Johannes Kepler Research Paper

Task 1

Part A

In 1599 Tycho Brahe, a Danish astronomer, had called for Johannes Kepler, a clever mathematician, to become his assistant. Kepler embraced the heliocentric model of the universe and created a desire to prove it mathematically. When Brahe died in 1601, Kepler inherited his work and data on the motion on planets. Kepler worked for many years, using Brahe’s data and Copernicus’s ideas, how planets move about the sun, and he eventually produced an improved heliocentric model of the universe. His model stated that planets rotate around the sun, and in ellipses. He proved this using his three laws (Kahni Burrows, 2008).

Kepler’s first law: The law of ellipses

Kepler’s first law states that planets move in an ellipse with the sun

His reasoning behind this was that if you threw a rock it would go some distance before landing on the ground, but if you throw it fast and hard enough it could go into orbit of the earth and follow its gravitational force as it goes from one side to the other. So using this we can see that for any given altitude there is a specific orbital velocity required for any object to achieve a stable circular orbit. Applying Newton’s law of universal gravitation to the orbital motion of a satellite allows us to find a formula for orbital velocity.

This can be shown as: Fg=G mems/r^2 where me is the mass of the Earth (5.97 ×〖10〗^24 kg), ms is the mass of the satellite (kg), r is the radius of the orbit (m), and G is the universal gravitational constant (6.67×〖10〗^(-11) Nm^2 kg^(-2)). This gravitational force also serves as the centripetal force for the circular orbital motion, hence: Fg = Fc. Therefore we can equate the formula for Fg with that for Fc: G mems/r^2 =(msv^2)/r therefore v=√Gme/r, note that the radius of the orbit r is the sum of the Earth and the altitude of the orbit. v=√Gme/(re+altitude) Where v is the orbital velocity (ms^(-1)) G is the universal gravitation constant (6.67×〖10〗^(-11) Nm^2 kg^(-2)), me is the mass of the Earth (5.97 ×〖10〗^24 kg), re is the radius of the Earth (6.38×〖10〗^6 m), and altitude is the height of the orbit above the ground

If the orbit is over the equator, it allows a satellite to stay over a fixed point on the surface of the Earth throughout the day and night. From the Earth such a satellite appears to be stationary in the sky, regardless of time of day. This is particularly useful for communications satellites, because a receiving dish need only point at a fixed spot in the sky in order to remain in contact with the satellite. This can be calculated from Kepler’s law of periods.

If a satellite at this height is not positioned over the equator, but at some other latitude, it will not remain fixed at one point in the sky. Instead, the satellite will appear to trace out a figure 8 path each 24 hours. It still has a period of 24 hours, so this orbit is referred to as geosynchronous.

Part B

As stated before, geostationary orbit is useful for communications satellites. This is because a receiving dish will only need to point at one continuous spot instead of having to find another link every hour or so. Ground antennas can be aimed towards the satellite without having to follow the satellite motion, this makes it cheaper to make them as they can require less parts. This type of orbit can also decrease the amount of money needed on things like ground equipment (Wikipedia,