Theoretically on day 0 you could jump at the following height on Mars (Y), if you can jump on the earth at height X.
Y = X*GravityOnMars/GravityOnEarth
Note above that it is simply a ratio of the force of gravities that dictates this. You could do the same with any planet-including earth itself.
Y = X*GravityOnEarth/GravityOnEarth=X
However, we need to include muscle deterioration.
Muscle deterioration will change with the following differential equation:
d MuscleStrength / dt = -k* MuscleStrength
Note that the solution of MuscleStrength as a function of time is:
MuscleStrength(t) = MuscleStrengthInitial*exp(-k*t)
Note at time 0 you have:
MuscleStrength(t=0)=MuscleStrengthInitial and at time infinite we have:
MuscleStrength(t=infinite)=MuscleStrengthInitial*exp(-k*infinite)= 0.
It makes sense, however, that our muscle strength on Mars would not go to zero. We thus need to correct for that. So we add a term for the baseline on Mars:
MuscleStrengthMarsBaseline
But we need to ensure that the initial muscle strength is not also including the baseline so that the MuscleStrength at time zero is higher than the initial value itself.
So:
MuscleStrength(t) = (MuscleStrengthInitial-MuscleStrengthMarsBaseline)*exp(-k*t)+ MuscleStrengthMarsBaseline
Let’s put it all together now:
Y = X*GravityOnMars* ((MuscleStrengthInitial-MuscleStrengthMarsBaseline)*exp(-k*t)+ MuscleStrengthMarsBaseline )/GravityOnEarth
Now you can calculate how high you can jump on Mars in time incorporating a muscle deterioration rate constant, k.
One thing that I would like to pony out regarding the deterioration rate constant is that if you were to fit a curve to the muscle deterioration rate constant with biexponential degradation (and it will definitely fit better than monoexponential) then you could use this k below:
k= k1*k2/(k1+k2)
The equation above combines both. I am not suggesting you should fit a biexponential curve to it. Monoexponential may be perfectly fantastic.
Best wishes,
Pharmacoengineering.com