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430500/66300A satellite in a circular orbit 1250 kilometers above Earth makes one complete revolution every 190 minutes. What is its linear speed? Assume that Earth is a sphere of radius 6400 kilometers. (Round your answer to two decimal places.)

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Answer:

4216,22 m/s

Explanation:

Linear speed can be calculated with:

V = 2Ï€r / T

Where  V is linear speed, r is radius of the circular path which the object takes and T is period of the rotational motion. We want to calculate the speed in SI units, so:

1250 kilometers = 1250  * [tex]10^{3}[/tex]  meters

6400 kilometers = 6400 * [tex]10^{3}[/tex] meters

We will take r as the sum of radius of the Earth and height of the satellite because we need the distance between object and its centre of rotation. (Figure 1).  

r= 1250 * [tex]10^{3}[/tex]  + 6400 *  [tex]10^{3}[/tex] = 7650 *  [tex]10^{3}[/tex] meters

T=190 minutes * 60 = 11400 seconds.  

We will take pi as   3.1415, let us place values to the equation:

[tex]V = \frac{2*3,1415*7650*10^{3} }{11400}[/tex]

V= 4216,22  m/s

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