is (F=mg) equal to (F=GmM/r^2)? And what's the difference between them?
Newton's Law of Universal Gravitation tells us that the potential energy of object in a gravitational field is $$U ~=~ -\frac{GMm}{r}.\tag{1}$$  The experimentally verified near-Earth gravitational potential is  $$U ~=~ mgh.\tag{2}$$  The near-Earth potential should be an approximation for the general potential energy when $r\approx r_{\text{Earth}}$, but the problem I'm having is that they scale differently with distance.  $(1)$ scales as $\frac 1r$.  So the greater the distance from the Earth, the less potential energy an object should have.  But $(2)$ scales proportionally to distance.  So the greater the distance from the Earth, the more potential energy an object should have.  How is this reconcilable?
facl ignoring the "x" permission but only on files Let $f_n, f \colon [a,b] \to \mathbb{R}.$ Show that, if $f_n \to f$ uniformly, then the Lebesgue integrals are equal, i.e. $\int f = \lim \int f_n$.  This is clearly true for continuous functions, but how do I handle the case of non-continuous functions? VLC will not install 14.04 I tried to watch a dvd and found that VLC was not installed. When I went to the Ubuntu software center I got this message when trying to install VLC. I do not know why. Here's the ouput:  "This error could be caused by required additional software packages which are missing or not installable. Furthermore there could be a conflict between software packages which are not allowed to be installed at the same time.  The following packages have unmet dependencies:  vlc: Depends: vlc-nox (= 2.1.4-0ubuntu14.04.1) but 2.1.4-0ubuntu14.04.1 is to be installed      Depends: libc6 (&gt;= 2.15) but 2.19-0ubuntu6.4 is to be installed      Depends: libfreetype6 (&gt;= 2.2.1) but 2.5.2-1ubuntu2.2 is to be installed      Depends: libgcc1 (&gt;= 1:4.1.1) but 1:4.9.1-0ubuntu1 is to be installed      Depends: libqtcore4 (&gt;= 4:4.8.0) but 4:4.8.5+git192-g085f851+dfsg-2ubuntu4 is to be installed      Depends: libqtgui4 (&gt;= 4:4.8.0) but 4:4.8.5+git192-g085f851+dfsg-2ubuntu4 is to be installed      Depends: libstdc++6 (&gt;= 4.6) but 4.8.2-19ubuntu1 is to be installed      Depends: zlib1g (&gt;= 1:1.2.3.3) but 1:1.2.8.dfsg-1ubuntu1 is to be installed"   Thanks in advance for any help. Someone from Mars is called a Martian. Someone from America is called an American. Someone from Europe is a European. Someone from Hyrule is a Hyrulean.  We are not far from colonizing other planets, let alone Earth's only moon. Obviously you wouldn't call someone from the moon a lunatic right? What if you are on Mars? What would I call a foreigner from Earth? Earthean doesn't sound right. Why am I still being asked for an ssh password when I'm using private key authentication? I am trying to disable the touchscreen on my Asus S200E laptop.  In Ubuntu 16.04, I was able to do so with this:  Section "InputClass"     Identifier "evdev touchscreen catchall"     MatchIsTouchscreen "on"     MatchDevicePath "/dev/input/event*"     Driver "libinput" EndSection   How can this be done on Ubuntu 18.04? Is it OK to add [Solved] to the title of a question? I was confused when I saw this Question with [Solved] in its title where I couldn't find a solution: .  Then I found out that the "[Solved]" part was put there by someone else through editing like these:       I was feeling that this was not how it's supposed to be done and searched meta to find, that it is indeed discouraged to do so:        But on the other hand, the user who did the edits has a whole lot of reputation so I might me missing something here. So what's the correct thing to do? Should those edits be reverted? Sponsor Gallery, Page, or Something Like That? nhj asked the other day,  In , I wrote:     We were thinking of putting together a "sponsor gallery" of sorts that listed out all of the sponsors with a brief description. Kinda in the vein of what we have on the every-now-and-then  at TDWTF.   So, consider this the discussion for that topic.   What would you like to see in a sponsor gallery? Should it be one big, long list? How does the order get determined? Should we organize the sponsors based on tags (and if so, what about sponsors who target like java and .net)? What about former sponsors... do they get removed from the list, or go on like the "past sponsors" section/page?  Keep in mind that this is not something we intend to charge extra for... this is more a service to the you, the members, than anything else. I've gotten postive feedback from the TDWTF "gallery", as a lot of readers come through RSS, have adblock running, etc. Dynamic pivot: sum of sales per month My table is like this:  Sitecode    Month  Amount --------    -----  ------ XX          Jan     1000 XX          Jan     3000 XX          Apr     3000 XX          Apr     1000   What I want is, to show the result something like this:  Sitecode    MonthJAN   MonthAPR --------    --------   -------- XX          4000       4000 I need to get the size of directory in terminal for signing purposes. I'm using following command:  du -s /path/to/dir   I'm multiplying the result by traditional UNIX block size (512 bytes) and get actual directory size in bytes. However, the Finder's "Get Info" dialog window shows the size slightly smaller than the one calculated with the terminal command. And it seems that it's reproducible on any folder/bundle. What am I missing? How $\frac{1}{n}\sum_{i=1}^n X_i^2 - \bar X^2 = \frac{\sum_{i=1}^n (X_i - \bar X)^2}{n}$ How   $\frac{1}{n}\sum_{i=1}^n X_i^2 - \bar X^2 = \frac{\sum_{i=1}^n (X_i - \bar X)^2}{n}$   i have tried to do that by the following procedure:  $\frac{1}{n}\sum_{i=1}^n X_i^2 - \bar X^2$  =$\frac{1}{n}(\sum_{i=1}^n X_i^2 - n\bar X^2)$  =$\frac{1}{n}(\sum_{i=1}^n X_i^2 - \sum_{i=1}^n\bar X^2)$  =$\frac{1}{n} \sum_{i=1}^n (X_i^2 - \bar X^2)$  Then i have stumbled. Macbook Air - how to install a dual bootable Ubuntu 14.04 LTS On my Macbook Air 10.9.2 without crashing the current setup, how can i install as dual boot: Ubuntu 14.04 LTS ? (can anyone please advise some step by step please) Where are my Minecraft saves located? How do I sort a two-dimensional (rectangular) array in C#? I have a two-dimensional array (of Strings) which make up my data table (of rows and columns).  I want to sort this array by any column.  I tried to find an algorithm for doing this in C#, but have not been successful.  Any help is appreciated. Allowed memory size of 33554432 bytes exhausted (tried to allocate 43148176 bytes) in php This error message is being presented, any suggestions?     Allowed memory size of 33554432 bytes exhausted (tried to allocate   43148176 bytes) in php My girlfriend forgot the password for her MacBook (don't ask), so I reset it using the Utilities.  But as you may be aware, this doesn't reset the Keychain password. Her MacBook is now virtually unusable due to Keychain demands continuously provoking a request for the Keychain password.  There are lots of other people asking this around the web and they're usually pointed towards this , which doesn't apply in our situation because the support advice involves entering the old password.  We don't know the old password.  Is it possible to reset Keychain without knowing the old password? We're not bothered about what's in there and what will be lost, we just want the continuous password prompts to go away. Hey guys I am currently working on a scene with multiple objects.  There is one object (a wheel) that I need to rotate the origin for without rotating the wheel itself.  I have the exact angle that I need so a manual solution would be the best.  Any help would be much appreciated. I am used to parametric modeling and Blender is a little out of my depth. Do there exist two primes $p<q$ such that $p^n-1\mid q^n-1$ for infinitely many $n$? We can prove that there is no integer $n&gt;1$ such that $2^n-1\mid 3^n-1$. This leads to the following question:     Is it true that for every pair of primes $p&lt;q$ there are only finitely many integers $n$ such that $p^n-1\mid q^n-1$? Are there two primes $p&lt;q$ and an integer $n&gt;p+q$ such that $p^n-1\mid q^n-1$?Is it true that if $n&gt;6$ then $2^n-1\nmid 5^n-1$?   Edit: Here are some examples: \begin{array}{ll} 2^{36}-1\mid 41^{36}-1, &amp;3^{12}-1\mid 97^{12}-1,\\ 5^{6}-1\mid 37^{6}-1, &amp;7^{4}-1\mid 151^{4}-1. \end{array}  Now I prove there is no integer $n&gt;1$ such that $2^n-1\mid 3^n-1.$  Proof: Denote $A=2^n-1$ and $B=3^n-1$. If $n$ is even then $3$ divides $A$ but not $B$, a contradiction.  If $n$ is odd then $A\equiv -5\pmod {12}$. Since every prime greater than $3$ is $\equiv \pm1,\pm5 \pmod {12}$, some prime factor $p$ of $A$ must be congruent to $\pm5 \pmod {12}$. As $p \mid B$, we have $3^{n+1}\equiv 3 \pmod p$. Since $n+1$ is even, we get $(\frac{3}{p})=1$, a contradiction again. Apache error "Could not reliably determine the server's fully qualified domain name" When I restart my Apache server using the command  sudo /etc/init.d/apache2 restart   I get the following error:  Restarting web server apache2 apache2: Could not reliably determine the server's fully qualified domain name, using 127.0.1.1 for ServerName ... waiting apache2: Could not reliably determine the server's fully qualified domain name, using 127.0.1.1 for ServerName   Is the server using 127.0.1.1 instead of 127.0.0.1? What causes this error? Show that   $$\displaystyle1 - \frac{x^2}2\leq\cos x\quad\forall x \in \mathbb{R}$$   Let $f(x) = \cos x - 1 + \frac{x^2}2$; then we need to show that $f(x) \geq 0\quad\forall x \in \mathbb{R}$.  Since $$\lim_{x \to \pm\infty} f(x) = +\infty$$ and $$\operatorname{im}f'' = [0, 2]$$ (as $f''(x) = 1 -\cos x$), it remains to show that $x = 0$ is a minimum for $f(x)$, but I cannot do that because I would need to solve $f'(x) = x -\sin x = 0$, which I cannot. I think it would suffice to show that plugging in $x = 0$ we have that $f'(x) = 0$, and combining it with the fact that the function $f$ is convex, we would be done.    Is it enough? What's a better proof?  EDIT: It just occurred to me that there is a way to solve $f'(x) = 0$. Just observe that $f'(x)$ is an odd function and that $f''(x) \geq 0$ implies that $f'(x)$ is monotonically increasing. Since an odd function, if defined at $x = 0$, must be $0$ there, $x = 0$ is a solution to the equation and it's the only one.