This is an interesting problem involving astronomy, in fact, simple physics.
Let r=distance of sun to mars, in metres
<span>Mars had an orbital period of 1.88 years.
=>
tangential velocity, v, of the planet, in m/s is
![v=\frac{2\pi{r}}{T}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B2%5Cpi%7Br%7D%7D%7BT%7D)
</span>
![=\frac{2\pi{r}}{1.88*365.25*86400}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B2%5Cpi%7Br%7D%7D%7B1.88%2A365.25%2A86400%7D)
m/s, accounting for leap years
![=3.371*10^{-8}\pi{r}](https://tex.z-dn.net/?f=%3D3.371%2A10%5E%7B-8%7D%5Cpi%7Br%7D)
m/s
The centripetal force, Fc, generated is
where m=mass of mars = 6.39*10^(24) kg
![=\frac{mv^2}{r}](https://tex.z-dn.net/?f=%3D%5Cfrac%7Bmv%5E2%7D%7Br%7D)
![=\frac{6.39*10^{24}v^2}{r}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B6.39%2A10%5E%7B24%7Dv%5E2%7D%7Br%7D)
![=7.26168*10^9\pi^2r](https://tex.z-dn.net/?f=%3D7.26168%2A10%5E9%5Cpi%5E2r)
The gravitation pull from the sun, Fg, is given by
![Fg=\frac{GMm}{r^2}](https://tex.z-dn.net/?f=Fg%3D%5Cfrac%7BGMm%7D%7Br%5E2%7D)
where G=grav. const., =6.67408*10^(-11) m^3<span> kg^(</span>-1)<span> s^(</span>-2)
M=mass of sun=1.989*10^(30) kg
![=\frac{6.67408*10^{-11}1.989*10^{30}6.39*10^{24}}{r^2}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B6.67408%2A10%5E%7B-11%7D1.989%2A10%5E%7B30%7D6.39%2A10%5E%7B24%7D%7D%7Br%5E2%7D)
![=\frac{8.4826*10^44}{r^2}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B8.4826%2A10%5E44%7D%7Br%5E2%7D)
Since the radial distance is in equilibrium, the average distance, r can be found by equation Fc=Fg and solving for r:
Fc=Fg
=>
![7.26168*10^9\pi^2r=\frac{8.4826*10^44}{r^2}](https://tex.z-dn.net/?f=7.26168%2A10%5E9%5Cpi%5E2r%3D%5Cfrac%7B8.4826%2A10%5E44%7D%7Br%5E2%7D)
Solving for the real root:
![r^3=\frac{8.48256*10^44}{7.26168*10^9*%pi^2}](https://tex.z-dn.net/?f=r%5E3%3D%5Cfrac%7B8.48256%2A10%5E44%7D%7B7.26168%2A10%5E9%2A%25pi%5E2%7D)
![=\frac{1.1681263*10^{35}}{\pi^2}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1.1681263%2A10%5E%7B35%7D%7D%7B%5Cpi%5E2%7D)
=
2.279*10^11 m