Question

A satellite is launched to orbit the Earth at an altitude of 2.20 * 10^7 m for use in the Global Positioning system (GPS). Take the mass of the Earth to be 5.97 * 10^24 kg and its radius 6.38 * 10^6 m. (a) What is the orbital period of this GPS satellite? (b) With what speed does it orbit the Earth?

Answer #1

Period of the satellite is given by,

t^2 = (4π^2* r^3)/(G * Me)

where

G = 6.67 * 10^-11 m^3/(kg-s^2),

Me = 5.97 * 10^24 kg and

r = 6.38 * 10^6 m + 2.20 * 10^7 m = 2.838 * 10^7 m

Substitung Values,

t^2 = (4*π^2* (2.838 * 10^7)^3)/(6.67 * 10^-11 * 5.97 *
10^24)

**t = 47604.5 s**

Converting to days,

**t = 0.55 days**

b)

velocity of the satellite is given by,

v^2 = G(Me)/r

v^2 = (6.67 * 10^-11 * 5.97 * 10^24)/(2.838 * 10^7)

**v = 3745.8 m/s
v = 3.746 km/s**

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