$T^{2}=\frac{4 \pi^{2}}{G M_{m}} R^{3}$

$M _{m}=\frac{4 \pi^{2}}{G} \frac{R^{3}}{T^{2}}$

$=\frac{4 \times(3.14)^{2} \times(9.4)^{3} \times 10^{18}}{6.67 \times 10^{-11} \times(459 \times 60)^{2}}$

$M _{m}=\frac{4 \times(3.14)^{2} \times(9.4)^{3} \times 10^{18}}{6.67 \times(4.59 \times 6)^{2} \times 10^{-5}}$

$=6.48 \times 10^{23} kg$

$(ii)$ Once again Kepler's third law comes to our aid.

$\frac{T_{M}^{2}}{T_{E}^{2}}=\frac{R_{M S}^{3}}{R_{E S}^{3}}$

where $R_{\text {us }}$ is the mars -sun distance and $R_{E S}$ is the earth-sun distance.

$\therefore T_{M}=(1.52)^{3 / 2} \times 365$

$=684$ days

We note that the orbits of all planets except Mercury, Mars and Pluto* are very close to being circular. For example, the ratio of the semi-minor to semi-major axis for our Earth is, $b / a=0.99986$