The Explorer VIII satellite, placed into orbit November 3, 1960, to investigate the ionosphere, had the following orbit parameters: perigee, 459 km; apogee, 2 289 km (both distances above the Earth's surface); period, 112.7 min. Find the ratio vp/va of the speed at perigee to that at apogee

Apogee: is the maximum distance for an object orbiting the Earth. For this case the distance for the apogee would be the sum of radius for the earth and 2289 km.

Perigee: Minimum distance of an object orbiting the Earth. For this case the distance for the perigee would be the sum of radius for the earth and 459 km.

A good approximation for these terms are related to the figure atached.

Isolated system: That happens when the system is not subdued to an external torque, on this case the change of angular momentum would be 0.

2) Notation and data

[tex]r_{earth}[/tex] radius for the Earth, looking for this value on a book we got [tex]r_{earth}=6.37x10^{3}km[/tex]

[tex]r_p =459km +r_{earth}=459km +6.37x10^3 km=6.829x10^3 km[/tex], represent the radius for perigee

[tex]r_a =2289km +r_{earth}=2289km +6.37x10^3 km=8.659x10^3 km[/tex], represent the radius for perigee

[tex]T=112.7min[/tex], period

[tex]v_p[/tex] velocity of the perigee

[tex]v_p[/tex] velocity of the apogee

[tex]r=\frac{v_p}{v_a}[/tex] is the variable of interest represented the ratio for the speed of the perigee to the speed of the apogee

3) Formulas to use

We don't have torque from the gravitational force since is a centered force. So the change of momentum is 0

[tex]\Delta L=0[/tex] (1)

[tex]L_a=L_p[/tex] (2)

From the definition of angular momentum we have [tex]L_i=mv_i r_i[/tex], replacing this into equation (2) we got:

[tex]mv_a r_a =mv_p r_p[/tex] (3)

We can cancel the mass on both sides

[tex]v_a r_a =v_p r_p[/tex] (4)

And solving [tex]\frac{v_p}{v_a}[/tex] we got:

[tex]\frac{v_p}{v_a}=\frac{r_a}{r_p}[/tex] (5)

And replacing the values obtained into equation (5) we got:

[tex]\frac{v_p}{v_a}=\frac{8.659x10^3 km}{6.829x10^3 km}=1.268[/tex]

And this would be the final answer [tex]\frac{v_p}{v_a}=1.268[/tex]