“An astronaut is on board the International Space Station orbiting at an altitude of 4.00.10^2km above the Earth (the radius of the Earth is 6371km).

Calculate the orbital velocity and orbital period of the ISS, assume that g=9.81 m/s^2.”

I’ve calculated the orbital velocity as 8150m/s however I’m a little stuck at trying to calculate the orbital period. At first I decided to let the orbital period equal to time so using the angular velocity formula, time should be the change in angle divided by the angular velocity. I know that a full circle is 2 pi so the end formula to get my orbital period should be 2 pi divided by the angular velocity.

But I can’t seem to calculate the latter? Any clues on how to start?

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Renato Sinclair is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.

put on hold as off-topic by ACuriousMind Dec 9 at 1:26

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  • Hi and welcome to physics.SE! Please note that homework-like questions and check-my-work questions are generally considered off-topic here. We intend our questions to be potentially useful to a broader set of users than just the one asking, and prefer conceptual questions over those just asking for a specific computation. – ACuriousMind Dec 9 at 1:26
  • I’ve done that but I’m still getting a different answer from what it’s suposed to be – Renato Sinclair Dec 9 at 1:28

First of all your value for orbital velocity is incorrect.

Consider Newton's general gravitational equation:

Where $G$ is the gravitation constant, and $m_{1}$ and $m_{2}$ are your two bodies that you care about. In this case, they are the earth and the ISS, which will be denoted as $m_{e}$ and $m_{s}$, respectively.

To find the gravitational acceleration the earth produces on another body use $F=ma$


To find the acceleration of the satellite, divide through by $m_{s}$, obtaining


At this point we know that this acceleration is equal to the centripetal acceleration on the satellite. We can thus set this equal to the centripetal acceleration equation and solve for the velocity of the space station.


This yields the following:


Note that this distance $r$ is equal to the center-center distance.

To find out the time it takes for the space station to orbit once, use

$v=\frac{\delta x}{\delta t}$

Think about what your $\delta x$ is for a single orbit.

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Daniel Agramonte is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
  • I used the rearranged formula for the centripetal acceleration but I still get approximately 8150m. I got the square root of the gravitational acceleration multiplied by the radius of the earth plus the distance of the ISS to the Earth and yeah, it’s still the same. Am I doing something wrong? – Renato Sinclair Dec 9 at 1:21
  • You are correct in your radius, but G stands for the universal gravitational constant and $m_{e}$ stands for the mass of the earth. These are values you'll have to look up... – Daniel Agramonte Dec 9 at 1:26
  • Oh - I understand! Thank you very much! It’s just - this was question I missed during a lecture and a friend gave me notes on it and apparently, we were supposed to calculate it without looking up the mass of the earth. And her notes are telling me to divide the gravitational acceleration by the orbital velocity (which she calculated as 8150) to get the angular velocity? And she somehow got 1.20m rad/s and changed it into milli radians to get 1.20 x 10^-3 rad/s. So calculating the orbital period was just a matter of dividing 2 pi by that - except I don’t understand why. – Renato Sinclair Dec 9 at 1:35
  • I think your way is definitely easier but I’m a little paranoid about what I’d do in an exam if they’ve omitted some information, you know? Is it still possible to do it my mate’s way or has she just over complicated things? – Renato Sinclair Dec 9 at 1:37
  • 2
    Daniel as a newish user please read about the community policy for homework type questions. – StephenG Dec 9 at 11:33

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