Question

Synchronous communications satellites are placed in a circular orbit that is 3.59 107 m above the surface of the earth. What is the magnitude of the acceleration due to gravity at this distance?

Answer 1

Answer: $0.223 m/s^{2}$

Explanation:

We can solve this with the Law of Universal Gravitation and knowing the acceleration due gravity $g$ 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:

$F=G\frac{m_{E}m}{r^2}$  (1)

Where:

$F$ is the module of the force exerted between both bodies

$G=6.67(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the gravitational constant

$m_{E}=5.98(10)^{24} kg$ is the mass of the Earth

$m$ are the mass of each communications satellite

$r=R_{E}+h$ is the distance between the center of the Earth and the satellite

$R_{E}=6.38(10)^{6} m$ is the radius of the Earth

$h=3.59(10)^{7} m$ 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:

$F=mg$  (2)

Combining (1) and (2):

$G\frac{m_{E}m}{r^2}=mg$  (3)

Isolating $g$:

$g=\frac{G M_{E}}{r^2}$  (4)

Remembering $r=R_{E}+h$:

$g=\frac{G M_{E}}{(R_{E}+h)^2}$  (5)

$g=\frac{(6.67(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24} kg)}{(6.38(10)^{6} m+3.59(10)^{7} m)^2}$  

Finally:

$g=0.223 m/s^{2}$