Answer
a) less than $54.6Km/s$
b) $v_a=783m/s$
Work Step by Step
(a) We know that when the comet travels from perigee to apogee, it loses kinetic energy while gaining gravitational potential energy. Hence, we conclude that its speed is less than $54.6Km/s.$
(b) We can find the required speed as follows:
$v_a^2=v_p^2+2GM(\frac{1}{r_a}-\frac{1}{r_p})$
$\implies v_a=\sqrt{v_p^2+2GM(\frac{1}{r_a}-\frac{1}{r_p})}$
We plug in the known values to obtain:
$v_a=\sqrt{(54.6\times 10^3)^2+2(6.67\times 10^{-11})(2.00\times 10^{30})(\frac{1}{6.152\times 10^{12}}-\frac{1}{8.823\times 10^{10}})}$
$v_a=783m/s$
