Answer:
[tex]\frac{a_{r,earth}}{a_{r,mars}} = 2.325[/tex]
Explanation:
The distance of Earth from the Sun is [tex]149.6\times 10^{9}\,m[/tex] and of Mars from the Sun is [tex]227.9\times 10^{9}\,m[/tex]. Let assume that both planets have circular orbits. The centripetal accelaration can be found by using the following expression:
[tex]a_{r} = \frac{v^{2}}{R}[/tex]
Since planet has translation at constant speed, this formula is applied to compute corresponding speeds:
[tex]v=\frac{2\pi\cdot r}{\Delta t}[/tex]
Earth:
[tex]v_{earth} = \frac{2\pi\cdot (149.6\times 10^{9}\,m)}{(365\,days)\cdot(\frac{24\,hours}{1\,day} )\cdot(\frac{3600\,s}{1\,h} )}[/tex]
[tex]v_{earth}=29806.079\,\frac{m}{s}[/tex]
Mars:
[tex]v_{mars} = \frac{2\pi\cdot (227.9\times 10^{9}\,m)}{(687\,days)\cdot(\frac{24\,hours}{1\,day} )\cdot(\frac{3600\,s}{1\,h} )}[/tex]
[tex]v_{mars}=24124.244\,\frac{m}{s}[/tex]
Now, centripetal accelarations can be found:
Earth:
[tex]a_{r,earth} = \frac{(29806.079\,\frac{m}{s} )^{2}}{149.6\times 10^{9}\,m}[/tex]
[tex]a_{r,earth} = 5.939\times 10^{-3}\,\frac{m}{s^{2}}[/tex]
Mars:
[tex]a_{r,mars} = \frac{(24124.244\,\frac{m}{s} )^{2}}{227.9\times 10^{9}\,m}[/tex]
[tex]a_{r,mars} = 2.554\times 10^{-3}\,\frac{m}{s^{2}}[/tex]
The ratio of Earth's centripetal acceleration to Mars's centripetal acceleration is:
[tex]\frac{a_{r,earth}}{a_{r,mars}} = \frac{5.939}{2.554}[/tex]
[tex]\frac{a_{r,earth}}{a_{r,mars}} = 2.325[/tex]
