Answer: [tex]0.223 m/s^{2}[/tex]
Explanation:
We can solve this with the Law of Universal Gravitation and knowing the acceleration due gravity [tex]g[/tex] of an object above the surface of the planet decreases with the distance (height) of this object from the center of the planet.
Well, according to the law of universal gravitation:
[tex]F=G\frac{m_{E}m}{r^2}[/tex] (1)
Where:
[tex]F[/tex] is the module of the force exerted between both bodies
[tex]G=6.67(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex] is the gravitational constant
[tex]m_{E}=5.98(10)^{24} kg[/tex] is the mass of the Earth
[tex]m[/tex] are the mass of each communications satellite
[tex]r=R_{E}+h[/tex] is the distance between the center of the Earth and the satellite
[tex]R_{E}=6.38(10)^{6} m[/tex] is the radius of the Earth
[tex]h=3.59(10)^{7} m[/tex] is the height of the satellite, measured from the Earth's surface
On the other hand, we know according to Newton's 2nd law of motion:
[tex]F=mg[/tex] (2)
Combining (1) and (2):
[tex]G\frac{m_{E}m}{r^2}=mg[/tex] (3)
Isolating [tex]g[/tex]:
[tex]g=\frac{G M_{E}}{r^2}[/tex] (4)
Remembering [tex]r=R_{E}+h[/tex]:
[tex]g=\frac{G M_{E}}{(R_{E}+h)^2}[/tex] (5)
[tex]g=\frac{(6.67(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24} kg)}{(6.38(10)^{6} m+3.59(10)^{7} m)^2}[/tex]
Finally:
[tex]g=0.223 m/s^{2}[/tex]
