- These satellites are placed into orbit at a distance of around 35,800 km from the earth’s surface.
- They
**rotate in the same direction as the earth and one revolution of such satellites**is the same as one day on earth (roughly 24 hours). - This means that, as seen from earth, these satellites will appear to be at the same spot throughout. Hence, the name “geostationary” satellites.
- These satellites are used as communication satellites and for weather-based applications.

- Polar satellites revolve around the earth in a north-south direction around the earth as opposed to east-west like the
**geostationary satellites.** - They are very useful in applications where the field vision of the entire earth is required in a single day.
- Since the entire earth moves below them, this can be done easily.
- They are used in weather applications where predicting weather and climate-based disasters can be done in a short time.
- They are also used as
**relay stations.**

The International Space Station (ISS) was launched into orbit in 1998. It is a habitable artificial satellite and sometimes can be seen on nights with a clear sky. It functions as a lab, observatory, and landing base for possible expeditions. |

- The main thing one can understand about a satellite is that at the end of the day, they are
**projectiles.** - Any object, that only has the force of gravity acting upon it, is known as a satellite.
- The gravity’s force is the only thing that affects a satellite once it is launched into orbit.
- To understand this concept clearly, we will use the example of launching a satellite from the top of Newton’s Mountain which is a hypothetical location well above the influence of the drag force of the air.
**Newton**was the first scientist to give the concept that if an object is launched with the adequate speed it will start orbiting the earth. This object would experience a gravitational pull that would try to pull it downwards when it travels in a horizontal direction tangentially to the earth.- If the launch speed is slower than the escape velocity it will fall back to the earth.

- Lines A and B of the diagram represent those types of projectiles.
- If a projectile is shot off at an escape velocity with the perfect speed it will fall into an orbit outside the earth and will start revolving around the earth; the dotted line C represents such an object.
- If launched at a higher speed, the object will still revolve around the earth but will now have an elliptical orbit; the dotted line D represents such an object.
- It can also be possible that the object is shot at such a speed that it escapes the gravitational pull of the earth and become a free body; the solid line E represents such an object.
**The objects C and D**never fall back to the earth even though being pulled towards it continuously, as our earth is a circular body.

**Velocity Needed for an Object to Orbit the Earth in a Circular Pattern:**

- This entire observation raises a very basic question, that how much velocity is necessary for shooting a body out of the earth’s lower atmosphere and establishing it into the outer one still in the range of the gravitational force.
- We get the answer by observing the most basic aspect of the earth, measuring its curvature. It has been measured that for every 8000 meters that one goes along the horizon of the earth, the surface dips down by about 5 meters.
- Thus, applying basic mathematics we get the assumption that if a projectile wants to orbit around the sun, it will have to be at such a speed that it travels 8000m for every 5m of downward fall. It was observed that if an object is launched horizontally it will fall by around 5 meters in the first second.

Thus, we get to the conclusion that an object that is launched with a velocity of around 8000m/s will orbit the earth in a circular pattern. This is only applicable when the object experiences an insignificant amount of atmospheric drag.

- The launched object will travel at a speed of around 8000 meters in a second and will drop around 5 meters but as the earth is spherical and has a curvature that drops 5 meters every 8000 meters, the object will never touch the ground.

**SIMPLE PROBLEMS ON SATELLITES**

**PROBLEM 1 **

**Calculate the velocity of an artificial satellite orbiting the earth in a circular orbit ****at an altitude of 150 miles above the earth’s surface.**

SOLUTION

Given: r = (3,960 + 150) x 5,280 = 21,700,800 ft

Equation (1.6),

v = SQRT[ GM / r ]

v = SQRT[ 1.408×10^{16} / 21,700,800 ]

v = 25,470 ft/s

**PROBLEM 2 **

**An artificial earth satellite is in an elliptical orbit which brings it to an altitude of ****250 km at perigee and out to an altitude of 500 km at apogee. Calculate the velocity of ****the satellite at both perigee and apogee.**

SOLUTION

Given: Rp = (6,375 + 250) x 1,000 = 6,625,000 m

Ra = (6,375 + 500) x 1,000 = 6,875,000 m

Vp = SQRT[ 2 x GM x Ra / (Rp x (Ra + Rp)) ]

Vp = SQRT[ 2 x 3.986×10^{14} x 6,875,000 / (6,625,000 x (6,875,000 + 6,625,000)) ]

Vp = 7,828 m/s

Va = SQRT[ 2 x GM x Rp / (Ra x (Ra + Rp)) ]

Va = SQRT[ 2 x 3.986×10^{14} x 6,625,000 / (6,875,000 x (6,875,000 + 6,625,000)) ]

Va = 7,544 m/s

**PROBLEM 3**

**A satellite in earth orbit passes through its perigee point at an altitude of 200 km above the earth’s surface and at a velocity of 7,850 m/s. Calculate the apogee altitude of the satellite.**

SOLUTION,

Given: Rp = (6,375 + 200) x 1,000 = 6,575,000 m

Vp = 7,850 m/s

Equation (1.18),

Ra = Rp / [2 x GM / (Vp^{2} x Rp) – 1]

Ra = 6,575,000 / [2 x 3.986×10^{14} / (7,850^{2} x 6,575,000) – 1]

Ra = 6,795,000 m

Altitude @ apogee = 6,795,000 / 1,000 – 6,375 = 420 km

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