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Use the ratio form of Kepler’s third law, [tex]( \frac{Ta}{Tb} )^2[/tex] = [tex]( \frac{Da}{Db} )^2[/tex], and the data provided to determine the time it takes Mars to orbit the Sun. Round your answer to the nearest tenth.
Earth’s orbital period = 1.0
Earth year Earth’s distance from the Sun = 1.0 AU
Mars’s distance from the Sun = 1.5 AU

Respuesta :

the awnser is that it takes 1.8 earth years for mars to orbit the sun

Answer: Time period taken by Mars to orbit round Sun is 1.8 earth years

Explanation:

Earth’s orbital period =[tex]P_E= 1.0[/tex]  Earth year

Earth’s distance from the Sun =[tex]a_E =1.0 [/tex]AU

Mars’s orbital period = ? [tex]=P_M[/tex]

Mars’s distance from the Sun =[tex]a_M= 1.5 [/tex]AU

According to Kepler third law:

[tex]P^2\propto a^3[/tex]

P = time period taken by the planet to orbit Sun

a = semi major axis between the planet's orbit

So, we can rewrite it as:

[tex]\frac{(P_E)^2}{(a_E)^3}=\frac{(P_M)^2}{(a_M)^3}[/tex]

[tex]P_M=1.83\approx 1.8[/tex] earth years

Time period taken by Mars to orbit round Sun is 1.8 earth years

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