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Solution :

(i) We employ `T^2 = K (R_E + h)^3` ( where `K = 4pi^(2)// GM_E`) with the Earth.s mass replaced by the Martian mass `M_m` <br> `T_2 = (4pi^2)/( GM_m) R^3` <br> `M_m = (4pi^2)/( G) (R^3)/( T^2)` <br> ` = (4 xx (3.14)^2 xx (9.4)^3 xx 10^(18))/( 6.67 xx 10^(-11) xx (459 xx 60)^2)` <br> `=6.48 xx 10^(23)` kg. <br> (ii) Once again Kepler.s third law comes to our aid, <br> `(T_M^2)/( T_E^2) = (R_(MS)^3)/( R_(ES)^3)` <br> Where `R_("MS")` is the Mars Sun distance and `R_("ES")` is the Earth-Sun distance. <br> `therefore T_M = (1.52)^(3//2) xx 365` <br> `=684` days <br> For example, the ratio of the semi-minor to semi-major axis for our Earth is, b/a `=0.99986`.