This is an interesting problem involving astronomy, in fact, simple physics.
Let r=distance of sun to mars, in metres
Mars had an orbital period of 1.88 years.
=>
tangential velocity, v, of the planet, in m/s is
[tex]v=\frac{2\pi{r}}{T}[/tex]
[tex]=\frac{2\pi{r}}{1.88*365.25*86400}[/tex] m/s, accounting for leap years
[tex]=3.371*10^{-8}\pi{r}[/tex] m/s
The centripetal force, Fc, generated is
[tex]Fc=\frac{mv^2}{r}[/tex]
where m=mass of mars = 6.39*10^(24) kg
[tex]=\frac{mv^2}{r}[/tex]
[tex]=\frac{6.39*10^{24}v^2}{r}[/tex]
[tex]=7.26168*10^9\pi^2r[/tex]
The gravitation pull from the sun, Fg, is given by
[tex]Fg=\frac{GMm}{r^2}[/tex]
where G=grav. const., =6.67408*10^(-11) m^3 kg^(-1) s^(-2)
M=mass of sun=1.989*10^(30) kg
[tex]=\frac{6.67408*10^{-11}1.989*10^{30}6.39*10^{24}}{r^2}[/tex]
[tex]=\frac{8.4826*10^44}{r^2}[/tex]
Since the radial distance is in equilibrium, the average distance, r can be found by equation Fc=Fg and solving for r:
Fc=Fg
=>
[tex]7.26168*10^9\pi^2r=\frac{8.4826*10^44}{r^2}[/tex]
Solving for the real root:
[tex]r^3=\frac{8.48256*10^44}{7.26168*10^9*%pi^2}[/tex]
[tex]=\frac{1.1681263*10^{35}}{\pi^2}[/tex]
=2.279*10^11 m
