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Review. A global positioning system (GPS) satellite moves in a circular orbit with period 11h 58 min.(b) Determine its speed.

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id1001265

The global positioning system (GPS) satellite that moves in a circular orbit with period of 11h 58 min has a speed of: 3872.9848 m/s

To solve this problem the formulas and the procedures we will use are:

  • r =[(G* m *T²)/(4 * π²)]
  • v = (2*π * r)/ T

Where:

  • r = orbital radius
  • G = Gravitational constant
  • m = Mass of the Earth
  • T = Time period of the satellite
  • π = mathematical constant
  • v = tangential velocity

Information about the problem:

  • T = 11h 58 min
  • G = 6.67 × 10⁻¹¹ m³/kgs²
  • m= 5.972 × 10²⁴ kg
  • π = 3.1416
  • r =?
  • v =?

By converting the time period of (h) and (min) from (s) we have:

T = (11h *3600 s/ 1 h) + (58 min * 60 s/1 min)

T = 39600 s +3480 s

T = 43080 s

Applying the orbital radius formula we get:

r =[(G* m *T²)/(4 * π²)]

r =[(6.67 × 10⁻¹¹ m³/kgs² * 5.972 × 10²⁴ kg *(43080 s)²)/(4 * (3.1416)²)]

r =[1.8725*10^22 m³]

r= 26.55465166*10^6 m

Applying the tangential velocity formula we get:

v = (2*π * r)/ T

v = (2 * 3.1416 * 26.55465166*10^6 m)/ 43080 s

v = 3872.9848 m/s

What is the orbital radius?

It is the maximum of the radial distribution curve of the outermost orbital.

Learn more about orbital radius at: brainly.com/question/14437688

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