Much of the Florida landscape is built upon limestone. Frequent rains seep down and erode the limestone as the rainwater travels
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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
</span>
m/s, accounting for leap years
m/s
The centripetal force, Fc, generated is
where m=mass of mars = 6.39*10^(24) kg
The gravitation pull from the sun, Fg, is given by
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
Since the radial distance is in equilibrium, the average distance, r can be found by equation Fc=Fg and solving for r:
Fc=Fg
=>
Solving for the real root:
=2.279*10^11 m
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
</span>
The centripetal force, Fc, generated is
where m=mass of mars = 6.39*10^(24) kg
The gravitation pull from the sun, Fg, is given by
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
Since the radial distance is in equilibrium, the average distance, r can be found by equation Fc=Fg and solving for r:
Fc=Fg
=>
Solving for the real root:
=2.279*10^11 m