Question

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.)

Answer 1

4216,22 m/s

Step-by-step explanation:

Linear speed can be calculated with:

V = 2πr / 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 *
`10^(3)` meters

6400 kilometers = 6400 *
`10^(3)` 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 *
`10^(3)` + 6400 *
`10^(3)` = 7650 *
`10^(3)` meters

T=190 minutes * 60 = 11400 seconds.

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

`V = (2*3,1415*7650*10^(3) )/(11400)`

V= 4216,22 m/s