Determining distance from semi-major axis and eccentricity I am trying to obtain the distance covered by an object in orbit around the earth within a specified amount of time past its perigee passage. The object is in an elliptical orbit with eccentricity 0.2 and has a semi-major axis of 9600km. Where would I look to find the objects position 90 mins after it passes its perigee? Thanks for any advice
Supposing that the mass of the object is negligible compared with the mass of the Earth, you can derive the orbital period $T$ from the 3rd Keplero's law:

$\frac{T^2}{a^3} = \frac{4\pi^2}{G(m_E + m_b)} \approx \frac{4\pi^2}{Gm_E},$

where $a$ is the semi-major. With $T$, for each time istant you also know the mean anomaly $M$, given by (suppose $t = 0$ at perigee):

$M(t) = \frac{2\pi}{T}t$.

Solving numerically the Keplero's equation for the eccentric anomaly $E$ (where $e$ is the eccentricity)

$M = E - e\sin E$

and then use the following equation to derive the true anonaly $\nu$, which is the angle between the direction of periapsis and the current position of the body, as seen from the Earth:

$ \cos \nu = \frac{\cos E - e}{1 - e\cos E}$
and
$ \sin \nu = \frac{\sqrt{1-e^2}\sin E}{1 - e\cos E}$.

The distance from the Earth is just given by the orbit equation

$ r = \frac{a(1-e^2)}{1 + e\cos \nu}$.

If I'm not wrong with calculation, it should be:

$ T = 9364 s = 2.6 h.$

$ M(90min) = 207.60°$

$ E(90min) = 203.11°$

$ \nu(90min) = 198.95°$

$ r = 11'366 km$

To get the covered distance you should calculate the line integral of the orbit equation.