# Tag Info

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The Sun is losing the equivalent of around 4.26 million metric tons per second of hydrogen, as it is converted into energy. The Sun does not have an exact mass.

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Humphreys & Larsen (1995) suggest, using star count information, a distance of $20.5 \pm 3.5$ pc above the Galactic plane; consistent with, but more precise than the Bahcall paper referred to by Schleis. Joshi (2007) is more guarded, investigating some systematic uncertainties in the estimation techniques and ends up with distances between 13 and 28 pc ...

1

http://ilrs.gsfc.nasa.gov/docs/2014/196C.pdf gives G*m for the Sun as 132712440041.939400 km^3/s^2. http://physics.nist.gov/cgi-bin/cuu/Value?bg gives G as 6.67384*10^-11 m^3*kg^-1*s^-2 If these numbers were exact (they're not), the sun's mass (after converting m^3 to km^3 above) would be: 165890550052424250000000000000000000/83423 kg or about ...

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The Galactic rotation period at the Sun's Galactocentric radius is about 230 million years (with a five percent uncertainty) and the Sun, as stated in user8's answer is 4.57 billion years old - giving $\sim 20\pm 1$ orbits. However, the idea of a Galactic year is misleading. For instance, the Milky Way rotation curve is quite flat between 1 and 10 kpc from ...

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The answer to your question is a very simple one but, a good one at that. The exact mass is 1.9891×1030 kg. Math and new advances in solar exploration have given us so much more information about our Sun. We have been able to see solar eruptions, (CMA's), and gained information about the Sun never known.

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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 ...

4

I thought I'd contribute an answer because there's a very recent paper on the subject: Measuring the solar radius from space during the 2012 Venus Transit It appeared in my RSS feeds this morning! A related writeup is online at the HMI website. To answer the question, this measurement uses the transit of Venus to fit the limb-darkening law of the Sun. ...

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Is there any resource that provides these current values? Yes. The JPL HORIZONS on-line solar system data and ephemeris computation service provides these values, and much more.

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According to wikipedia and other sources, a planet and a star always move in a circular orbit around the common center of mass of the both bodies ... This is not true. In the absence of other gravitational sources, a planet and a star move in elliptical orbits about the common center of mass. Ancient scientists assumed circular orbits, but only because ...

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The trajectory of Earth's orbit is shaped not only by Sun's gravity. Earth's orbit is being changed by different sources of gravity as well, for example center of our galaxy. This is why it's not perfectly round.

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Tracy Cramer is correct. Early in the life of the universe it is thought that star formation in "metal-free" gas favoured larger stars. These had short lives and very quickly enriched the interstellar and intergalactic medium with nucleosynthesis products. In fact the enrichment of the interstellar medium (ISM) in our own Galaxy is thought to have mainly ...

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Here is the very study you are looking for by Bailer-Jones (2014). Using the re-reduction of the Hipparcos astrometry, he has integrated orbits for 50,000 stars to look for objects that might come or might have come close to the Sun. The K-dwarf Hip 85605 is the winner on that timescale, with a "90% probability of coming between 0.04 and 0.20pc between ...

1

Solar flares are observed at wavelengths right across the electromagnetic spectrum, not just H alpha. The basic model for a solar flare starts with the magnetic field in the corona. You can think of the topology of the magnetic field to consist of loops that poke up out of the photosphere and extend into the corona. However, the photosphere of the Sun is ...

1

In the transition from a higher electron energy level to a lower one, say $m\mapsto n$, a hydrogen atom emits a photon of wavelength $\lambda$ satisfying $$\frac{1}{\lambda} = R_\infty\left[\frac{1}{n^2}-\frac{1}{m^2}\right]\text{,}$$ where $R_\infty = 1.09737315685\,\mathrm{m}^{-1}$ is the Rydberg constant. For $n=1$, i.e. the destination energy level is ...

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