Question

Mars had an orbital period of 1.88 years. In two or more complete sentences,explain how to calculate the average distance from mars to the sun, then calculate it

Answer 1

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

`v=\frac{2\pi{r}}{T}`

`=\frac{2\pi{r}}{1.88*365.25*86400}` m/s, accounting for leap years

`=3.371*10^(-8)\pi{r}` m/s

The centripetal force, Fc, generated is

`Fc=(mv^2)/(r)`
where m=mass of mars = 6.39*10^(24) kg

`=(mv^2)/(r)`

`=(6.39*10^(24)v^2)/(r)`

`=7.26168*10^9\pi^2r`

The gravitation pull from the sun, Fg, is given by

`Fg=(GMm)/(r^2)`
where G=grav. const., =6.67408*10^(-11) m^3 kg^(-1) s^(-2)
M=mass of sun=1.989*10^(30) kg

`=(6.67408*10^(-11)1.989*10^(30)6.39*10^(24))/(r^2)`

`=(8.4826*10^44)/(r^2)`

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=(8.4826*10^44)/(r^2)`
Solving for the real root:



`=(1.1681263*10^(35))/(\pi^2)`
=2.279*10^11 m