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Another answer to stress that the expansion of the universe a) is happening at every point in space by all points receding from all points b)it is a very weak effective "force", which means it is observable between clusters of galaxies receding from each other. The gravitational energy of galaxies, as well of all their stars and planets allows them to ...


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The other answers address mainly the solar system's motion relative to our galaxy. To answer the question in your (time varying) title at the time I saw it: Does our solar system rotate parallel or perpendicular to the direction of the expansion of the universe? The answer is neither. There is no "direction" to the universe's expansion. In technical ...


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Assuming our solar system is at a 60degree angle in respect to the galactic plane, the planets would still not trail behind the sun like they do in the video. The theory of relativity(as well as our observation of the planets in our solar system) show that the sun's gravity acting on the planets are enough to form a flat plane. If you want to know why things ...


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Earth is about 100x more massive than the moon, and since $F \propto M / r^2 $, the distance from Earth to the astronaut would have to be about $\sqrt{100}$ = 10x further than from the moon to the astronaut. Therefore, the astronaut falls "up" about 90% of the way to the moon. [The earlier answers go a lot more into detail (and are more technically ...


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Does our solar system rotate parallel or perpendicular to the galactic disc? Neither. You have created a false dilemma. There are lots and lots of other possibilities; in fact, there are an uncountably infinite number of other possibilities. The ecliptic plane is inclined at about a 60 degree angle with respect to the galactic plane. With the possible ...


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The cosmological principle is that the universe is both homogeneous and isotropic on large scales -the last meaning that it looks the same in every direction. This principle has been extensively tested and holds up pretty well (there are some debatable anomalies in the cosmic microwave background that are suggested anisotropies, but they are small). ...


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The distance I got was 346 084km. Here are the maths I used: ($E_m$) Earth mass = $5.9736\times10^{24}$ kg ($M_m$) Moon mass = $7.3477\times10^{22}$ kg ($D_{em}$) average Earth-Moon distance = 384 467km ($G$) gravitational constant = $6.67384\times10^{-11}$ ($W$) my weight = 85kg ($D_{fe}$) distance from earth = ? The attraction force between two objects ...


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The main plot below shows the potential energy of a mass in the Earth-Moon system under the unrealistic assumption that the system is not rotating. i.e. This mirrors (at present) all but one of the 4 answers given, in assuming that this point is defined where the gravitational force on a mass due to the Earth and the Moon are equal and opposite (i.e. at the ...


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Set the forces on the test particle from the Earth and Moon equal: $$F_E=F_M$$ $$G\frac{M_EM_{\text{ test particle}}}{R_E^2}=G\frac{M_MM_{\text{ test particle}}}{R_M^2}$$ The $G$s and $M_{\text{ test particle}}$s cancel, leaving you with $$\frac{M_E}{R_E^2}=\frac{M_M}{R_M^2}$$ but you know that $R_M$, the distance between the test particle and the Moon, is ...


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To calculate this by yourself, you need to know that gravity force exerted on an object (for exapmle You) is equal to $F=GMm/r^2$, where $G$ is gravity constant, $M$ is the mass of the big object ($M_m$ for moon, $M_e$ for earth), $m$ is the mass of small object. $r$ is the distance from the center of the mass. Now you need to know masses of earth and moon ...


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At Lagrange point L1. Specifically for Earth-Moon L1, these calculations show 326054 km.


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Just use an equation derogating from the two forces which pull the objects (universal gravitation)to get the equilibrium point, something like (already simplifyed): M/d^2 = m/(384000000 - d)^2 Where M is the mass of earth, m the mass of moon and d the distance from earth. As d gets bigger than this value, you start falling into the moon I get a value of ...


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First off, we should take the wisdom of Jack Wisdom to heart: "Calculation of the history of the lunar orbit is fraught with difficulties." While calculating the history may well be fraught with difficulties, calculating the future is hugely problematic. For the sake of argument, I'll ignore that issue. The OP suggests the end of the planet is about 5 ...


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I think I caught that show on the science channel too and I think they represented that particular point poorly. First, I'm quite sure the sun already has a healthy amount of iron in it already, because all the inner planets do. There's no logical reason why the sun wouldn't. Iron is reasonably plentiful in the universe. What happens when a star's ...


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The responses so far seem to suggest that, assuming known density, we can tell how far away an object is by the strength of its gravitational attraction. I'm quite sure this is wrong. Since we are in free-fall about the sun (as we are about the moon) I don't think there is any direct way to measure the force of its attraction. It is however possible to ...


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For the two orbs, we use Newtons universal law of Gravitation: $$ F_{1\to2}=G\frac{m_1m_2}{r_{12}^2}\tag{1} $$ which does not require any information about densities, purely the total mass of the objects. The data we need are: $m_1=5.972\times10^{24}$ kg $m_2=7.34767309 \times 10^{22}$ kg (moon) $m_2=1.99\times10^{30}$ kg (sun) $r_{12}=384,400$ km ...


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The apparent diameter of an object like the Sun or Moon varies as $r^{-1}$, that is an object twice as far away looks half the size. So the nearly perfect fit during an eclipse tells us that the Earth-Moon distance $d_M$ and the Earth-Sun distance, $d_S$, are related to the radius of the Moon, $r_M$, and the radius of the Sun, $r_S$, by: $$ \frac{d_M}{d_S} ...


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A recent overview can be found in Pireaux et al (2001). I quote the authors on page 3: However, the perihelion shift of planets, and hence Mercury, can not be measured directly because the perihelion is a Keplerian element whereas the motions of the planets are not exactly Keplerian due to mutual gravitational interactions and figure effects. So, only ...


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This question has received some excellent responses. As the person asking seems keen to get a larger variety of responses, I am going to give this question another twist by enquiring about the maximum speed relative to Earth: Earth is a planet, which means it cleans its orbit around Sun from material objects. What is the maximum speed at which such ...


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A comet doesn't need to impact the sun in order to come very close to solar escape velocity at perihelion. There is a class of comets known as sungrazers that pass very close to the sun. Although small ones evaporate on their first pass near the sun, larger ones can survive several orbits, and be considered periodic comets. There is a class of sungrazing ...


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Kepler's Three Laws of Planetary Motion are particularly helpful when addressing this question. They state that (in informal language) The shape of a planet's orbit in an ellipse, with the Sun at one focus of the ellipse. As planets move around their elliptical orbits, the imaginary line drawn from the planet to the Sun sweeps out equal regions of equal ...


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Depending on what you are looking for, here are some possible candidates for the fastest bodies in the solar system: Comets from outside the solar system that fall into the Sun, just before impact Comets with a periodic elliptical orbit around the sun, at the moment of their closest approach to the sun Mercury, with a mean orbital velocity of 47.9 km/sec ...


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As I understand the question, comets (or any thing coming from outside the solar system) can not be considered. This leaves only the asteroids and other debris that still circle the sun at distance r. If this mass starts "falling" towards the sun, it will attain a velocity given by Pulsar's equation, if it is corrected by replacing the term (1/Rsun) with ...


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Many planets have been found where their orbital axes do not align with the rotation axis of their star. This is achieved using measurements of the Rossiter-McLaughlin effect in transiting systems or by observing planets transit over spotted features on a star's surface. As the stellar rotation axis is highly likely to coincide with its protoplanetary disk ...


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I think what you mean is - is it possible for a planetary system to exist such that the planets do not orbit in a single plane, but the planets have a large scatter of inclination angles? Our solar system has a relatively modest range, providing you ignore Pluto, of orbital inclination values (and eccentricities); zero to 7 degrees (Mercury). This is ...


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The asteroid "1566 Icarus" has a perihelion distance of 0.187 au and a semi-major axis of $a=1.078$ au, an orbital period of 1.119 years and eccentricity $e=0.827$. Using $$v_{\rm peri} = \sqrt{\frac{GM}{a}\frac{(1+e)}{(1-e)}},$$ where $M$ is a solar mass, then its fastest speed is 93.5 km/s. So this does not come close to Comet Lovejoy (mentioned in other ...


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We don't know, but there what I gather is the typical pattern of hypothesis, accretion disc theory, which attempts to find a model dynamic that could explain what we see, particularly in our own solar system, which is the only one we have very complete information about. In addition to seeing that our own solar system's planets all orbit in the same plane, ...


2

Your Brazilian Professor is partly right: the high temps at the surface of Venus are partly due to the high pressures at its surface. But that's not sufficient to explain the total temp. Other factors are that Venus is a lot closer to the sun than Earth is, and of course the greenhouse effect. As a thought experiment, if you took two planets that were ...


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Well, to clarify some things first In atmospheric science, or more correct: If you do the math... your only to free Variables are Density and Temperature. The equation of state which gives you the pressure, is a material property. The equations for the atmospheric variables are interconnected at any moment, it is nonsense to say P causes T or T causes P. ...


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(I have only read the first paragraph of the question) The pressure of Venus's atmosphere is about 90x greater than that on earth. It also happens to be about 90x more dense that that on earth. Coincidence? No. The density is the reason the pressure is so high. If you were to descend 1 km into one of our oceans, the pressure would be comparable to that of ...


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When there aren't comets falling into the sun, Mercury is hard to beat. This NASA fact sheet lists Mercury's orbital velocity around the sun as varying from $38.86$ to $58.98$ km/sec, not so much greater than Earth (less than a factor $2$, even at maximum).


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The fastest moving object that does not get destroyed by crashing into the sun would be the apollo asteroids that get very close to the sun. For example Icarus gets going pretty fast at perihelion, (0.18665203 AU from the sun) at just under 100 km/sec.


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The maximum speed of an object that orbits the Sun at a certain distance $r$ is known as the escape velocity: $$ v_\text{esc} = \sqrt{\frac{2GM_\odot}{r}}, $$ where $M_\odot$ is the mass of the Sun. If the object would have a greater speed, it would eventually leave the solar system. So I'd say that the absolute maximum possible speed of any object in the ...


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What book on Voyager were you referencing? In any case, they started developing the Voyager spacecraft in the 1960's. The first measurement that a solar wind even existed wasn't reported until 1960-1961 [K. Gringauz using the Lunik 2 spacecraft]. They were able to determine a flux of particles (i.e., number per area per time), but did not determine speed ...


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Some texts indicate that a large hunk of matter crashed into the earth and splashed to its orbit the material composing our moon. They also suggest this event could be one of the mean reasons for earth's axis tilt.



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