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16

You're right that the Sun being 4.5 billion years old makes observations difficult. The Sun goes around the Galaxy about once every 225 million years, so since the Sun formed it has gone around the Galaxy perhaps 20 times. The trouble is that the Galaxy is not like the Solar System: stars don't go around on nice nearly circular orbits, everything is a bit ...


11

You certainly could define your origin of coordinates to be the center of the Earth. It would be a little tricky, because this would no longer be an inertial frame of reference, so there would be fictitious forces (or Coriolis forces). That is, your equations of motion would no longer look the same. One reason the standard barycenter frame of reference is ...


10

Back before Copernicus (Or rather, before his view was accepted), we used to think the earth was the center of our solar system. Therefore, if you search for those models, you can find examples such as: This is, of course, based on observations rather than calculations, but it represents the complication of the solution nonetheless. (Image taken from ...


8

You seem to be groping towards the fact that the gravitational three-body problem is, in general, not solvable. We can get away with saying "the sun is at one focus of an elliptical orbit" in the solar system because the sun is so much larger than anything else around. The sun is 1000 times more massive than Jupiter, so the sun-Jupiter barycenter is about ...


5

Good answer from Kyle. I will just add that there is a great deal of effort going into trying to discover "solar twins". These are stars with such similar parameters (including age inferred from the HR diagram or asteroseismology, which can be good to about 10% in the best cases) and photospheric compositions to the Sun, that it is thought likely they must ...


5

We only have precise measurements of the magnetic field for one star i.e. the Sun. It turns out the Sun's magnetic field flips every eleven years i.e. the North and South poles switch over. This means there is no connection between the Sun's magnetic North and the orbits of the planets. Since we have no reason to suppose the Sun is special this means it is ...


3

I suggest that it doesn't make much sense to say that the planets orbit the barycenter of the solar system. Beware that you are going very much against the grain of the best models of the solar system in writing that. All three of the leading ephemeris models (JPL's Development Ephemeris, the Russian Institute for Applied Astronomy's Ephemerides of the ...


3

Since a new moon has to have the sun shine on the entire surface facing away from the Earth, the time between 2 new moons is the time for the Earth Moon system to complete a synodic period. The period of two masses around the common center of mass is always the same.


2

At L4 and L5, the object would stay in the same relative position because it is in the same orbit around the Sun as Earth. Any other points on the circle would not be in the same orbit as Earth. They would be in orbits of differing inclination and so, in the course of the orbit of that object about the Sun, it would necessarily leave those points. The L4 and ...


1

This circle is the set of locations where the gravity of the sun and earth cancel out. Uhm. No. They are neither co-linear nor the same magnitude. To get a zero "force" at the Lagrange points you have to work in a rotating frame of reference, implying a centrifugal pseudo-force which closes the triangle. But the points on the circle you defined other ...


1

In general, polymers are not a good idea as atomic oxygen and charged particles will eat them up. High density polyethylene, however, may form a protective oxidative coating against atomic oxygen (source). I would argue that it's still worth depositing a thin film of a metal that forms a nice protective oxide layer, e.g. Al, which would also prevent charge ...


1

For an Hohmann transfer you want the phase angle to be correct, namely you want to aim for where planet 2 is going to be. You cover an angle of $\pi$ radians, starting as planet 1, which takes $T_H$ seconds, so planet 2 will have moved $T_H n_2$ radians, so the phase angle, $\Delta\theta$, between the two planets at the start of the transfer should be, $$ ...


1

For the sake of simplicity assume the binaries' motion to be circular. Then you can use the formalism of the circular restricted 3-body problem (CR3BP) to model the motion of the test particle $m_3$. Your Lagrangian will be time-independent and the conserved quantity (Jacobi constant) can be evaluated at infinity to give an equation for conditions of escape, ...


1

I agree with everything John Rennie said but let me just take a slightly different direction. Note that Schwarzschild space-time has a time-like Killing field $\xi^{\mu}$. In a stationary space-time such as this, one can define the Newtonian analog of gravitational potential by $\phi \equiv \frac{1}{2}\log(-\xi_{\mu}\xi^{\mu})$. One can then easily show ...


1

But we also know that a perpendicular force always causes an acceleration according to the rule of addition of forces. All forces cause acceleration. Perhaps you mean specifically tangential acceleration (changes in speed)? If the centripetal force is greater the resulting vector is near the perpendicular, if the centripetal force is in perfect ...


1

In a reference frame where the center of mass of the Earth-Sun system is at rest, we will have $$ m_\text{Sun} \vec{v}_\text{Sun} + m_\text{Earth} \vec{v}_\text{Earth} = 0 \quad \Rightarrow \quad \vec{v}_\text{Sun} = - \frac{m_\text{Earth}}{m_\text{Sun}} \vec{v}_\text{Earth}. $$ In particular, if this is true at some initial time, then it's true at all ...


1

Here's how I demonstrated this concept to my son when he was younger. Take a plastic bottle and put some pebbles or little toys in it. Then toss it in the air and catch it. If you look in mid-flight, you can see the little toys just floating around inside the bottle. But they're still at 1G. And when you look at this example, it's totally obvious what's ...


1

Fortunately, you don't need any special relativity or general relativity. All you need to describe orbital motions (to a high enough degree for NASA to use) is classical mechanics. The key in classical mechanics is that rotating reference frames are not inertial. The laws of physics as you know them only hold in an inertial frame. Over time scales of a ...


1

The Earth and the Moon revolve about their barycenter (center of mass of the Earth/Moon system), which is inside the Earth and is about 4,670 Km from the center of the Earth on a line connecting the center of the Earth with the center of the Moon (http://en.wikipedia.org/wiki/Barycentric_coordinates_(astronomy)). The apocenter of the Moon is its center when ...


1

You can use other parameters, which define an orbit, such as angular momentum, $$ h = \omega r^2 = \sqrt{GMa(1-e^2)}, $$ where $\omega$ is the angular velocity, thus $\dot{\nu}$. By rewriting this equation to $\omega$ and substituting in the expression for $r(\nu)$ you obtain, $$ \omega(\nu) = \sqrt{\frac{GM}{a^3(1-e^2)^3}}(1+e\cos(\nu))^2. $$ The ...


1

The polar angle of the plane (the angle of the plane with respect to the symmetry axis) should relate to the eccentricity of the orbit. For $90^o$, the section is a circle and the eccentricity is zero. For an angle between the $90^o$ and the angle of the cone, you will get an ellipse with an eccentricity $0<\epsilon<1$. In both of these cases, the ...



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