# Tag Info

62

What is important for tidal forces is not the absolute gravity, but the differential gravity across the planet, that is, how different the gravity is at a point on the surface near the sun relative to a point on the other side. Because the Sun is far away, gravity doesn't change much between the two extremes on earth. However, if you compare it with the ...

56

The Earth's climate isn't quite as stable as you think. The Earth's climate has toggled back and forth between a greenhouse Earth and an icehouse Earth for the last 600 million years or so. During the icehouse Earth phases, the climate can enter an ice age, an extended period of time during which the climate in oscillates between glaciations and ...

49

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 ...

43

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 ...

38

Neither of those statements are true. It's an easy approximation to make: a neutron star has all of that 'space' removed from between nucleons --- so we just need to know how big a neutron star of mass equal to the solar system would be. Well, the only significant mass is the sun (jupiter is about 1% the mass of the sun---negligible). If the sun were ...

37

Imagine two donut-shaped spaceships meeting in deep space. Further, suppose that when a passenger in ship A looks out the window, they see ship B rotating clockwise. That means that when a passenger in B looks out the window, they see ship A rotating clockwise as well (hold up your two hands and try it!). From pure kinematics, we can't say "ship A is ...

36

Tides are caused by the gradient of the gravitational field - so the tidal "force" experienced drops with the third power of the distance. This means that the relative strength of the tides should go as $$ratio = \frac{M_{moon} \cdot D_{sun}^3}{M_{sun} \cdot D_{moon}^3}\\ =\frac{7 \cdot 10^{22}\cdot (1.5 \cdot 10^{11})^3}{2\cdot 10^{30}\cdot (3.7\cdot ... 35 We haven't ironed out all the details about how planets form, but they almost certainly form from a disk of material around a young star. Because the disk lies in a single plane, the planets are broadly in that plane too. But I'm just deferring the question. Why should a disk form around a young star? While the star is forming, there's a lot of gas and dust ... 33 The answer kind of depends on how old you are. At a very introductory level, say, maybe middle school or younger, it's "okay" to refer to Jupiter as a failed star to get the idea across that a gas giant planet is sort of similar to a star in composition. But around middle school and above (where "middle school" refers to around 6-8 grade, or age ~12-14), I ... 27 You've used the gravitational constant with only three significant digits. So it's no surprise that your answer isn't accurate to five significant digits. Instead of G and M_\odot separately, you should use the product GM_\odot, known as the standard gravitational parameter. Its value is known very accurately: in the link, you'll find$$ GM_\odot = ...

23

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

A slightly simpler version of David Hammen's (as usual excellent) answer: Earth is "big enough" to have sufficient pull on the atmosphere: gravity stops it from escaping Earth is "close enough" to the sun to keep liquid water (and liquid core) Core is sufficiently magnetic that it acts to protect against solar wind (which would otherwise strip the ...

20

An orbit is stable because of conservation of angular momentum. Suppose we start with an object in an exactly circular orbit and slow it down slightly. That means it is moving at less than orbital velocity so it starts to fall inwards. However as its distance to the Sun decreases the tangential component of its velocity has to increase to conserve angular ...

19

When you're trying to understand the mechanics of a system it's usually convenient to choose coordinates that reflect the symmetry of the system. The solar system is roughly centrally symmetric because the Sun is by far the largest mass in it, and the coordinates that reflect this symmetry are polar coordinates with the Sun at the centre. For example in ...

19

The simple answer to Why does the Pluto's orbit crosses the Neptune's orbit is to just say that's the way it is. For any object orbiting in a central inverse square law field, like the gravitational field of the Sun, the stable orbits are ellipses with the Sun at one focus. The ellipses can be almost circular like the Earth's orbit or wildly eccentric like ...

19

yes, you may describe the motion from any reference frame, including the geocentric one, assuming that you add the appropriate "fictitious" forces (centrifugal, Coriolis, and so on). But the special property of the reference frame associated with the Sun - more precisely, with the barycenter (center of mass) of the Solar System, which is just a solar radius ...

18

There are a few things that keep Saturn's rings roughly the way they are. First, Saturn's D ring actually is "raining" down on Saturn currently. But, the phenomenon of shepherd moons prevents the vast majority of material from leaving the other rings: "The gravity of shepherd moons serves to maintain a sharply defined edge to the ring; material that ...

18

When we say that the Moon rotates, we don't mean relative to an observer on Earth, because we're also rotating. Maybe best is to think of it from the perspective of the Sun. If you were at the centre of the solar system, looking at the Earth, you'd see the Moon rotates once every 28 days or so. That also happens to be the amount of time it takes for the Moon ...

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 ...

16

This correlation is known as Titius-Bode's law, which is often stated as $$d=0.4 + 0.3 \cdot 2^n$$ where d represents planet's mean distance from the Sun in Astronomical Units and n = -∞, 0, 1, 2... for Mercury, Venus, Earth, Mars, asteroid belt, Jupiter and so on. The rule is not satisfied exactly with Neptune's orbit (n=7) ...

16

Non-physical/philosophical answer (see comment of @annav): We orbit Sun because we called the star we're orbiting "Sun". (I think that's not actually as "stupid" as it may sound initially. See also Anthropic principle ) Physical answer: The Earth formed from matter near to the Sun, so it ended up near the Sun, and is orbiting it, because gravity depends ...

16

There are three main reasons. 1) While Venus is orbiting the Sun at 35.02 Km/s, the Earth is also orbiting the Sun in the same direction at 29.78 Km/s. This factor will decrease the relative transit velocity of Venus as seen from earth. 2) Venus is travelling at 35.02 Km/s an elliptical orbit. Hence the actual distance traveled by Venus during the transit ...

16

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 ...

15

The leading theory is that at a distant point in its past, Uranus was struck by a very large object, which knocked it to its side, and current tilt. Imagine if you took a top, and smacked it with a rock. The top might be turning perfectly alright at first, but after it had been hit, the top would most likely be wobbling significantly. Similarly, after an ...

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Space is big. You just won't believe how vastly, hugely, mind- bogglingly big it is. I mean, you may think it's a long way down the road to the chemist's, but that's just peanuts to space. Douglas Adams, Hitchhikers Guide to the Galaxy

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The night sky would certainly be spectacular from within a globular cluster! To get an estimate for the total brightness of the stars, let's say there are 6000 stars visible to the naked eye1. Let's make the globular cluster a big one, like Omega Centauri, with ten million stars2. The roughest approximation would be to say that those ten million stars ...

14

A brief history of the misapplication of magnetohydrodynamics to the analysis of the solar wind: 1959: Soviet satellite Luna 1 directly observed the solar wind for the first time and measured its strength. http://en.wikipedia.org/wiki/Luna_1 So as of 1959, by direct experimental observation, it was known that the heliopause was at least the radius of the ...

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 ...

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