Answer
The difference in the periods is $~~1.5\times 10^3~s$
Work Step by Step
In part (a), we found that the period of the circular orbit is $~~2.15\times 10^4~s$
In part (g) we found that the semimajor axis of the elliptical orbit is $~~a = 4.01\times 10^7~m$
We can find the period of the elliptical orbit:
$T^2 = \frac{4\pi^2~a^3}{GM}$
$T = \sqrt{\frac{4\pi^2~a^3}{GM}}$
$T = \sqrt{\frac{(4\pi^2)(4.01\times 10^7~m)^3}{(6.67\times 10^{-11}~N~m^2/kg^2)(9.50\times 10^{25}~kg)}}$
$T = 2.00\times 10^4~s$
We can find the difference in the periods:
$\Delta T = (2.15\times 10^4~s)-(2.00\times 10^4~s)$
$\Delta T = 1.5\times 10^3~s$
The difference in the periods is $~~1.5\times 10^3~s$
