Answer
a) $L_p=4.7\times 10^{30}Kg.\frac{m^2}{s}$
b) $L_a=4.7\times 10^{30}Kg.\frac{m^2}{s}$
Work Step by Step
(a) The angular moment at perihelion is given as
$L_p=mv_pr_p$
We plug in the known values to obtain:
$L_p=9.8\times 10^{14}(54,600)(8.823\times 10^{10})$
$L_p=4.7\times 10^{30}Kg.\frac{m^2}{s}$
(b) The angular moment at aphelion is given as
$L_a=mv_ar_a$
We plug in the known values to obtain:
$L_a=9.8\times 10^{14}(783)(6.152\times 10^{12})$
$L_a=4.7\times 10^{30}Kg.\frac{m^2}{s}$