Why does an aircraft descend quicker when it is lighter?

[jimmygill]

For an equilibrium flight

RoD = TAS / (L/D) (For idle power glide)

This is derived not from aerodynamics but from simple consideration of equilibrium.

Code:

RoD@Vmp = Vmp / (L/D@Vmp)

Rod@Vmd = Vmd / (L/D@Vmd)

RoD@Vxx = Vxx / (L/D@Vxx)  where Vxx is a speed greater than Vmd

1. Vmd > Vmp also L/D@Vmd > L/D@Vmp
   The increase in the numerator (TAS goes up from Vmp to Vmd) more than compensates for increase in the denominator (L/D)
   
   Hence we get RoD@Vmd > RoD@Vmp.
   
   
   
   
2. Vxx > Vmd, but L/D@Vxx < L/D@Vmd in order for RoD@Vxx > RoD@Vmp
   
   Here increase in the numerator and decrease in the denominator both favor an overall increase in RoD
   
   
3. In considering a speed limited descend the case to case variation in TAS is not permitted.
   Hence for cases where TAS is kept constant RoD is purely a function of the ratio L/D.
   
   This is the case for a speed restricted descent such as in a 777.
   For different weights and same TAS, the 777 will have different angle of attacks in a descent.
   
   The heavier 777 will always have higher AoA.
   But a higher AoA doesn't always translate into a better L/D, in some cases of TAS, an increase in AoA may worsen the L/D. The later cases are the ones which can't be explained by the argument of high momentum, high energy etc.. which prompts me to reject such arguments.
   
   Also there are several angle of attacks "or speeds" for which a certain increase in AoA will lead to no change in value of L/D.
   In such cases the changed in RoD is entirely due to change in the TAS of descend.