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You've used the gravitational constant with only three significant digits. So it's no surprise that your answer isn't accurate to five significant digits. Instead of $G$ and $M_\odot$ separately, you should use the product $GM_\odot$, known as the standard gravitational parameter. Its value is known very accurately: in the link, you'll find $$GM_\odot = ... 20 The effective gravity inside the ISS is very close to zero, because the station is in free fall. The effective gravity is a combination of gravity and acceleration. If you're standing on the surface of the Earth, you feel gravity (1g, 9.8 m/s2) because you're not in free fall. Your feet press down against the ground, and the ground presses up against your ... 18 This is not possible. The lowest possible mass for a main sequence star (sustaining H-1 fusion; it's the regular kind of star) is around 80 Jupiter masses. Just below this, objects are referred to as Brown Dwarfs, which are technically not stars. Whereas the highest possible mass for a terrestrial planet is about 5-10 Earth masses (as per here). Above this ... 16 This web page has a nice discussion on it: http://archive.ncsa.illinois.edu/Cyberia/NumRel/EinsteinTest.html Basically the orbit's eccentricity would precess around the sun. Classical stellar mechanics (or Newtonian gravity) couldn't account for all of that. It basically had to do with (and forgive my crude wording) the sun dragging the fabric of ... 15 To some extent the universe exhibits something called self-organized criticality where a dynamic, non-linear system with many degrees of freedom (the gas after the Big Bang but before the emergence of structure) eventually forms a system with a notable degree of scale invariance (moons orbiting planets, planets orbiting stars, stars orbiting galactic ... 14 This is not that there is no exact solution, only the exact solutions for x(t) and y(t) use elliptic functions. The problem whether elliptic functions (which are defined by inverse of some integrals) are "good" functions is a bit philosophical one; one can on one hand state that sine is not a real function because one must integrate or sum a infinite ... 14 The excitement behind various claims is somewhat excessive. First, the Mayan astronomers, see e.g. Mayan astronomy at this page, didn't use any armillary spheres or sextants as others did. Their observations were made with naked eye and they were depicting positions of planets with crosses. The accuracy of the Venus' position after a synodic 584-day cycle ... 13 The Moon moves at about a thousand metres per second, but it's a long way away so it only appears to move slowly. Most of the apparent movement of the Moon is actually due to the rotation of the Earth. We see it appearing to go round the Earth once a day, but it actually takes about 28 days to complete an orbit. The Wikipedia article on the Moon's orbit has ... 13 First, you state a few things that aren't quite right in your question. While the view that's generally talked about is that Phobos and Deimos are likely captured asteroids, dynamically it's a pretty difficult problem (you generally need a third (in this case fourth?) body to take away the extra energy, and it's hard to get a circular orbit around the ... 13 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 ... 12 Set the forces on the test particle from the Earth and Moon equal:$$F_E=F_MG\frac{M_EM_{\text{ test particle}}}{R_E^2}=G\frac{M_MM_{\text{ test particle}}}{R_M^2}$$The Gs and M_{\text{ test particle}}s cancel, leaving you with$$\frac{M_E}{R_E^2}=\frac{M_M}{R_M^2}$$but you know that R_M, the distance between the test particle and the Moon, is ... 11 To expand on Prahar's answer, let me run some numbers to try and convince you this is reasonable. Your answer is correct to within one part in 104:$$ \frac{365.256363004}{365.2075}\approx 1.000133795. $$The main perturbing influence on Earth's orbit is the gravitational pull of Jupiter, whose mass is about 1000 times smaller than the Sun, and which orbits ... 11 Is it possible for a star to have the same mass and radius as e.g. the Moon and orbit a planet like Earth at the same distance (at which Moon orbits Earth in actuality)? No. The lowest mass type of star is a Brown Dwarf, which still has a mass greater than that of Jupiter. Even brown dwarfs have too little mass to fuse light hydrogen. Neutron stars ... 11 It's possible, but it seems like it'd be rare. The planets with the most moons are giant, and very far away from the sun. That means the moons will be very strongly bound to the planet and not get disturbed much by the sun. If our moon had a moon, it'd have to be just the right distance from it that it wouldn't collide with it (the moon's gravity is far from ... 11 The horseshoe orbit shape does occur only in the reference frame of the Earth’s orbit. It is a manifestation of a third body problem, and the orbit is in an accelerated reference frame. The loop, which is this distended horseshoe shape, has no central gravitational source inside the loop. As a result the orbit is a “pseudo-orbit.” From the perspective of an ... 11 There are two elements to why the universe appears to be so orderly: the physical laws of that govern the universe are the same everywhere, and astronomical objects are very, very, very far from each other. Consider two objects, one much larger than the other, and both very far from anything else. Because of gravity (which works the same everywhere), the ... 10 The spiral arms don't mean that the mass is getting sucked to the center. They're just wave-like density patterns. The bodies in orbit around the center of the galaxy are in stable orbit; just like the Earth around the Sun and the Moon around the Earth. What happens is that gravity accounts for the centripetal force (in the orbiting frame, gravity is ... 10 Kepler's 3rd law assumes that the Earth travels in a perfect ellipse with the only gravitational force on it being from the Sun. Further, Kepler's laws are derived from Newtonian gravitation. In reality, the orbit of the Earth is affected by the gravitational pull of other planets, and by the effects of General Relativity and is therefore not quite ... 9 In his lecture The Cosmic Distance Ladder (video), mathematician Terence Tao describes the history of how mankind has successfully mapped the solar system and beyond. In particular, he describes why Copernicus put the sun in the center (reason: he discovered that the sun is dozens of times bigger than the earth) and how Kepler found his laws of planetary ... 9 No "large bodies" that I know of. Certainly it is physically possible for something to orbit a moon; lots of spacecraft have been orbited around the Moon and other moons in the solar system. As long as we're simply discussing hierarchies of orbits, the Sun orbits the galactic center and the Milky Way is gravitationally bound to the Local Group. [EDIT: ... 8 The force you experience is of the form \vec{F} = - Gmr\vec{u_r}, and we also know that in the surface, r=R, it is \vec{F}=- gm\vec{u_r}, so$$\vec{F} = -gm\frac{r}{R}\vec{u_r}$$This is a conservative force that can be derived from a potential$$U = \frac{1}{2}gm\frac{r^2}{R}$$Because this is a central force, angular momentum will be conserved, ... 7 Mercury's orbit is elliptical. The orientation of this ellipse's long axis slowly rotates around the sun. This process is known as the "precession of the perihelion of Mercury" in astronomical jargon. It's a total of 5600 arcseconds of rotation per century. The precession is mostly a result of totally classical behavior; almost all of the movement of the ... 7 The main problem with being near a large planet are the tidal forces produced by its gravity. The Moon is pretty small compared to the Earth, and it's a quarter of a million miles away, but it still produces large movements in sea level i.e. the tides. If you imagine the moon getting bigger and closer the tidal effects would increase. The land feels the same ... 7 This was previously a comment to space_cadet's answer but became long (down-vote wasn't me though). I don't understand space_cadet's talk about unstable orbits. Recall that two-body system with Coulomb interaction has an additional SO(3) symmetry and has a conserved Laplace-Runge-Lenz vector which preserves the eccentricity. Because interactions between ... 7 At Lagrange point L1. Specifically for Earth-Moon L1, these calculations show 326054 km. 6 A simpler example is Kepler's laws of planetary motion. In a spherically-symmetric gravitational field, the planets follow elliptical orbits. The orbits certainly do not have the full spherical symmetry of the potential. They may be extremely lopsided. :-D 6 The faster you go, the less velocity you theoretically can gain from a gravity assist. The reason for this is that the faster you go the harder it is to bend the orbit. To proof this we have to use the patched conics approximation, which means that while within a sphere Kepler orbits can be used. The sphere can be simplified to be infinitely big, since the ... 6 One can get an order of magnitude estimate of the maximum speed attainable by gravitational slingshots without doing any real calculation. The 'rough physics' reasoning goes as follows: The gravitational field of the planets used for slingshots needs to be strong enough to "grab" the speeding spaceship. As a planet cannot "grab" a spaceships moving faster ... 6 "before gravity stopped holding it together" is the same (pretty much) as "so that apparent gravity at the surface is zero". This means that$$m \omega^2 r = \frac{GM_{moon}m}{r^2}$$with G=6.7\cdot 10^{-11}, M_{moon}=7.3\cdot 10^{22} kg, r_{moon}=1740 km, we find$$\omega=\sqrt{\frac{GM_{moon}}{r^3}}=0.0092 rev/min = 0.55 rev/hour It is ...