For telecommunications, satellites are better served by rotating around the earth at the same speed as the earth rotates

For telecommunications, satellites are better served by rotating around the earth at the same speed as the earth rotates


Asked by 3 years ago
398.7k points

For telecommunications, satellites are better served by rotating around the earth at the same speed as the earth rotates, or 24.0 h.  What is the distance from the center of the Earth that the satellite must be placed to achieve this geosynchronous orbit?  What is the orbital speed of the satellite?

For telecommunications
maddy

1 Answer

Answered by 3 years ago
389.5k points

Kepler's laws of planetary motion are as follows, 1. The orbit of every planet is an ellipse with the Sun at one of the two foci
2. A line drawn from the Sun to any planet sweeps out equal areas in equal time intervals 3. The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit, or

$$ T^2=\frac{4 \pi^2}{GM} a^3 $$

where T is the period of orbit, M is the mass which the object orbits, and a is the distance from the center of the orbit, G is the gravitation constant,

$ G = 6.67 \times 10^{-11} \, m^3 /\, kg\, s^2 $

this is especially useful for finding the mass of an object, once its period and average distance are known.

given,

$ M_E =5.98 \times 10^24 \, kg $ (mass of the earth)

$ G = 6.67 \times 10^{-11} \, m^3 / \,kg\, s^2 $

$ T = 24.0 \, h = 86400 \, s $

plugging these values into Kepler's third law, and solving for r give,

$ r = 4.23 \times 10^7 $

this is the distance from the center of the earth to the satellite.

The orbital speed of the satellite can be found by using,

$ v=\frac{d}{t}=\frac{2 \pi r}{t} = \frac{2 \pi (4.23 \times 10^7 \,m)}{86400 \,s}=3100 m/s $

or about 7000 mph.

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Asked: 3 years ago

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