Understanding MS SQL Server Date Types

Question

Consider the following:

declare @dt datetime, @dt2 datetime2, @d date
set @dt  = '2013-01-01'
set @dt2 = '2013-01-01'
set @d   = '2013-01-01'

select convert(varbinary, @dt) as dt,
       convert(varbinary, @dt2) as dt2,
       convert(varbinary, @d) as d

Output:

dt                    dt2                     d
------------------    --------------------    --------
0x0000A13900000000    0x07000000000094360B    0x94360B

Now, I already understand from the documentation that datetime has a smaller range, and starts from 1753-01-01, while datetime2 and date use 0001-01-01 as their start date.

What I don't understand though, is that datetime appears to be little-endian while datetime2 and date are big-endian. If that's the case, how can they even be properly sortable?

Consider if I want to know how many integer days are represented by a date type. You would think you could do this:

declare @d date
set @d = '0001-01-31'
select cast(convert(varbinary, @d) as int)

But due to the endianness, you get 1966080 days!

To get the correct result of 30 days, you have to reverse it:

select cast(convert(varbinary,reverse(convert(varbinary, @d))) as int)

Or, of course you can do this:

select datediff(d,'0001-01-01', @d)

But that means internally somewhere it is reversing the bytes anyway.

So why did they switch endianness?

I only care because I'm working on a custom UDT in SQLCLR and the binary order of the bytes does seem to matter there, but these built-in types seem much more flexible. Does SQL Server have something internal where each type gets to provide it's own sorting algorithm? And if so, is there a way I can tap into that for my custom UDT?

See also, a related (but different) question on StackOverflow.