|
GR0177 #71
|
|
|
|
|
Alternate Solutions |
| There are no Alternate Solutions for this problem. Be the first to post one! |
|
|
Comments |
Camoph
2010-02-18 08:24:02 | I like your idea about using common, and it works I think, however answers B and c are very close. In the moment of the test, under pressure, is really hard try to deicide between them. There is a way to make a basic caculation to get the right answer between B and C ? |  | istezamer
2009-11-01 16:21:53 | Actually I solved this a little bit different!! I used No calculations!!
Since the wavelength has elongated then one must assume that the velocity is away from earth "Red-Shift" This eliminates choices A & B
Choice E is certainly out because it is a speed larger than c..
The Elongation is obviously large !! 6 times the initial length.. This means that the speed must be closer to c.. which yields choice D.
I Try to solve most of the problems using common sense and it works!! I guess the trick to score good in this exam is to go quick!
physicsman
2010-02-17 15:41:57 |
So 2.8 is close but 2.4 isnt?
|
Camoph
2010-02-18 08:22:12 |
I like your idea about using common sense, and it works, however, in this case the answers B and C are very close, in the moment of the test, under pressure is very hard try to deicide about one of them. There is an alternative way to do some basic calculation to give the right answer, beteween B and C?
|
Camoph
2010-02-18 08:26:13 |
Sorry, I mean answers C and D.
|
| | Poop Loops
2008-11-01 20:35:17 | What the hell is a "radial velocity"? I had the answer perfect and then I decided to multiply it by  because it was "radial" and got E.
Jeremy P
2008-11-03 14:36:57 |
In spherical coordinates, as opposed to Cartesian coordinates, radial is toward or away from the earth (which is spherical).
|
jmracek
2009-10-20 13:16:24 |
(E) could not have been the right answer because that would mean the planet is moving faster than the speed of light, obviously impossible.
|
Albert
2009-10-28 17:20:04 |
The planets and galaxies can indeed move near or even faster than the speed of light. Getting a signal from such a galaxy is not only possible but rather practised regularly for various things in Astronomy. Refer to the link to get more details. Thanks
http://curious.astro.cornell.edu/question.php?number=575
|
| | petr1243
2008-04-07 16:52:42 | Source is moving away from the reciever, if:
 > or f < f_0
Source is moving toward the observer, if:
< or f>f_0
| | hisperati
2007-10-27 01:37:06 | This question implies the interpretation that distant objects in the universe are just receding from us, but getting an answer faster than the speed of light is ok! It is cosmology. Ned Wright's page, "Acutally However, if we assume that the distance of an object at time t is the distance from our position at time t to the object's position at time t measured by a set of observers moving with the expansion of the Universe, and all making their observations when they see the Universe as having age t, then the velocity (change in D per change in t) can definitely be larger than the speed of light. This is not a contradiction of special relativity because this distance is not the same as the spatial distance used in SR, and the age of the Universe is not the same as the time used in SR." Basically distant objects have a redshift, or Z , and that should be interpretted as what it is (which depends on the assumed properties of the universe, like curvature density, etc.) not necessarily a velocity. ETS doesn't ask perfect questions. | | grep
2006-10-31 20:29:29 | If you happen to've forgotten the relativistic doppler shift (like me) and are reduced to using the non-relativistic version, you get an answer like 4c. Obviously this has some level of inaccuracy when used at these velocities, but it is accurate enough at least to say that v is darn near the speed of light and take a guess at D.
Gaffer
2007-10-23 14:42:42 |
The 4c answer is option E to trip up those of us not clever enough to realize that  is greater than c.
|
| | grep
2006-10-31 20:26:56 | | | Andresito
2006-03-08 18:47:38 | Yosun, the ratio as pointed out above is exactly 5, even with an approximation one can say lamba/lamba0=61/12=5
Furthermore, this gives beta=4/4.2 ~~0.9
Thus v = beta*c = c = 2.8 E+8 m/s
Cheers, and thank you for the site |  | ee7klt
2005-11-11 12:01:22 | actually,  exactly. then  . | |
|
| Post A Comment! |
|
|
Bare Basic LaTeX Rosetta Stone
|
LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces  .
|
| type this... |
to get... |
| $\int_0^\infty$ |
 |
| $\partial$ |
 |
| $\Rightarrow$ |
 |
| $\ddot{x},\dot{x}$ |
 |
| $\sqrt{z}$ |
 |
| $\langle my \rangle$ |
 |
| $\left( abacadabra \right)_{me}$ |
_{me}) |
| $\vec{E}$ |
 |
| $\frac{a}{b}$ |
 |
|
|
|
|
|
