Question

Earth is 149.6 million meters from the Sun and takes 365 days to make one complete revolution around the Sun. Mars is 227.9 million meters from the Sun and has an orbital period of 687 days. What is the ratio of Earth’s centripetal acceleration to Mars’s centripetal acceleration

Answer 1

Answer:

$\frac{a_{r,earth}}{a_{r,mars}} = 2.325$

Explanation:

The distance of Earth from the Sun is $149.6\times 10^{9}\,m$ and of Mars from the Sun is $227.9\times 10^{9}\,m$. Let assume that both planets have circular orbits. The centripetal accelaration can be found by using the following expression:

$a_{r} = \frac{v^{2}}{R}$

Since planet has translation at constant speed, this formula is applied to compute corresponding speeds:

$v=\frac{2\pi\cdot r}{\Delta t}$

Earth:

$v_{earth} = \frac{2\pi\cdot (149.6\times 10^{9}\,m)}{(365\,days)\cdot(\frac{24\,hours}{1\,day} )\cdot(\frac{3600\,s}{1\,h} )}$

$v_{earth}=29806.079\,\frac{m}{s}$

Mars:

$v_{mars} = \frac{2\pi\cdot (227.9\times 10^{9}\,m)}{(687\,days)\cdot(\frac{24\,hours}{1\,day} )\cdot(\frac{3600\,s}{1\,h} )}$

$v_{mars}=24124.244\,\frac{m}{s}$

Now, centripetal accelarations can be found:

Earth:

$a_{r,earth} = \frac{(29806.079\,\frac{m}{s} )^{2}}{149.6\times 10^{9}\,m}$

$a_{r,earth} = 5.939\times 10^{-3}\,\frac{m}{s^{2}}$

Mars:

$a_{r,mars} = \frac{(24124.244\,\frac{m}{s} )^{2}}{227.9\times 10^{9}\,m}$

$a_{r,mars} = 2.554\times 10^{-3}\,\frac{m}{s^{2}}$

The ratio of Earth's centripetal acceleration to Mars's centripetal acceleration is:

$\frac{a_{r,earth}}{a_{r,mars}} = \frac{5.939}{2.554}$

$\frac{a_{r,earth}}{a_{r,mars}} = 2.325$