Using only weighings with one coin on each side, one can prove that the ordering can be determined with seven weighings:
Weigh coins $A$ versus $B$, then $C$ versus $D$, and then the heavier in each pair against each other. We can re-label the heaviest found as $D$, its original partner as $C$, and the loser of the "weigh-off" as $B$. These three weighings have led knowledge that $D>B>A$ and $D>C$.
Next weigh the coin $E$ (which has thus far not been touched) against $B$,
and then against $D$ if $E>B$, or against $A$ if $E<B$. These fourth and fifth weighings tell you the exact order of $D, E, B$ and $A$. For example, you might find that $D>B>A>E$. It does not matter which of the possibilities you find; the important facts are that you know the order and that the second lightest and the lightest are not $D$.
For the sixth weighing, weigh coin $C$ against the second lightest of the other coins. In our example, we would weigh $C$ versus $A$. Then the only doubt left is the relative weight of $C$ and just one other coin. In our example, if $C>A$ then we can complete the full ordering by weighing $C$ versus $B$ because we already know that $D>C$.
Thus the seventh weighing determines the complete order.
The way to figure this out was:
Without loss of generality, one can assume that the first weighing tells you that coin $B$ is heavier than coin $A$ (written as $B>A$). Then there are only two classes of choices of next weighings:
(1) Weigh coin $C$ against coin $A$ (or any equivalent weighing against an already-weighed coin): Here, a possible result is $C>A$ which leaves 40 allowed orders. Since each weighing provides only one bit of information, discerning among $40$ orders requires at least six weighings, for a total of eight.
(2) Weigh coin $C$ against coin $D$ (or any equivalent weighing of two not-yet-weighed coins): Then there are three classs of next weighing choices available.
(2a) If the third weighing is $E$ versus $D$ (or any of the already-weighed coins) then one possibility says that $E<D$, which leaves $20$ allowed orders, requiring at least $5$ further weighings, for a total of eight.
(2b) If the third weighing is $D$ versus $A$ (or equivalently $B$ versus $C$) then the answer $D>A$ leaves $25$ possible orderings (slotting $E$ into any of $5$ positions, with the first four ordered $DCBA, DBCA, DBAC, BDAC$ or $BDCA$)
which again requires at least $5$ further weighings, for a total of at least eight.
(2c) If the third weighing is $D$ versus $B$ (or equivalently $A$ versus $C$)
then the answer must be equivalent to $D>B$. So $D>B>A$ and you can slot $C$ into any of three positions (but not into $C>D$) and then $E$ into any of $5$ positions. With $15$ allowed possibilities, it might still be possible to discern the ordering in only four more weighings.
So assume we have (in three weighings) $D>B>A$ and $D>C$. The next weighing needs to involve coin $E$ because if we weigh $C$ against $B$ or $A$ there will be some chance of leaving two remaining positions possible for $C$, thus two times five remaining allowed orderings, which cannot be disambiguated in only three further weighings.
3a) Say the next choice (the 4th weighing) is to weigh $E$ versus $B$ (this is the obvious one to try because $D>B>A$). Then if we find $E<B$ we will end up knowing that $D>B>E>A$ or $D>B>A>E$. We spend our fifth weighing on $A$ versus $E$ to fully order those four. Since we know $D>C$ we can now find where $C$ belongs in only two more weighings (starting from weighing $C$ against the second lightest of the others), for a total of seven weighings.
Similarly, if we find $E>B$ we will end up knowing that $D>E>B>A$ or $E>D>B>A$.
We spend our fifth weighing on $D$ versus $E$ to fully order those four. And again we can properly place $$ in two more wiehings, for a total of seven.
(3b) If the fourth weighing were to be $E$ versus $A$ or $E$ versus $D$, then we would require eight weighings in the end.