Hot answers tagged

69

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. This is faint (especially at the faint end), but certainly not out of reach of today's big telescopes. I understand that various parts of the sky are ...


48

Let's assume mass of the person plus spacesuit to be $m_1$=100kg Asteroid density: $\rho=$2g/cm$^3$ (source) that is 2 000kg/m$^3$ 15km/hour is a good common run. That's roughly v=4m/s The orbital height is negligible comparing to the radius, assume 0 over surface. Linear to angular velocity (1): $$ \omega = {v \over r } $$ Centripetal force (2): $$ F = ...


47

One thing to keep in mind is that objects that are bound gravitationally actually revolve around each other around a point called a barycenter. The fact that the earth looks like its revolving around the sun is because the sun is much more massive and its radius is large enough that it encompasses the barycenter. This is a similar situation with the Earth ...


41

The other answers provide a first-order approximation, assuming uniform density (though Adam Zalcman's does allude to deviations from linearity). (Summary: All the mass farther away from the center cancels out, and gravity decreases linearly with depth from 1 g at the surface to zero at the center.) But in fact, the Earth's core is substantially more dense ...


37

And you jumped in. No, I'd refuse, You should do that Yourself. :=( Do You want to "solve" this with or without friction by air? Without friction, You would fall and reach maximum speed in the center of earth, going on until You reach the antipods, where You would stand still for a fraction of a second, then You would go down again. Very boring indeed ...


36

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 ...


36

Anything the mass of a star is going to get hot like a star and fuse hydrogen like a star. In other words it will be a star not a planet! While it's technically possible to have a rocky planet the mass of a star, in practice when stellar systems form there aren't enough metals available to build such a large object. Large objects are invariably built from ...


29

If the mass/charge is symmetrically distributed on your sphere, there is no force acting on you, anywhere within the sphere. This is because every force originating from some part of the sphere will be canceled by another part. Like you said, if you move towards on side, the gravitational pull of that side will become stronger, but then there will also be "...


28

Correct. If you split the earth up into spherical shells, then the gravity from the shells "above" you cancels out, and you only feel the shells "below" you. When you are in the middle there is nothing "below" you. Refrence from Wikipedia Gauss & Shell Theorem. {I am using some simplistic terms, but I don't want to break out surface integrals and ...


28

Forget about force. Force is a bit much irrelevant here. The answer to this question lies in energy, thermodynamics, pressure, temperature, chemistry, and stellar physics. Potential energy and force go hand in hand. The gravitational force at some point inside the Earth is the rate at which gravitational potential energy changes with respect to distance. ...


28

Has Musk done his homework? With regard to the basic idea of using nuclear weapons to release CO2 and thereby warm Mars, no, he hasn't. I suspect this was either Bored Elon Musk speaking, or perhaps the Elon Musk who didn't quite deny being a super villain ( 1-900-MHA-HAHA Elon Musk?) in that interview with Colbert. CO2's enthalpy of sublimation is about ...


28

I live close to Lake Erie and often see scenes like in this picture. Note that the bottom of the cargo ship cannot be seen due to the curvature of the Earth.


25

Pluto is now classified as a dwarf planet. The main difference between a planet and a dwarf planet has to do with the requirement that a planet clear out the material in and near its orbit. Planets do this, dwarf planets do not. The reclassification was triggered by the discovery of many additional object (the Edgeworth-Kuiper Belt) out beyond the orbit ...


25

The answer is that inside a spherically symmetric shell of matter (your hollow earth or massive beach ball) there is no gravitational force anywhere - you will not "fall" in any direction, whether you are at the centre or not, regardless of the radius of the sphere. This is a classic result of both Newtonian Gravity, and Einstein's General Theory of ...


24

Yes, this is possible. It is perfectly fine for a mass configuration to produce, for points outside a sphere of radius $R$ centred at $\mathbf r_0$, a gravitational field identical to that of a point mass at $\mathbf r_0$, and still be completely empty inside a smaller sphere of radius $a$ around $\mathbf r_0$. The spherical-shell model you describe is ...


24

It depends what you mean by day and night. The day and night are not of equal lengths now, where I live at latitude 53N. The tilt of the Earth's rotation axis with respect to the ecliptic plane means that this is generally true. The situation you describe would have to be considerably more extreme. If the planet was in a highly eccentric orbit and had a ...


24

I know that the opposite could happen. There is an old book called "Night Fall" about a planet that had three stars. Because there was always a star shinning on all sides it would never get dark. About every 500 years everything lines up just right so that all the stars were on one side. Then as the planet rotated on its normal spin, night fall would come ...


24

The reason why we can see exoplanets 13,000 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. The ...


23

Are we increasing the gravity of Earth with our population No, we don't increase the total mass of the Earth, because our bodies are made of things that were already on Earth (food, water, air, minerals etc). Do we, other species and things have our own gravity? Yes everything with mass has gravity (and things without but that's complicated) How ...


22

This is a gravitational phenomenon known as tidal lock. It is closely related to the phenomenon of tides on Earth, hence the name. Tidal locking is an effect caused by the gravitational gradient from the near side to the far side of the moon. (That is, the continuous variation of the gravitational field strength across the Moon.) The end result is that the ...


21

No, not by jumping. Jumping gives you an acceleration only from the location on the surface. As soon as you leave the surface, you have no way of adjusting your orbit. Either you reach escape velocity, or you will return to your initial location after exactly one orbit. The only way to prevent this would be to have an additional acceleration once you ...


21

Because orbits are general conic sections. Why this is true is another fascinating question in and of itself, but for now I'll just assume it. The point is that circular orbits are special examples of general orbits. It's perfectly possible to get a circular orbit, but the relationship between the bodies' velocities and separation needs to be exactly right. ...


21

This is a really rough calculation that doesn't take into account the realistic direction of the bow shock, or calculation of the drag force. I just take the net momentum flow in the solar wind and direct it so as to produce the maximum decceleration and see what happens. Apparently the solar wind pressure is of the order of a nanoPascal. As I write this it'...


21

Related to the Andrea di Biagio answer. Here is a typical flight path for the Beunos Aires-Auckland route. The distance is approximately 10,300 km by the shortest route along a sphere. Direct flights are offered by Air New Zealand and take 11h40m - an average speed of 882 km/h. Looking at the map you present, the distance from Auckland to Buenos Aires looks ...


20

The simplest way to think about it is that there is mass all around you in the center of the Earth so you get an equal gravitational "pull" from all directions. The pulls cancel out so you get no acceleration. If one assumes constant density for the Earth (which isn't strictly speaking true but it is close enough for this illustration) the gravitational ...


20

I think you are confused as to what the 'surface' of Jupiter or Saturn are. They have a large liquid hydrogen centre, but this is surrounded by an incredibly thick layer of atmosphere, which has clouds, gases, liquids etc. So you would first pass through the outer layers of atmosphere, falling through denser and denser gas until you float at a height which ...


19

There are a lot of factors that go into whether or not a planet has an atmosphere. First, the mass and size of the planet. Really what it comes down to is the escape velocity. The higher the escape velocity (ve), the easier it is for a planet (or moon) to retain any atmosphere it gets as the gases that make up the atmosphere have to be moving faster to ...


19

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

This correlation is known as Titius-Bode's law, which is often stated as \begin{equation} d=0.4 + 0.3 \cdot 2^n \end{equation} 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) ...



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