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5

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 ...


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

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 ...


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, ...


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.


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 ...


3

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 ...


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 ...


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 ...


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 ...


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 ...


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 ...


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 ...


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

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 ...


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 ...


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.



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