What is the true weight of an object in a geostationary satellite that weighed exactly $10 \cdot 0 \

Question

What is the true weight of an object in a geostationary satellite that weighed exactly $10 \cdot 0 \mathrm{~N}$ at the north pole?

Professor PC Thomas Classes · Answer

The true weight of an object at the North Pole is purely due to Earth's gravitational pull (since there is no centrifugal force from rotation at the poles):
$
W=\frac{G M m}{R^2}=10.0 \mathrm{~N}
$
For a geostationary satellite, the orbital radius $r$ from the center of the Earth is approximately 42400 km (radius of Earth $R \approx 6400 \mathrm{~km}+$ height $h \approx 36000 \mathrm{~km}$ ).
The "true weight" in orbit is the actual gravitational force acting on the object at that distance (not the apparent weight, which is zero due to free fall).
$
\begin{gathered}
W^{\prime}=\frac{G M m}{r^2}=W\left(\frac{R}{r}ight)^2 \
W^{\prime}=10.0 	imes\left(\frac{6400}{42400}ight)^2 \
W^{\prime}=10.0 	imes\left(\frac{8}{53}ight)^2 \approx 0.228 \mathrm{~N}
\end{gathered}
$
Rounding it off, the true weight is approximately 0.23 N .