# Tag Info

## Hot answers tagged solar-system

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.

85

The problem with finding a new planet in our solar system is not that it is too faint, but knowing where to look in a big, big sky. This putative planet 9 is likely to be in the range 20-28th magnitude (unless it is a primordial, planet-mass black hole, in which case it will be invisible except for any accretion luminosity). This is faint (especially at the ...

82

Voyager $1$ is headed away from the Sun at around $17$ km per second at an angle to the ecliptic of around $35$º. The orbital velocity of the Earth is $29.8$ km per second, and multiplying by $\cos 35$º to get the component of the velocity in Voyager's direction gives a shade over $24$ km per second. So at the point in its orbit where Earth is moving ...

68

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 ...

67

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 the ...

66

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 ...

65

The concept you're looking for is that of a planet's Hill sphere. If a planet of mass $m$ is in a roughly circular orbit of radius $a$ about a star of mass $M$, then the radius of this "sphere" is given by $$r_H = a \sqrt[3]{\frac{m}{3M}}.$$ For the Sun-Earth system, this yields $r_H \approx 0.01 \text{ AU}$, or about 1.5 million kilometers. The ...

62

You can't calculate the length of a mean solar day just by taking the mean of the shortest & longest apparent solar days. That would work if the apparent day lengths varied in a simple linear fashion, but that's not the case. From Wikipedia's article on the Equation of Time, The equation of time describes the discrepancy between two kinds of solar ...

46

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 ...

44

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 ...

41

Yes, you can construct classical orbits from the Schrodinger equation, as long as you take the right limit. For example, consider the hydrogen atom. While the lower energy levels, like the $1s$ or $2p$ look nothing like classical orbits, you may construct wavefunctions that do by superposing solutions with high $n$ appropriately. Such solutions obey the ...

38

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 ...

34

According to the caption for that picture on the same Wikipedia article, it is Francesco Fontana’s drawing of the supposed satellite(s) of Venus. Woodcuts from Fontana’s work (1646). The fringes of light around Venus are produced by optical effects.$^1$ Fontana lived from around 1580 to around 1656. He was an Italian lawyer at the University of Naples ...

33

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 ...

31

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 $$GM_\odot = 132\... 30 The limitation to hit the sun is that the object has to have very little angular momentum. The reason for this is that as the distance to the sun gets smaller, the velocity in a direction perpendicular to the sun gets larger, thanks to conservation of angular momentum:$$ L = mv_\perp r\rightarrow v_\perp={L\over mv} A good first-order approximation can ...

29

From Wikipedia, the sidereal rotation period is 3.91 hrs, so the angular frequency of its rotation is $2\pi/(3.91 \, \mathrm{hrs}) = 4.48 \times 10^{-4} \, \mathrm{rad/s}$, or $7.12 \times 10^{-5} \, \mathrm{Hz}$. Its "semimajor axis" is about 1000 km, so this corresponds to a surface speed of $2\pi (1000 \, \mathrm{km})/(3.9 \, \mathrm{hrs}) \approx 1600 \,... 28 The reason why we can see exoplanets thousands of light years away but not a planet 200 AU away (about 30 light-hours) is because these planets are found using different techniques. The planet discussed in the article I linked was discovered using a technique known as "microlensing," which requires a star to pass behind another star with a planet around it. ... 27 One point, the difficulty of seeing colors in dim light is due to properties of the human vision system. Most cameras will not have the same effect and will be able to show vivid colors in even dim light (as long as the light is sufficient for imaging). But as a good guess, with accommodation, you can read (to some extent) under a full moon. The sun (... 26 Dark matter would affect planetary motion, but the influence of dark matter on planets in our solar system is too small to detect even currently due to the low concentration of dark matter compared to ordinary matter in our solar system. See Constraints on Dark Matter in the Solar System. The density of dark matter is very low,$ <~10^{-19} grams/cm^3$,... 25 A slightly simpler version of David Hammen's (as usual excellent) answer: Earth is "big enough" to have sufficient pull on the atmosphere: gravity stops it from escaping Earth is "close enough" to the sun to keep liquid water (and liquid core) Core is sufficiently magnetic that it acts to protect against solar wind (which would otherwise strip the ... 25 The main plot below shows the potential energy of a mass in the Earth-Moon system under the unrealistic assumption that the system is not rotating. i.e. This mirrors (at present) all but one of the 4 answers given, in assuming that this point is defined where the gravitational force on a mass due to the Earth and the Moon are equal and opposite (i.e. at the ... 24 The habitable zone is typically defined as the range of orbits where liquid water can persist on a planet. This is somewhat ill-defined, since climate modelling of planets with various properties (atmosphere density and composition, rotation, etc.) changes distances where water can persist. But to a first approximation the habitable zone scales as$R_{hz}=\...

23

This one is tricky unless you know the magic term: ephemeris. An ephemeris gives the position of celestial bodies over time. Once you know that one, finding out information about their uncertainties is easier. The uncertainties are actually rather interesting in that they are planet specific. For example, the dominating factor for Mercury's uncertainty ...

23

The Voyager probes passed escape velocity for the solar system using gravitational boosts off of Jupiter, so they're almost certainly never coming back to us. To see this, we just calculate the total energy (divided by the probe mass) using the formula \begin{align} \frac{E}{m_{\mathrm{probe}}} & = \frac{1}{2} v_{\mathrm{probe/sun}}^2 - G \frac{M_{\...

23

The reason GR is praised much more so than those approximation methods you refer to is that the approximations are derived from the theoretical framework given by GR. Post-Newtonian approximation is simply a set of techniques used to find approximate solutions to the Einstein field equations. The reason people care more about GR as a whole is that GR gives a ...

23

This is an interesting question, since it raises the problem of the reference frame where Kepler's laws are true, which is often neglected. As a consequence of Newton's laws, in the inertial reference frames where the center of mass (c.m.) is fixed (there is a triple infinity of them, differing only with respect to the position of the c.m.) both planet and ...

22

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 ...

22

A lot (to put it mildly) of elements are created in stars and supernovae. These elements then travel through space until they fall to Earth (or, to be exact, some microscopic portion of them reach us). Earth itself wouldn't exist if stars hadn't generated elements which then clumped into dust, into minerals, and so on until a big ball of matter started to ...

21

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 ...

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