\(g=-{g}_{ma}e^{-\frac{x}{{r}_{ma}}}+{g}_{p}e^{(-\frac{Orbs}{{r}_{p}}+\frac{x}{{r}_{p}})}\)
\(g\) = -0.003711*e^(-x/3390)+0.0000057*e^(x/11.27-9377/11.27) in kms-2.
A plot is shown below
At an orbital radius of 9377 km, phobos is still under the gravity influence of Mars. The gravity on the satellite, from the equation is -0.00023 kms-2. But based on measurements,
Phobos orbital speed 2*pi*9377/(7.66*60*60) = 2.137 kms-1
Centripetal force, \({F}_{ma}\) = (2.137)^2/9377 = -0.000487 kms-2.
Therefore a new estimate for the mass of Mars based on centripetal force on Mars orbit around the Sun is
9.942779860*10^(27)*0.00000254/ (0.007833 - 0.00000254) = 3.22518*10^(24) kg from
\({M}_{ma} = {M}_{s}. \frac{{F}_{ma} }{({g}_{ma}-{F}_{ma})}\),
where \({M}_{ma}\) is the mass of Mars and \({M}_{s}\) is the mass of the Sun.