GR9277 #19



Alternate Solutions 
wahoo1 20111031 11:55:12  you can also solve it pretty easily using the fact that g~10, G=6.67x10^11.
g= Me G / Re^2 >> Me= gRe^2/G
using order of magnitudes: Me~ (10)(10^6)^2 / (10^11)
this becomes (10^1)(10^12)(10^11)= 10^24 which is closest to choice (A).   dinoco 20101107 06:23:30  You can use = noting that a day is 86400 seconds and solve for M.   SlickAce21 20051111 11:05:34  This problem is pretty easily solved by a usage of Kepler's law, if one remembers that g = (GM)/R^2, then one gets 10 = (6E11)M/36E12, multiplying through you have M = 60E23 or 6E24  

Comments 
justacomment 20140930 16:51:36  If you can't think of anything but the volume of a sphere (like myself), this is how I approach the answer:
You are given the radius of the Earth, so do the math then one shall find V~10^21 (m^3), using 2^10~10^3 as approximation.
Then remember the density of most elements on Earth lies in the scale of 10^3 (kg/m^3), and there you go, A is the only possible answer.   TMFMende 20121026 23:10:20  If you apply Gauss's law to gravity... you will know that
d = 4 GM
or rather
g 4 r = 4 GM
ETS gives you r in the problem and G at the beginning of the test; you already know g (~10 m/s). So solve for M:
M = gr/G
...Gauss law will also be helpful if you need to calculate the force of gravity inside a mass (e.g. if there were a magical shaft going toward the center of the Earth in which you could send a test mass).
Thank you Yosun so much for this site! It's great to see the different ways people go about tackling the same problem.   wahoo1 20111031 11:55:12  you can also solve it pretty easily using the fact that g~10, G=6.67x10^11.
g= Me G / Re^2 >> Me= gRe^2/G
using order of magnitudes: Me~ (10)(10^6)^2 / (10^11)
this becomes (10^1)(10^12)(10^11)= 10^24 which is closest to choice (A).   wahoo1 20111031 11:54:50  you can also solve it pretty easily using the fact that g~10, G=6.67x10^11.
g= Me G / Re^2 >> Me= gRe^2/G
using order of magnitudes: Me~ (10)(10^6)^2 / (10^11)
this becomes (10^1)(10^12)(10^11)= 10^24 which is closest to choice (A).   dinoco 20101107 06:23:30  You can use = noting that a day is 86400 seconds and solve for M.
The Bridge 20181026 19:15:17 
That won\'t work. The formula on the left is for centripetal acceleration of someone standing on the Equator due to the rotation of the Earth, and works out to about .88 m/s2. This is in fact many times less than , the acceleration due to gravity of something on the Earth\'s surface, which is the right side. If you knew the orbital period of an object at the Earth\'s surface off the top of your head (84 min), then this method would work, but it gives you 10 m/s2 in the table of information at the top of the test anyway, so the left side of the equation is completely unneeded.

  theevilmachines 20090906 20:53:13  You can easily solve this problem by assuming that the earth is entirely water (most of it is), so the density is . Then the mass is about .
pavamanacs 20090916 02:03:34 
volume of sphere is (4/3) pi r^3 is it not..........

justguessing 20091005 10:33:25 
i like this! he says the Earth is a cube filled with water! that would be so awesome.

Albert 20091103 03:15:47 
Dude! By treating the sphere like a cube, you are off by at least . You are just plain lucky that in the question other options are simply greater than or else you would have been badly confused. This is simply not the best analogy to solve this problem.

Crandor 20100407 22:19:23 
By neglecting the , the exponent is off by . Before the last approximation, the answer was . Rounding to 24 took that missing factor into account. Also, the density approximation is good enough for government work.

ashestofeonix 20100831 01:54:18 
uhh... I hate to be a stickler about something as silly as correct units but.... the density of water is 1 g/cm, that is 1kg/dm^3... And most of the earth is not water... most of the SURFACE of the earth is covered by water, but most of the volume is iron & nickel. There really is a big difference.

his dudeness 20100903 19:29:40 
Not really. Water has a density of 1000 kg/m^3, while Iron has a density of ~8000 kg/m^3. By slightly overestimating the volume of the earth (cube instead of sphere), and slightly underestimating its density, you come reasonably close to the real answer.

Almno10 20101112 12:58:34 
First of all, 4/3*pi will NOT change the factor of ten, which is the only thing you need to get the answer right.
The iron/nickel argument still wont change the factor of ten. 8000 = 8(10^3). Same factor of ten. At most, this will shift your answer by one power of ten...but all the choices are separated by 10^3, so whats the problem? Give the guy a break, cool answer.

  jax 20051204 20:24:47  You need to use curly brackets {} around superscripts or else only the first number of your exponent will show up as an exponent!
And I agree that SlickAce21's solution is much simpler. I would not have remembered the mass of the sun, or the density of the earth on the exam!
yosun 20051204 23:17:42 
jax: thanks for the typoalert; it's been corrected. SlickAce21: thanks for the alternate solution.

  SlickAce21 20051111 11:05:34  This problem is pretty easily solved by a usage of Kepler's law, if one remembers that g = (GM)/R^2, then one gets 10 = (6E11)M/36E12, multiplying through you have M = 60E23 or 6E24   SlickAce21 20051111 11:03:44   




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