(fork) returns ... 0 to the child process,

So this code is creating three processes:

• Parent: first fork returns nonzero $pid, parent executes waitpid($pid). This returns very quickly, since the child exits almost immediately (though not the grandchild).
• Child: First fork returns zero in child, second fork returns PID of grandchild, child exits.
• Grandchild: First fork returns zero in child, second fork returns zero in grandchild, grandchild either exec's or die's.

The OP's problem is that he's waiting for the child but running the long-running process in the grandchild. The fix is to eliminate the second fork, and just do the work in the child; the purpose of a second fork is to do something in the background without having to wait for it, which is what the OP is trying to avoid.

Here's a small example:

warn "PID $$ (parent) forking\n";
unless ($pid = fork) {
  warn "  PID $$ (child) forking again\n";
  unless (fork) {
    warn "    PID $$ (grandchild) exec'ing\n";
    exec "sleep 5";
      die "Couldn't run getSite.pl";
    exit 0;
  }
  warn "  PID $$ (child) exiting\n";
  exit 0;
}
warn "PID $$ (parent) executing waitpid\n";
waitpid($pid,0);
warn "PID $$ (parent) done.\n";
[download]

PID 6083 (parent) forking
  PID 6084 (child) forking again
    PID 6085 (grandchild) exec'ing
  PID 6084 (child) exiting
PID 6083 (parent) executing waitpid
PID 6083 (parent) done.

Here's a version which does what I believe the OP wants.

warn "PID $$ (parent) forking\n";
unless ($pid = fork) {
    warn "  PID $$ (child) exec'ing\n";
    exec "sleep 5";
      die "Couldn't run getSite.pl";
}
warn "PID $$ (parent) executing waitpid\n";
waitpid($pid,0);
warn "PID $$ (parent) done.\n";
[download]

But, this is completely equivalent to system, so why not just use that?