Answer
The angular speed at perihelion is $~~9.24\times 10^{-5}~rad/s$
Work Step by Step
We can find the semimajor axis:
$T^2 = \frac{4\pi^2~a^3}{GM}$
$a^3 = \frac{T^2~G~M}{4\pi^2}$
$a = \sqrt[3] {\frac{T^2~G~M}{4\pi^2}}$
$a = \sqrt[3] {\frac{(8.00\times 10^4~s)^2~(6.67\times 10^{-11}~N~m^2/kg^2)~(7.00\times 10^{24}~kg)}{4\pi^2}}$
$a = 4.23\times 10^7~m$
We can find the distance from the planet at perihelion:
$r_p = 2a-r_a$
$r_p = (2)(4.23\times 10^7~m)-(4.5\times 10^7~m)$
$r_p = 3.96\times 10^7~m$
We can use conservation of angular momentum to find the angular speed at perihelion:
$m~v_p~r_p = m~v_a~r_a$
$\omega_p~r_p^2 = \omega_a~r_a^2$
$\omega_p = \frac{\omega_a~r_a^2}{r_p^2}$
$\omega_p = \frac{(7.158\times 10^{-5}~rad/s)~(4.5\times 10^7~m)^2}{(3.96\times 10^7~m)^2}$
$\omega_p = 9.24\times 10^{-5}~rad/s$
The angular speed at perihelion is $~~9.24\times 10^{-5}~rad/s$
