An orbiting space station is observed to remain always vertically above the same point on the earth. Where on earth is the observer? Describe the orbit of the space station as completely as possible (radius).

I considered the orbit ideally circular. Therefore, we can consider the translational velocity as follows v=ωR=2πR/T We know the space station has two types of acceleration while it orbits: centripetal and tangential. We can assure the following: R= space station orbit’s radius. R’= Earth’s radius Fg=ma(centripetal)=mv^2/R=GMm/(R^2). Therefore: v^2=GM/R We know as well: mg=GMm/(R’^2). Therefore: GM=g(R’^2) Therefore: v^2=g(R’^2)/R Therefore: (2πR/T)^2=g(R’^2)/R Therefore: R^3=(g(R’^2)(T^2))/4(π^2) My doubt is if I can improve my answer. Is it possible describe better the orbit of the space station? Thanks

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The math is correct. Just put in the all the data for the time period as the time period of rotation of the earth and the value of $g$ at the equator and you'll get the orbital radius. The reason we are taking the equator is that it is suitable to launch a geostationary satellite orbiting in a circular path which is concentric to the equatorial circle. Any other circular path will be geosynchronous and not geostationary meaning that it'll come to the same spot within a day but would not seem as if remaining in the same spot all the time. The geostationary satellite if observed from the equator will always remain stationary.

Tapas Mazumdar - 3 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$