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I mean total inertial on my body siting here at my computer "California US." Ok so the earth is rotating on its axis and in turn around the sun and the sun around the galaxy. The object the question how much kinetic energy is there in a litter of water "at rest" on my desk.

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Why stop at the galactic centre? The galaxy moves through the Local Group, which has a drift velocity relative to the Virgo super cluster, etc... If you start down this road, the end point is to measure things relative to the Cosmic Microwave Background, which defines (astronomically speaking) a preferred time-slicing of spacetime. In that case, the dipole moment from the CMB gives the answer: about 370 km/s towards Virgo. Source: phy.duke.edu/~kolena/cmb.htm –  genneth Jun 18 '11 at 12:35
    
@genneth: seems like a fairly "slow" speed compared to the CMB, I would like to see an example of some faster runners :) –  BjornW Jun 19 '11 at 21:45

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The standard value for the speed of the solar system relative to the Galactic center is 220 km/s. If you want to start digging into the literature on this, here's one article. This value is found by measuring radial velocities (via Doppler shifts) of many objects around the Galaxy and fitting to a model of Galactic rotation.

To get your velocity at any given location and time of year / day, you have to add in the velocity of the Earth relative to the solar system center of mass (essentially, the velocity of Earth relative to the Sun), and the velocity of you relative to the center of the Earth.

The magnitudes of those are easy to calculate, at least approximately. The Earth-Sun velocity is roughly $2\pi R/T$, where $R=1.5\times 10^{11}$ m is the radius of Earth's orbit and $T$ is a year. Similarly, the velocity of a point on Earth's surface relative to Earth's center is $2\pi r\cos l/t$, where $l$ is your latitude, $r=6400$ km is Earth's radius, and $t$ is a sidereal day (4 minutes less than a regular day). The two speeds work out to about 30 km/s and (460 m/s)$\cos l$.

To add them up, you need to know the directions of the various vectors. That's a mildly annoying calculation. The Sun-Galaxy velocity is simple in Galactic coordinates (where the direction to the Galactic center is one of the coordinate directions). The Earth-Sun velocity is simple in ecliptic coordinates. The you-Earth velocity is simple celestial coordinates. The rotation matrices required to convert all these into each other are known and can be looked up somewhere, no doubt.

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"with respect to center of the galaxy" we do not move at all :=) –  Georg Jun 18 '11 at 17:35
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I have no idea what you're talking about. We certainly are moving with respect to the center of the galaxy. The center of the galaxy has a world line. We have a world line. Those world lines are not parallel. Our radial velocity with respect to the center of the galaxy is zero, but there is a nonzero tangential velocity. –  Ted Bunn Jun 18 '11 at 18:14
    
Right, but motion relative to a point is when distance is changed by that, isn't it? –  Georg Jun 18 '11 at 19:43
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When you drive in your car past a lamp post, at the moment of closest approach, do you think that you're not moving relative to it? That's certainly not consistent with my understanding of the word "motion." –  Ted Bunn Jun 18 '11 at 20:26
    
@ Ted, this is semantics, but not free from importance to physics. The wording is moving "around" usually when a circular motion is the topic. The lamppost is "passed" when I move along in a straight line. –  Georg Jun 18 '11 at 20:48

Easy to remember guide by the Monty Python Institute of Astronomy

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