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inysia [295]
3 years ago
6

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

ars to the sun, then calculate it
Biology
1 answer:
Vanyuwa [196]3 years ago
5 0
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}
</span>=\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=\frac{mv^2}{r}   
        where m=mass of mars = 6.39*10^(24) kg
=\frac{mv^2}{r}
=\frac{6.39*10^{24}v^2}{r}
=7.26168*10^9\pi^2r

The gravitation pull from the sun, Fg, is given by
Fg=\frac{GMm}{r^2}
    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}
=\frac{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=\frac{8.4826*10^44}{r^2}
Solving for the real root:
r^3=\frac{8.48256*10^44}{7.26168*10^9*%pi^2}
=\frac{1.1681263*10^{35}}{\pi^2}
=2.279*10^11 m


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