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GR9277 #19
Problem
 GREPhysics.NET Official Solution Alternate Solutions
\prob{19}
Which of the following is most nearly the mass of the Earth? (the radius of the Earth is about 6.4E6 m.)

1. 6E24kg
2. 6E27kg
3. 6E30kg
4. 6E33kg
5. 6E36kg

Mechanics$\Rightarrow$}Mass of Earth

If one does not remember the mass of the earth to be on the order of $10^{24} kg$, one might remember the mass of the sun to be $10^{30} kg$. Since the earth weighs much less than that, the answer would have to be either (A) or (B). The problem gives the radius of the earth, and one can assume that the density of the earth is a few thousand $kg/m^3$ and deduce an approximate mass from $m=\rho V$. The answer comes out to about $10^{22}$, which implies that the earth is probably a bit more dense than one's original assumption. In either case, the earth can't be, on average, uniformly $10^9 kg/m^3$ dense. Thus (A) is the best (and correct) answer.

An alternate solution is provided by the user SlickAce21. Equating the mass of some object with the gravitational force, one has $mg=GmM/r^2\Rightarrow g = GM/r^2 \Rightarrow M = gr^2/G \approx 6 E24$.

Alternate Solutions
 wahoo12011-10-31 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).Reply to this comment dinoco2010-11-07 06:23:30 You can use $\frac{4\pi^2r}{T^2}$ = $\frac{GM}{r^2}$ noting that a day is 86400 seconds and solve for M.Reply to this comment SlickAce212005-11-11 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 = (6E-11)M/36E12, multiplying through you have M = 60E23 or 6E24Reply to this comment
justacomment
2014-09-30 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
2012-10-26 23:10:20
If you apply Gauss's law to gravity... you will know that

$\oint$ $\vec{g}$ $\cdot$ d$\vec{A}$ = 4 $\pi$ GM

or rather
g 4 $\pi$ r$^2$ = 4 $\pi$ GM

ETS gives you r in the problem and G at the beginning of the test; you already know g (~10 m/s$^2$). So solve for M:

M = gr$^2$/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
2011-10-31 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
2011-10-31 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
2010-11-07 06:23:30
You can use $\frac{4\pi^2r}{T^2}$ = $\frac{GM}{r^2}$ noting that a day is 86400 seconds and solve for M.
 The Bridge2018-10-26 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 $g$, 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 $g\\approx$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
2009-09-06 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 $1000 kg/m3$. Then the mass is about $M\approx(1000 kg/m^3)(6.4 \times 10^6)^3 \approx 10^{24}$.
 pavamanacs2009-09-16 02:03:34 volume of sphere is (4/3) pi r^3 is it not..........
 justguessing2009-10-05 10:33:25 i like this! he says the Earth is a cube filled with water! that would be so awesome.
 Albert2009-11-03 03:15:47 Dude! By treating the sphere like a cube, you are off by at least $10^3$. You are just plain lucky that in the question other options are simply greater than $10^24$ or else you would have been badly confused. This is simply not the best analogy to solve this problem.
 Crandor2010-04-07 22:19:23 By neglecting the $\frac{4}{3} \pi$, the exponent is off by $log_10 \left(\frac{4}{3}\pi\right)=0.622$. Before the last approximation, the answer was $M\approxeq\left(1000kg/m^{3}\right)\left(6.4\times 10^{6}\right)^{3}=10^{23.419}$. Rounding to 24 took that missing factor into account. Also, the density approximation is good enough for government work.
 ashestofeonix2010-08-31 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 dudeness2010-09-03 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.
 Almno102010-11-12 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
2005-12-04 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!
 yosun2005-12-04 23:17:42 jax: thanks for the typo-alert; it's been corrected. SlickAce21: thanks for the alternate solution.
SlickAce21
2005-11-11 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 = (6E-11)M/36E12, multiplying through you have M = 60E23 or 6E24
SlickAce21
2005-11-11 11:03:44

LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$