Wednesday, May 7, 2014

Phobos

Phobos is close to Mars at 9377 km and has a radius of 11.27 km.  An equation for the gravity profile is

\(g=-{g}_{ma}e^{-\frac{x}{{r}_{ma}}}+{g}_{p}e^{(-\frac{Orbs}{{r}_{p}}+\frac{x}{{r}_{p}})}\)

\(g\) = -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.

The negative sign indicates that it is pointed towards Mars. The minimum value of that part of the curve is -0.00023; the theoretical value and calculated value will tally only if the gravity of Mars is increased to \({g}_{ma}\) = 0.007833 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.