radius of circle?


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As pointed out in one of the comments, you have already declared the mass $M$ (in the first case) to be fixed. Asking to prove that it indeed is fixed rather than moving is like asking to prove a definition: No need, it is already a given by power of your mind as author. I suspect you are seeming some sort of paradox. There is none: The two cases you ...


0

The first case is more of a thought experiment, since gravity would move the mass M. To apply it no real life, something must be attached to M. This attached thing would move to so it must be attached to something else. This goes on ad infinitum. This is why a ball gravitationally attracts a another ball on the earth, but also the earth. Situation 1 is ...


4

Why does no object end up at the center of their circular paths? If the path is circular then, by definition, the particle maintains a constant distance from the center of the path. Perhaps you're asking why the particle has a circular path? Would someone have a rigorous proof that a constant centerwards force produces a circular orbit One ...


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It is possible to find $t(r)$ and $t(\theta)$ easily, however inverting to find $r(t)$ and $\theta(t)$ is hard to do in general (unless you use Fourier series etc..) \begin{align*} t(r) &= \sqrt{\frac{m}{2}} \int \frac{dr}{\sqrt{\frac{k}{r} - \frac{l^2}{2mr^2} + E}} \\ t(\theta) &= \frac{l^3}{mk^2} \int \frac{d\theta}{\left( 1 + e \text{cos }(\theta ...


1

But I am wondering if it is possible to solve the two equation of r with respect to t and θ with respect to t. Yes, you can. In the case of an elliptical orbit with non-zero angular momentum, you need to introduce the concepts of eccentric anomaly, mean anomaly, and mean motion. Eccentric anomaly is related to true anomaly (your theta) via ...


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A great amount of planetary data is available at this NASA website. There doesn't appear to be a strong correlation, but you could use this NASA data to do the statistics.


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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 ...


0

If they both were in orbit, then they would already be accelerating down towards the earth, so there wouldn't be anything sudden about it - they have the exact same acceleration before and after impact.


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To a first approximation the spacetime curvature around the Sun is indeed spherically symmetric. I say to a first approximation because the masses of the planets (particularly Jupiter) also produce curvature and this breaks the spherical symmetry. However let's ignore this for now because I don't think it's relevant to your question. If I understand you ...


1

I'm eggrobin, I happened across this while doing research for my mod. Regarding the question: The simulations by Matt Roesle that showed Vall being ejected from the Jool system were computed by interpreting KSP's orbital elements as body-centric, which means that when the system was integrated, the bodies ended up in significantly different orbits around ...


1

In respect to above answer, although you are correct but there is a little discrepancy in your last line " a bigger leap was to try the Sun's location at a focus and not the center." A circle is a ellipsoid with a special case having the two foci at same point, so there could not be so chance putting in centre because it was observed that planets near ...


2

In principle in Newtonian mechanics the rest frame can be any of the bodies in a gravitational complex. The geocentric system is one possible rest frame and a one to one mathematical transformation exists going from a heliocentric to a geocentric system. It is when one introduces the concept gravitation, a theory that explains the orbits, that the ...


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I feel that in general there is a lot of confusion around what black holes are and what they do. in reality, it only gets very different from other objects such as stars very close to or inside the event horizon; in most circumstances it is simply a heavy object. therefore objects would simply orbit around it like any other massive object, as long as the ...


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In the comments you mention Susskind's use of a metaphor involving water flow 7 minutes into this video, but this shouldn't be understood in terms of spacetime behaving fundamentally differently around a black hole as opposed to any other gravitating body. Rather, I suspect Susskind is just referring to the analysis of a black hole in a particular type of ...


1

For systems involving central forces, the orbit can be any of the conic sections. And which conic it will be is determined by the total energy of objects revolving around the centre of force. If the total energy is negative ( as in the case of planets), the eccentricity of the orbit will be either zero or one. If eccentricity is equal to zero, it will be a ...


0

At the moment you are on the physics part of this site. As I'm sure you know, some computer languages are written especially for solving maths problems and nothing else. If you also posted your question on this site http://scicomp.stackexchange.com/ I think you will find someone who can help you. There is probably code already written that might solve ...


1

The simplest solution is to find the center of mass of the two objects - at any moment in time, if the stars are a distance $d$ apart, and their respective masses as $m_1$ and $m_2$, then the center of mass is found at a distance $x$ from $m_1$ where $$x = d \frac{m_2}{m_1+m_2}$$ From this you can see that if $m_1 >> m_2$, the center of mass will be ...


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I found in an italian book (Barone) that using special relativity precession is 1/6 of the 43''/century observed. The book has merit to treat an argument usually ignored (orbit precession in special relativity) but I find the treatment very concise so I tried to be more explicit and I connect result to a specific initial condition. I find delightful this ...


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Comets are coming from almost infinite distance towards Sun. Due to Sun's Gravitational force they should be pulled into Sun rather than forming Elliptical or Hyperbolic trajectory. If object is starting travel from closer distance at considerable speed, we can visualize formation of Elliptical orbit. For object coming from infinite distance getting ...


3

The Moon's orbit would be nearly Keplerian were it not for the perturbing effects of the Sun. The time from perigee to perigee and from apogee to apogee wouldn't change, and the time from perigee to apogee would be exactly half the orbital period. What you are seeing are perturbing effects of the Sun on the Moon's orbit. If you use push the site you found, ...


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I guess you are asking how the process of rotation has come about in the first place. Cosmologists believe that many billions of years ago, our solar system was nothing but a big cloud of dust and gas, which resulted from the explosion of nearby stars. We can deduce this because right now, at this very instance, astronomers are observing how new solar ...


0

Firstly, there is no centrifugal force here. Centrifugal force is a fictitious force that you observe if you are rotating, in this case we don't need to have a rotating observer, so there's no centrifugal force. The only force acting on the planets is a centripetal (inwards) one, gravity. Now, it's true that if the planets were stationary, then the ...


1

Interesting question. The earth/moon system around the sun certainly changes speed as the earth gets closer to are farther from the sun, but the Earth/Moon orbit the sun together so the direct effect on the Moons orbit regarding the earth's Apogee and Perigee would probobly be small. I think a larger effect is the tidal effects on the moon's orbit around ...


3

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 ...


0

There are at least 8 more planets in the solar system, besides the Earth (and some more were discovered). When Copernicus decided to place the Sun in the center of the solar system, instead of the Earth, that was mainly because this arrangement simplified drastically the form of the orbits of the other planets. With the Geocentric model of the solar system ...


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.


0

I have an answer, but I was hoping to see confirmation from another source. I used the centripetal force formula to come up with: $$ v = \sqrt{Gm/r-ar} $$ Where v is the tangential velocity of the orbit and a is the radial acceleration away from Earth. Is this right?


1

Since the force is radial, you are not changing the angular momentum, but you are adding potential energy: this tells you what must happen to the tangential velocity (decreases) and radial velocity (increases). I will leave it up to you to figure out by how much.


1

No, that's not possible. Even if the two bodies could be compressed to be just larger than their Schwarzschild radius (they can't really, without collapsing further to black holes), their combined Schwarzschild radius, which grows linearly with mass, is twice their individual Schwarzschild radii. That means that even if they effectively rolled on each other ...



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