# Use Kepler s law to determine the period of a satellite that has an altitude of one earth radius...

## Question:

Use Kepler's law to determine the period of a satellite that has an altitude of one earth radius above the surface of the earth.

## Kepler's Third Law

The German astronomer and mathematician Johannes Kepler is known primarily for his three laws of motion of the planets in their orbit around the Sun. The Newtonian formulation of Kepler's third law allows us to describe the movement of satellites around the planet Earth.

The orbital motion of terrestrial satellites can be studied using Newton's formulation of Kepler's third law for circular orbits.

Kepler's third law:

{eq}\displaystyle \frac{T_{1}^{2}}{T_{2}^{2}} = \frac{r_{1}^{3}}{r_{2}^{3}} {/eq}

where {eq}T {/eq} is the period (time for one orbit) and r is the average radius (valid only for comparing two small masses orbiting the same large one).

For circular orbitals, using the concepts of centripetal force and universal gravitational force, we can obtain from Kepler's third law, the equation:

{eq}\displaystyle T = 2\cdot \pi \cdot \sqrt{\frac{r^{3}}{G\cdot M}} {/eq}

where:

• {eq}T {/eq} is the period of the satellite
• {eq}r {/eq} is the orbit radius
• {eq}G {/eq} niversal gravitational constant
• {eq}M {/eq} central body mass

In this exercise, we are asked to calculate the period of a satellite with an altitude ({eq}r = 6378\cdot 10^{3}~m {/eq}) above the earth's surface. From the above equation we can calculate the orbit period as follows:

{eq}\displaystyle T = 2\cdot \pi \cdot \sqrt{\frac{\left (6378\cdot 10^{3}~m \right )^{3}}{6.67\cdot 10^{-11}~m^{3}~kg^{-1}~s^{-2} \cdot 5.97\cdot 10^{24}~kg}} \\ \displaystyle T = 2\cdot \pi \cdot \sqrt{\frac{2.59\cdot 10^{20}~m ^{3}}{3.98\cdot 10^{14}~m^{3}~s^{-2}}} \\ \displaystyle T = 5065 ~s \sim 84 ~min {/eq}

Answer -) The period of the satellite is, approximately, {eq}\displaystyle \color{black}{ 84 ~min} {/eq} 