# Tag Info

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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 ...

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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 ...

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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 ...

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If the centripetal force is greater [than the tangential force] the resulting [force] vector is near the perpendicular [or centripetal] [force], if the centripetal force is in perfect balance [with the tangential force] the resulting [force] vector should point exactly at 45°. Can you please explain why this doesn't apply to an orbiting body? Who ...

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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 ...

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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 ...

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Your question doesn't have an answer because it isn't possible to split energy into a potential part and a kinetic part, or at least not in an observer independant way. The Schwarzschild metric is time independant and spherically symmetric, and these symmetries mean there are two conserved quantities that you can think of as total energy, $E$, and angular ...

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When Orbit 2 makes rendezvous with Orbit 1, it is travelling more slowly than Orbit 1 (which is why, if it misses rendezvous, it will then drop to a lower altitude and reenter). So there are two possibilities for the rendezvous. 1) Just before joining up with #1, #2 fires rockets to gain speed to match with #1. In this case, the two conjoined ships will ...

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I think the previous answers misunderstood the question. It's rather obvious that a balloon cannot have buoyancy in vacuum, and therefore cannot float into vacuum at steady speed. However, one needs to also consider the possibility that a balloon would achieve enough speed in lower atmosphere to bring it to 2nd orbital velocity. Imagine a huge light ...

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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, ...

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I can't help but think: "Why would you want to do that? Using a bomb to deflect an asteroid seems to me such a waste of energy. We know already how to land on a comet (philae shown that it can work), so if you ignore the way to "get" to the comet (which is quite frankly the hardest problem, not the deflection itself). Why would you try to blow it away? ...

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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, $$... 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 ... 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 ... 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 ... 0 Orbital velocity is measured relative to the object that it is orbiting. If you went at the same orbital velocity to observe something, while I would see to have no speed, neither would you (in your own reference frame) and therefore there is nothing paradoxical about it. 0 What you are basically asking is how to convert initial conditions, position and velocity, to orbital elements. In this case both position and velocity are 2D vectors, with a reference frame positioned at the center of the celestial body, with gravitational parameter \mu, and the orbital elements: "longitude of the ascending node" and "inclination" do not ... 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 ... 0 Your problem is that you are trying to do this in Cartesian coordinates (x,y). This makes the math much harder than it needs to be. It is more natural, when dealing with circular motion, to use polar coordinates - we express a position relative to the origin by its distance (r) and the angle relative to some reference axis (\theta). The relationship ... 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 ... 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 ... 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 ... 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 ... 0 Ellipses, parabolas and hyperbolas are both: Defined as the conic sections you speak of: work out the intersection between the cone \vec{R}.(\cos\phi\,\hat{X} + \sin\phi\,\hat{Z}) = \sin\theta\,|\vec{R}| and the plane z=const where \theta is your polar angle and \phi the angle between the cone's axis of symmetry and the slicing plane and you'll ... 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 ... 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 ...

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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 ...

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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.

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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 ...

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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 ...

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