Imagine you were on a spherical spacecraft that had equal mass all the way around and you were deep in space, away from the gravitational pull of any planetary bodies or stars. If you released a marble at arm's length, would it be gravitationally pulled toward your body? If you flicked it in one direction, would it begin to orbit you? (If a marble is too big, then maybe a grain of sand?)
Since humans are not large in mass, do we exert enough gravitational force for a small object to orbit around us? (Assuming that the only gravitational force acting on the object is from the human). Using Newton's law of gravity and using myself as the human mass, I exert a 10^-8 force on a 1 kg object .5 meter away, but I am not sure if even this small of a force will be acceptable. What is the baseline "force" between two objects for 1 to orbit another?  Edit: I am not sure if this is a SpaceExchange question.
Installing Office 2010 on Ubuntu 15.04 using Wine I had installed Office 2010 32bit on Ubuntu 12.04 before using guides such as . However, due to package changes, I was not able to replicate the same on Ubuntu 15.04.  How does one install Office 10 using Wine? Can I do it with reflection or something like that? number of ways to make $2.00 I am using extensively trigonometric functions when an angle is given in degrees. Some of these functions like sine or cosine have rational values, for example, the well   known example is that $\cos(\theta) =0.6 $ and $\sin(\theta) =0.8 $.       However besides the case of multiplicity of $90^\circ$ it seems there are no rational numbers $\theta$  with simultaneously rational  values of  sine and cosine.      Is it possible somehow to prove that for rational values of an angle given in degrees there are  no values simultaneously rational of sine and cosine functions, beside obvious case of multiplicities of $90^\circ$? How should I use a possessive apostrophe when a name is between commas? What is the difference between "I've been to China" and "I've gone to China"? Both are in the present perfect, but one uses the verb 'to go' and the other 'to be'. Is there a rule for this?  Is there any chance that the differences between "I have been" vs. "I have gone", are differences in English dialect (I don't know if this is the right word).  Edit Could I say that North American English speakers use "I have gone", and maybe U.K speakers say "I've been"? Or vice versa. Is there ever a right time to upgrade to full frame camera body? Context  I've been shooting with an a6000 (APS_C mirrorless) and I plan to move on up to full frame, preferably mirrorless as well. I'm debating on when to upgrade, and probably don't plan to buy until next year.  I know I'll need to buy new full frame lens.  Question  Is there a "right" time to buy camera bodies, both mirrorless in my case as well as DSLRs? (I've seen the a7rII and like it but there might also be an a9 looming.) I feel that this is like smartphones in which the best time to buy is now as you might always be waiting until the next upgrade.   Related, are full frame bodies discounted during the holiday season or is that limited only to entry level bodies and older full frame bodies?  Related Force JSF to process, validate and update readonly/disabled input components anyway I've been a Stack Exchange contributor for a couple of years now, currently most active on . One thing that keeps coming up is questions that would appeal to audiences on two different Stack Exchange sites but there are almost no features taking advantage of these questions. On travel.SE we have crossover questions most often with  and possibly next most often with  - but it can happen with almost any SE site. Today alone we had  with  and  had . I am absolutely sure that this must happen all the time between arbitrary pairs of Stack Exchange sites but the only feature that takes advantage of it is the possibility of migrating off-topic questions which leaves the original question acting as a redirect across the two sites. I think there is a missed opportunity to build the network between sites where users of one site don't realize they could make valuable contributions to a sister site they have had so far little to do with. At the very least there could be a way to &quot;advertise&quot; a question on a second site that could be interested in it. Probably by listing it in the sidebar where there are also links to meta questions, linked questions, related questions, new blog posts, and But there could also be some kind of features that make some sort of acceptable crossposting. There could be gamification/badges added around building cross-site questions or whatever else you can think up. I think this is a great way for people to spread their expertise from their main site onto other sites, to make the network stronger and more cohesive, to bring more attention to questions that would have a divided audience between two sites. I can think only of positives and no negatives. Does anyone else have some ideas to add or think this is a bad idea for any reason?  Example crossover questions: Here are some links to some crossover questions I've been finding in the Stack Exchange network here. (This was previously posted as an answer, but more people wanted it to be part of the question.)   ▶  :   ▶  :   ▶  :   ▶  :   ▶  :   ▶  :   ▶  :   ▶  :   ▶  :   ▶  :   ▶  :   ▶  :   ▶  : Condition for a tensor to be decomposable How to call a function in a Twig file Let's say I create a function in a module file:  function _get_test() {   return 'test'; }   How can I call this in any Twig file? For example *.html.twig, page.html.twig, node.html.twig etc.   OR   How can I pass a variable from PHP to Twig and display in any Twig file.  For example *.html.twig, page.html.twig, node.html.twig etc. I'm trying to re-learn my undergrad math, and I'm using Stephen Abbot's Understanding Analysis. In section 2.7, he has the following exercise:  Exercise 2.7.5 (a) Show that if $\sum{a_n}$ converges absolutely, then $\sum{a_n^2}$ also converges absolutely. Does this proposition hold without absolute convergence?  I'm posting about this here because my answer to that last question about whether the proposition holds without absolute convergence is "Yes", but my suspicions are raised by him merely asking the question. Usually questions like this are asked to point out that certain conditions are necessary in the statement of propositions, theorems, etc. I just want to see if I'm missing something here.  Anyway, here's how I prove the absolute convergence of $\sum{a_n^2}$, and note that I never use the fact that $\sum{a_n}$ is absolutely convergent:  Proof: Let $s_n = \sum_{i=1}^n{a_n^2}$. I want to show that $(s_n)$ is a Cauchy sequence. So, let $\epsilon &gt; 0$ and $n &gt; m$, and consider, \begin{equation}   \begin{aligned}     |s_n - s_m| &amp;= |(a_1^2 + a_2^2 + \cdots a_n^2) - (a_1^2 + a_2^2 + \cdots a_m^2)|\\                 &amp;= |(a_{m+1})^2 + (a_{m+2})^2 + \cdots + a_n^2|\\                 &amp;= |a_{m+1}|^2 + |a_{m+2}|^2 + \cdots + |a_n|^2\\   \end{aligned} \end{equation} Now, since $\sum{a_m}$ converges, the sequence $(a_m)$ has limit 0. Therefore I can choose an $N$ such that $|a_m| &lt; \sqrt{\epsilon/n}$ for all $m &gt; N$. So for all $n &gt; m&gt; N$,  we have, \begin{equation}   \begin{aligned}     |s_n - s_m| &amp;= |a_{m+1}|^2 + |a_{m+2}|^2 + \cdots + |a_n|^2\\                 &amp;&lt; \left(\sqrt{\frac{\epsilon}{n}}\right)^2 + \left(\sqrt{\frac{\epsilon}{n}}\right)^2 + \cdots + \left(\sqrt{\frac{\epsilon}{n}}\right)^2\\                 &amp;&lt; \left(\sqrt{\frac{\epsilon}{n-m}}\right)^2 + \left(\sqrt{\frac{\epsilon}{n-m}}\right)^2 + \cdots + \left(\sqrt{\frac{\epsilon}{n-m}}\right)^2\\                 &amp;= \epsilon   \end{aligned} \end{equation} Therefore $(s_n)$ is Cauchy and $\sum{a_n^2}$ converges. And since $\sum{a_n^2} = \sum{|a_n^2|}$, the series $\sum{a_n^2}$ is absolutely convergent. &#x220E;  Am I doing something wrong here? Am I right in thinking that $\sum{a_n}$ need not be absolutely convergent for $\sum{a_n^2}$ to be absolutely convergent? Is there a proof from the first principle that the Lagrangian L = T - V? Is there a proof from the first principle that for the Lagrangian $L$,    $$L = T\text{(kinetic energy)} - V\text{(potential energy)}$$  in classical mechanics? Assume that Cartesian coordinates are used. Among the combinations, $L = T - nV$, only $n=1$ works. Is there a fundamental reason for it?  On the other hand, the variational principle used in deriving the equations of motion, Euler-Lagrange equation, is general enough (can be used to to find the optimum of any parametrized integral) and does not specify the form of Lagrangian. I appreciate for anyone who gives the  answer, and if possible, the primary source (who published the answer first in the literature).      Notes added on Sept 22: - Both answers are correct as far as I can find. Both answerers were not sure about what I meant by the term I used: 'first principle'. I like to elaborate what I was thinking, not meant to be condescending or anything near to that. Please have a little understanding if the words I use are not well-thought of. - We do science by collecting facts, forming empirical laws, building a theory which generalizes the laws, then we go back to the lab and find if the generalization part can stand up to the verification. Newton's laws are close to the end of empirical laws, meaning that they are easily verified in the lab. These laws are not limited to gravity, but are used mostly under the condition of gravity. When we generalize and express them in Lagrangian or Hamiltonian, they can be used where Newton's laws cannot, for example, on electromagnetism, or any other forces unknown to us. Lagrangian or Hamiltonian and the derived equations of motion are generalizations and more on the theory side, relatively speaking; at least those are a little more theoretical than Newton's laws. We still go to lab to verify these generalizations, but it's somewhat harder to do so, like we have to use Large Hadron Collider. - But here is a new problem, as @Jerry Schirmer pointed out in his comment and I agreed. Lagrangian is great tool if we know its expression. If we don't, then we are at lost. Lagrangian is almost as useless as Newton's laws for a new mysterious force. It's almost as useless but not quite, because we can try and error. We have much better luck to try and error on Lagrangian than on equations of motion. - Oh, variational principle is a 'first principle' in my mind and is used to derive Euler-Lagrange equation. But variational principle does not give a clue about the explicit expression of Lagrangian. This is the point I'm driving at. This is why I'm looking for help, say, in Physics SE. If someone knew the reason why n=1 in L=T-nV, then we could use this reasoning to find out about a mysterious force. It looks like that someone is in the future. Subscript labels in QGIS legends Is it possible to include subscript and superscript text in the title of a legend item in QGIS ?  I have tried using &lt;sub&gt;some text&lt;/sub&gt; but this just appears in the title with no change to the contents.  I am looking to have layer titles such as   Annual NO2 concentration (μg/m3)  The superscript is not a big issue, as I can use the character map to get a 3, but I don't know how to get the subscript. Proving $\tan(\theta_1+\theta_2+\cdots+\theta_n)$ has the form $\frac{S_1-S_3+S_5+\cdots}{1-S_2+S_4+\cdots}$ $$\tan(\theta_1+\theta_2+\cdots+\theta_n)=\frac{S_1-S_3+S_5+\cdots}{1-S_2+S_4+\cdots}$$  where  $S_i$ denotes the sum of product of tangent of angles taken $i$ at a time.     For example,$$\tan(\theta_1+\theta_2)=\frac{S_1}{1-S_2}=\frac{\tan\theta_1+\tan\theta_2} {1-\tan\theta_1\tan\theta_2}$$   (This formula is given  in my textbook with no derivation or background)  How to derive this? How do I force Excel (and other Microsoft Office products) to stop opening files in the same application? Whenever I "double click" on an Excel file and another Excel file is open, the newly opened file automatically opens in the same application window as the previously opened Excel file. This isn't limited to just Excel, as I've seen Word do this as well. This poses a problem when wanting to compare documents side by side.    The current solution I have for this is to actually open another Excel or Word instance, and then open the file from within that application window itself. Is there a way to force Office to open a new instance of the application when double clicking on the file icons?  I'm currently using Office 2007 and Windows XP, but I've seen this on Office 2010 and Windows Vista and 7. I'm looking for an overall solution if possible. bash if -a vs -e option Both about -a and -e options in  is said:  -a file     True if file exists. -e file     True if file exists.    Trying to get what the difference is I ran the following script:  resin_dir=/Test/Resin_wheleph/Results  if [ -e ${resin_dir} ] ; then     echo "-e "; fi  if [ ! -e ${resin_dir} ] ; then     echo "! -e"; fi  if [ -a ${resin_dir} ] ; then     echo "-a"; fi  if [ ! -a ${resin_dir} ] ; then     echo "! -a"; fi   /Test/Resin_wheleph/Results exists and is a directory. And this is what I get:  -e -a ! -a   which seems to be a little strange (notice -a and  ! -a). But when I use double brackets (e. g. if [[ -e ${resin_dir} ]]) in the similar script it gives reasonable output:  -e -a   So:    What is a difference between -a and -e options? Why -a produces a strange result when used inside single brackets? Experimental observation of matter/antimatter in the universe How prove this diophantine equation $x^2+y^2+z^3=n$ always have integer solution show that: For any postive ineteger $n$,then the equation    $$n=x^2+y^2+z^3$$   always have integer solution   My idea: such as $n=1$,then we have $$1=0^2+0^2+1^3$$ $$2=0^2+1^2+1^3$$ $$3=1^2+1^2+1^3$$ $$4=2^2+0^2+0^3$$ $$5=1^2+2^2+0^3$$ $$6=1^2+2^2+1^3$$ $$7=2^2+2^2+(-1)^3$$ $$8=0^2+0^2+2^3$$ $$9=1^2+0^2+2^3$$ $$10=1^2+1^2+2^3$$ $\cdots\cdots\cdots$  But for  general $n$, How prove it? This seems to be a basic issue, but I just realized that I actually don't know how to test equality of coefficients from two different regressions. Can anyone shed some light on this?   More formally, suppose I ran the following two regressions:  $$ y_1 = X_1\beta_1 + \epsilon_1 $$ and  $$ y_2 = X_2\beta_2 + \epsilon_2 $$ where $X_i$ refers to the design matrix of regression $i$, and $\beta_i$ to the vector of coefficients in regression $i$. Note that $X_1$ and $X_2$ are potentially very different, with different dimensions etc. I am interested in for instance whether or not $\hat\beta_{11} \neq \hat\beta_{21}$.   If these came from the same regression, this would be trivial. But since they come from different ones, I am not quite sure how to do it. Does anyone have an idea or can give me some pointers?   My problem in detail: My first intuition was to look at the confidence intervals, and if they overlap, then I would say they are essentially the same. This procedure does not come with the correct size of the test, though (i.e. each individual confidence interval has $\alpha=0.05$, say, but looking at them jointly will not have the same probability). My "second" intuition was to conduct a normal t-test. That is, take   $$ \frac{\beta_{11}-\beta_{21}}{sd(\beta_{11})} $$  where $\beta_{21}$ is taken as the value of my null hypothesis. This does not take into account the estimation uncertainty of $\beta_{21}$, though, and the answer may depend on the order of the regressions (which one I call 1 and 2).   My third idea was to do it as in a standard test for equality of two coefficients from the same regression, that is take  $$ \frac{\beta_{11}-\beta_{21}}{sd(\beta_{11}-\beta_{21})} $$   The complication arises due to the fact that both come from different regressions. Note that   $$ Var(\beta_{11}-\beta_{21}) = Var(\beta_{11}) + Var(\beta_{21}) -2 Cov(\beta_{11},\beta_{21}) $$ but since they are from different regressions, how would I get $Cov(\beta_{11},\beta_{21})$?  This led me to ask this question here. This must be a standard procedure / standard test, but I cound not find anything that was sufficiently similar to this problem. So, if anyone can point me to the correct procedure, I would be very grateful!