Assuming that earth and mars move in circular orbits around the Sun, with the Martian orbit being 1.52 times the orbital radius of the earth. The length of the Martian year in days is

Question

Assuming that earth and mars move in circular orbits around the Sun, with the Martian orbit being 1.52 times the orbital radius of the earth. The length of the Martian year in days is (A) (1.52)(2/3) × 365 (B) (1.52)(3/2) × 365. (C) (1.52)2 × 365 (D) (1.52)3 × 365

Tardigrade Admin · Accepted Answer

Using Kepler's IIIrd law, (TM2/TE2)=(RM s3/RE3) ⇒ TM=((RM s/RE s))(3/2) × TE=(1.52)3 / 2 × 365 days.

Q. Assuming that earth and mars move in circular orbits around the Sun, with the Martian orbit being $1.52$ times the orbital radius of the earth. The length of the Martian year in days is

$(1.52)^{\frac{2}{3}} \times 365$

$(1.52)^{\frac{3}{2}} \times 365$ .

$(1.52)^{2} \times 365$

$(1.52)^{3} \times 365$

Using Kepler's $II I^{r d}$ law, $\frac{T _{M}^{2}}{T _{E}^{2}} = \frac{R _{M s}^{3}}{R _{E}^{3}}$
$\Rightarrow T_{M} = (\frac{R _{M s}}{R _{E s}})^{\frac{3}{2}} \times T_{E} = (1.52)^{3/2} \times 365$ days.

Q. Assuming that earth and mars move in circular orbits around the Sun, with the Martian orbit being 1.52 times the orbital radius of the earth. The length of the Martian year in days is

(1.52)32​×365

(1.52)23​×365.

(1.52)2×365

(1.52)3×365