# Tag Info

61

There is no alignment between the Sun or the Solar System's net angular momentum and the "spin axis" of the Galaxy. Think for a moment about whether the line of the ecliptic (which marks the "equatorial line" of the Solar System) and the Milky Way (which roughly marks the plane of the Galaxy) are lined up? If this were so, then you would always see the ...

57

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 ...

52

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 ...

48

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 ...

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 ...

36

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 ...

35

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 ...

32

This is very rough and based on eyeballing without careful measurements: I've got a four-watt nightlight. I can read by it (not comfortably) at a distance of about a meter. The sphere of radius 1 meter has a surface area of about 12 square meters, so it appears that 1/3 of a watt per square meter will (barely) suffice for reading. The earth gets about ...

32

The Solar wind does indeed exert a force on the planets, however it turns out that the force is so small that it has no measurable effect. The force can be calculated using the fact that force is equal to the rate of change of momentum. Suppose the total mass of all the Solar wind particles hitting the Earth per second is $M$, and the average velocity of ...

28

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 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 ... 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 ... 22 One point, the difficulty of seeing colors in dim light is due to properties of the human vision system. Most cameras will not have the same effect and will be able to show vivid colors in even dim light (as long as the light is sufficient for imaging). But as a good guess, with accommodation, you can read (to some extent) under a full moon. The sun ... 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 ... 18 The unit of illumination is the lux, lumens per square meter. What is the minimum lux required for reading? How many lux does the Sun provide at distance D? What is the minimum lux required for reading? You can plug all sorts of numbers into this depending on how good your eyes are, how big the print is, and how close you hold it to your face. I'm going ... 18 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 ... 18 Dark matter would affect planetary motion, but the influence of dark matter on planets in our solar system is too small to detect even currenlty, due to the low concentration of dark matter compare to ordinary matter in our solar system. See Constraints on Dark Matter in the Solar System. The density of dark matter is very low,  <~10^{-19} grams/cm^3, ... 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 ... 17 This correlation is known as Titius-Bode's law, which is often stated as $$d=0.4 + 0.3 \cdot 2^n$$ where d represents planet's mean distance from the Sun in Astronomical Units and n = -∞, 0, 1, 2... for Mercury, Venus, Earth, Mars, asteroid belt, Jupiter and so on. The rule is not satisfied exactly with Neptune's orbit (n=7) ... 16 This web page has a nice discussion on it: http://archive.ncsa.illinois.edu/Cyberia/NumRel/EinsteinTest.html Basically the orbit's eccentricity would precess around the sun. Classical stellar mechanics (or Newtonian gravity) couldn't account for all of that. It basically had to do with (and forgive my crude wording) the sun dragging the fabric of ... 16 Philosophical answer We orbit Sun because we called the star that is so important to our life "Sun". That's actually not as silly as it may look. Let's imagine we would orbit one of the other stars. We would see a bright star at the sky during day. We would call that star Sun. Maybe the day would be a little longer, and the star a little more ... 16 There are three main reasons. 1) While Venus is orbiting the Sun at 35.02 Km/s, the Earth is also orbiting the Sun in the same direction at 29.78 Km/s. This factor will decrease the relative transit velocity of Venus as seen from earth. 2) Venus is travelling at 35.02 Km/s an elliptical orbit. Hence the actual distance traveled by Venus during the transit ... 16 Set the forces on the test particle from the Earth and Moon equal:$$F_E=F_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 ...

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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