# Tag Info

54

As you said, the mass of the Moon is 1.2 percent that of the Earth. Now, if you mean the gravitational acceleration at the surface, it is calculated like this $G\frac{M}{R^2}$, where $M$ is the mass, and $R$ is the radius of the celestial body. The moon's mass is a hundred times smaller, but the radius is four times smaller, meaning its surface gravity will ...

50

Is the sun shrinking currently? It's the other way around: The Sun is slowly growing hotter and thus is slowly expanding. The Sun accumulates ever more helium in its core as it ages. This growing amount of fusion ash results in the core getting hotter, which in turn results in the Sun expanding in volume. This completely overwhelms what one would expect ...

41

There are a number of methods of detecting exoplanets, but all of them favour detection of larger planets over smaller ones, albeit for slightly different definitions of large: Radial velocity measurement — this detects the small movement of the star towards and away from us as the planet and the star orbit their mutual barycenter. This movement is fastest ...

35

There is basically an upper limit to the mass of a star because their luminosity is so great that the radiation pressure prevents the accretion of further mass. However, the upper limit depends on the composition of the accreting material. This is because the effect of the radiation depends on the opacity of the material - stuff that is more metal-rich is ...

31

How was the mass of Venus measured for the first time? In the mid 19th century, Urbain Le Verrier's predicted of the existence of a then unknown planet beyond the orbit of Uranus. He even predicted this planet's orbit. The discovery of Neptune based on his predictions was perhaps his greatest accomplishment. Le Verrier then went on to investigate Mercury. ...

25

There is a simple$^*$ way to know the total mass of the atmosphere: measuring the pressure it exerts on the surface, which necessarily integrate all of the atmosphere above ground level. If you take an atmospheric pressure of $1\cdot10^5$ Pa, it is equivalent to a force of $1\cdot10^5$ newton over one square meter. Multiply by the area of the planet in ...

19

Sun's mass is almost $M_\odot = 1.9*10^{30}$ kg. The mass-loss scale you are talking about is around $\delta M \approx 10^{10}$ kg/s. I'm guessing this mass-loss is due to both stellar winds and nuclear fusion. So, if the Sun evolves only using this mass-loss, it will last for $t=\frac{M\odot}{\delta M} = 10^{20} seconds.$ This time scale is almost 3 ...

18

Planets and stars, no. Globular clusters and galaxies, yes. Small scales To condense into such relatively compact objects as planets, stars, and even the more diffuse star-forming clouds, particles need to be able to dissipate their energy. If they don't do this, their velocities prohibit them from forming anything. "Normal" particles, i.e. atoms, ...

16

The mass of Venus was determined by weighing the Earth, or more precisely, by determining the ratio of the density of the Earth to the density of Schiehallion, and assuming Schiehallion to be typical rock of 2500 kg per cubic meter. Prior to that, Jérôme Lalande had worked out the relative masses of the major bodies of the Solar System as a byproduct of ...

14

Comet Shoemaker–Levy 9 crashed into Jupiter a few years back. As well as these molecules, emission from heavy atoms such as iron, magnesium and silicon was detected, with abundances consistent with what would be found in a cometary nucleus. Those heavy elements are consistent with the comet being at least being partially composed of rock. So Jupiter is ...

14

The confusion comes from the difference between the nucleus and the coma. The nucleus is a small icy body, only a few km across. The coma is the cloud of gas and dust released from the nucleus as it warms up. With not much gravity, the coma spreads out into space, and it can be hard to say exactly where the edge of the coma lies, however, a coma "the size ...

13

It doesn't matter if the body is made of gas, rocks, liquid or plasma, the four states of matter all have mass. So, as we know, mass create a gravitational field, and the more mass the stronger the gravity - and Jupiter has 317x Earth mass.

10

The mass of the Sun is determined from Kepler's laws: $$\frac{4\pi^2\times(1\,\mathrm{AU})^3}{G\times(1\,\mathrm{year})^2}$$ Each term in this component contributes to both the value of the solar mass and our uncertainty. First, we know to very good precision that the (sidereal) year is 365.256363004 days. We have also defined the astronomical unit (AU) to ...

10

Well, I wasn't going to answer but the other two answers are wrong, or at least incomplete. If you wish to make a black hole from a stellar-sized object, then there is no need to compress it to as small as the Schwarzschild radius (though that would certainly work and would certainly be the answer for smaller objects with negligible self-gravity). Instead, ...

10

I assume you're asking about central supermassive black holes (SMBHs, one per galaxy), not stellar-mass black holes. The answer is yes, but what actually happens is the two SMBHs have to merge first, and then the resulting combined SMBH can sometimes be ejected from the combined (merged) galaxy. [Edited to add: Since you've updated the question with a ...

10

Suppose the atmosphere has a density that decays exponentially with height. e.g. $$\rho = \rho_0 \exp[-h/h_0]\ ,$$ where $\rho_0$ is the density at some surface and $h_0$ is a characteristic height scale on which the density decreases. If we integrate this funcion from $h=0$ to $h = \infty$, then this gives a finite result. $$\int^{\infty}_0 \rho_0 \exp[-... 9 I concur with everyone else here (of course) that the gravity at the "surface" of Jupiter is entirely determined by the mass contained within that surface. The composition makes no difference. However I differ with some on the answer to the headline title question. We simply do not know whether Jupiter has a rocky core. A popular theory for the formation ... 9 I asked a somewhat similar question but just about the Earth. Here and in my question there's some links that you might be interested in. Jupiter for example is thought to have moved closer to the sun during the late heavy bombardment, then back outwards. So I suspect that through the wormhole didn't get that point quite right. Observations by the ... 9 A succinct summary of supernova types is given in the following image based on Heger et al. (2003): Image courtesy of Wikipedia user Fulvio 314 under the Creative Commons Attribution-Share Alike 3.0 Unported license. The graph is based on the graph in Fig. 1 of the linked paper. The pair instability realm is upwards of ~100 solar masses, though it is ... 9 There is no general consensus on this. Different evolutionary models give different results. The factors (in addition to the initial mass of the star) that effect the final black hole mass would be the rotation rate of the progenitor, its composition (or metallicity) and whether it was in a binary system or not and whether that binary system was able to ... 9 You are correct that the IAU definition of "clearing the orbit" has the problem of being not explicitly quantified. And a complete clearing was obviously never the intention behind the definition. I like this statement by Steven Soter: The IAU definition of a planet as a heliocentric body that "has cleared the neighborhood around its orbit” is problematic.... 8 Yes, the mass of the observable Universe is always increasing. Matter Even if you're only referring the "ordinary" matter (such as stars, gas, and bicycles) and dark matter, the mass of the observable Universe does increase, not because mass is being created, but because the size of the observable Universe increases. In a billion years from now, we ... 8 I'll shamelessly reference an answer I wrote on Worldbuilding to an almost identical question. Lammer et al. (2014) suggested that "super-Earths" with masses of 2-5M_\oplus1 could retain massive hydrogen/helium envelopes, up to \sim10^{25} kilograms. Above this, up to about 10M_\odot or more, "mini-Neptunes" exist, possibly composed of volatiles and ... 8 The gap appears because of pair instability supernovae. In short, as one looks at such massive stellar cores at increasing temperatures, an ever-larger fraction of the photons are sufficiently energetic to spontaneously form electron-positron pairs. True, they soon recombine, but there is nevertheless a loss in (radiation) pressure, which causes contraction, ... 8 Your approach is completely correct, just note three things: Logarithmic distribution First, since the distribution of masses is logarithmic in nature (as is most other things), be sure to bin them logarithmically. Otherwise you will oversample (undersample) the bins at the low-(high-)mass end. Comoving densities Second, to be able to compare mass ... 8 Your question body is different from your question title and it seems you really want to ask what you did in the question body so I'll address that. Short Answer: The simple power law which applies for larger asteroids and comets actually doesn't extend that well to smaller bodies and shouldn't be trusted too much in that range. Long Answer: You're right ... 8 The mass of an average galaxy appears to be totally dominated by dark matter, so your calculation would not give the galaxy mass. Even if all you wanted was the baryonic (non dark matter) mass then what you suggest will be very much a lower limit. For example you can look at this paper by Chabrier (2001), who estimates that gas forms less than half the ... 8 There are three factors that all play into this: The Jupiter/Sun mass ratio The Jupiter/Sun distance The Sun's radius The barycenter of any pair of orbiting masses lies on the line connecting their centers of mass, and its position depends on the masses of the two objects. If two objects have the same mass, their barycenter will lie halfway between them. ... 8 The answer is that there is a limited rate at which matter that can be crammed into a black hole of this size. That rate is small enough that the black hole will traverse through the Sun hardly picking up any mass (when expressed as a fraction of its original mass). Details: You could use the Bondi-Hoyle accretion rate:$$\dot{M} = \frac{dM}{dt} = \pi \rho \...

8

Red dwarfs, depending on your definition, can range from 2.5 to 150 times more dense than the Sun. What is the cause of this discrepancy? They give no calculations, so I can only guess. The article is from 1946 and we've gotten a lot better at science. It's 1946 and information exchange is limited. No internet, no TV, and long distance calls are expensive....

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