So it is more likely that given average orbital speed and average orbital distance the we have a good estimate of the planet mass, by calculating centripetal force, \({F}_{m}\)
\({O}_{avg}\) = ( aphelion + perihelion )/2 = ( 249 209 300 +206669000 )/2 = 227939150 km
\({F}_{m} = \cfrac{{V}^{2}_{avg}}{{O}_{avg}}\) = ((24.077)^2)/227939150 = 0.00000254 kms-2
And we move on to orbital speed at the aphelion,
(249 209 300*0.00000254 )^(0.5) = 25.15 kms-1
Orbital speed at the perihelion,
(206 669 000*0.00000254 )^(0.5) = 22.91 kms-1
And the orbital speed from considering \(\triangle GPE\) alone is,
\( v =\sqrt{3{g}_{m}{r}_{m}} \) = (3*0.003711*3390 )^(0.5) = 6.143 kms-1
The actual Mars orbital speed is measured as 24.077 kms-1
In a similar way based on the model for Earth where the final orbital velocity is the sum of two rotating velocities, \({V}_{m1}\) and \({V}_{m2}\), and Mars has a initial velocity \({V}_{i}\) when it was captured by the Sun,
\({V}_{m1} - {V}_{m2}\) = 25.15 kms-1
\({V}_{m1} + {V}_{m2}\) = 22.91 kms-1
\({V}_{m1}\) = ( 25.15 + 22.91 )/2 = 24.03 kms-1 and
\({V}_{m2}\) = 25.15 - 24.03 = 1.12 kms-1
The parallel component of the final orbiting velocity is the sum of the velocity the body would have considering a decrease in GPE alone and the parallel component of the initial approach velocity. To find the parallel initial velocity component,
\({V}_{m1}\) = \(\sqrt{2*\triangle GPE} + {V}_{ip} \) = 6.143 + \({V}_{ip}\) = 24.03 kms-1
\({V}_{ip}\) = 24.03-6.143 = 17.89 kms-1
The second smaller rotating velocity is the perpendicular component of the initial velocity, \({V}_{i}\).
\({V}_{ir}\) = 1.12 kms-1
\({V}_{i} =\sqrt{{V}^{2}_{ip} + {V}^{2}_{ir}}\) = ((1.12)^2 + (17.89)^2)^(0.5) = 17.92 kms-1
So, Mars approached the Sun with an initial velocity \({V}_{i}\) = 17.92 kms-1 and was captured by the Sun.
