# Suppose one of the Global Positioning System satellites has a speed of 4.50km/s, at perigee, and...

## Question:

Suppose one of the Global Positioning System satellites has a speed of 4.50km/s, at perigee, and a speed of 3.52km/s, at apogee. If the distance from the center of the Earth, to the satellite at perigee, is 2.21x10{eq}^4 {/eq}km, what is the corresponding distance at apogee?

## Satellite Motion

When a satellite moves around the earth two quantities are conserved. One is the total energy. The second is the angular momentum with respect to the center of the earth. As a result, a satellite in a typical elliptical orbit with the earth at the focus, will move faster when it is close to the earth and slow down when it is farther. Moreover, the conservation of the angular momentum vector entails that the motion is restricted to a plane.

The angular momentum of a particle of linear momentum {eq}\displaystyle {\vec{p}} {/eq} with respect to a point is defined as,

{eq}\displaystyle {\vec{L}=\vec{r}\times \vec{p}} {/eq} .

Here {eq}\displaystyle {\vec{r}} {/eq} is the position vector of the particle with respect to the reference point and {eq}\displaystyle {\vec{p}=m\vec{v}} {/eq} where {eq}\displaystyle {\vec{v}} {/eq} is the velocity vector.

The angular momentum of the satellite with respect to the center of the earth must be conserved. At the apogee (point of maximum distance) and the perigee (point of minimum distance), the velocity is perpendicular to the separation vector drawn from the center of the earth to the satellite. Therefore the angular momentum is simply the product of the linear momentum and the separation. Since the angular momentum is a conserved quantity it must be the same at the apogee and the perigee. So we have,

{eq}\displaystyle {L_a=L_p} {/eq}

That is,

{eq}\displaystyle {mv_ar_a=mv_pr_p}---------(1) {/eq}.

Here it is given that the speed at the perigee, {eq}\displaystyle {v_p=4.5\ km/hr} {/eq} and that at the apogee is {eq}\displaystyle {v_a=3.52\ km/hr} {/eq}

The perigee is given to be,

{eq}\displaystyle {r_p=2.21\times 10^4\ km} {/eq}

Plugging this into (1), we get,

{eq}\displaystyle { 4.5\times 2.21\times 10^{10}=3.52\times r_a} {/eq}

Therefore,

{eq}\displaystyle { r_a=2.83\times 10^4\ kms} {/eq}.