Hot answers tagged

95

Life forms are made up from materials already present in Earth. Thus, increasing population would not alter the overall mass of the planet, and can't impact its orbit.


55

On predicting planetary orbits A number of studies have shown that the inner solar system is chaotic, with a Lyapunov time scale of about 5 million years. This 5 million year time scale means that while one can somewhat reasonably create a planetary ephemeris (a time-based catalog of where the planets were / will be) that spans from 10 million years into ...


45

Incorrect Path I'm curious as to what does the moon's orbit around the sun looks like? One might think the orbit (in the sun's rest frame) follows the path of an epitrochoid. A (very) over exaggerated view of this motion (for unrealistic parameters, thus, not an accurate representation) can be seen in the following animation: Note that if you change ...


44

The Moon's orbit must be concave toward the Sun. The Moon's orbit with respect to the Sun is always convex. This is easily proven by comparing the minimum possible gravitational acceleration of the Moon toward the Sun (5.7 mm/s2) and the maximum possible gravitational acceleration of the Moon toward the Earth (3.1 mm/s2). The acceleration vector, and ...


38

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. (I don't know that "effective gravity" is a commonly used phrase, but it seems to me to be applicable here.) If you're standing on the surface of the Earth, you feel gravity (1g, 9.8 m/s2) ...


29

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 = 132\...


24

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 ...


22

When launching into a low Earth orbit only your velocity relative to the Earth matters, as seen from the not-rotating reference frame of the Earth. Your velocity relative to the sun does not matter, because once you are in the orbit your velocity vector relative to the Earth will oscillate between pointing towards and away from the velocity vector of the ...


21

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 space-...


20

You have to consider the limit of infinitesimally short time, in which the (vertical on the paper) component of velocity is infinitely short, and thus also the angle changes for an infinitesimal amount. In this limit, the correction to the length is quadratic in the time step and vanishes exactly in the physical limit of continuous time. Pythagoras: $$v_2=\...


19

To shine a light on Emilio's comment and develop Fabrice's answer, what will happen is the following. Given that the change in velocity can be considered instantaneous relatively to the period, the moon will find itself in a new orbit where its position (just after the slow down) is the apoapsis (the furthest point from the focus where lies the Earth) of a ...


19

Mass is conserved (up to whatever small contribution nuclear decay has to the overall loss of mass). All biological matter is just created from materials from the environment (we do eat, right?). With exception of an occasional space probe, shooting stars and solar wind effects, the earth can be considered a closed system in terms of matter exchange. ...


17

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 ...


17

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 ...


16

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 ...


16

Before answering let me mention that there is a terrific free applet showing the orbits, including the velocity vectors of the system Sun/Earth/Moon: https://phet.colorado.edu/en/simulation/gravity-and-orbits It is in java so pretty easy to download and use. The moon's orbit must be concave toward the sun. The Moon' orbit around the Sun is a ...


15

If you could take from orbital energy, then it would decrease, until at some point in the future it would zero. Hence, it can't be perpetual.


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

Here's another way of looking at it. Let M1, M2, M3 be our three masses. In the three body problem we're considering, the whole frame containing M1, M2 and M3 is rotating. You're right to think that if that frame was fixed then the points L4 and L5 would not be stable. After all if you perturb M3 from L4 or L5 then it should just roll down the potential ...


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 ...


14

Because the direction of the velocity changes. The velocity will start to point less and less 'towards' point A and when the distance between A and B is the smallest, the velocity will make a right angle with the radius, which means acceleration vector also makes a right angle with the velocity. At this point the radial component of the speed is zero and the ...


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 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

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 ...


12

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 ...


12

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 ...


12

This question points out the importance of symplecticity in physics. In an orbital simulation, suppose one simply advances state via $$\begin{align} \boldsymbol x(t+\!\Delta t) &= \boldsymbol x(t) + \boldsymbol v(t)\, \Delta t \tag 1 \\ \boldsymbol v(t+\Delta t) &= \boldsymbol v(t) + \boldsymbol a(t)\, \Delta t \end{align}$$ where $\Delta t$ is a ...


11

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 ...


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

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 ...



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