Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: philthewineguy on 03/05/2021 15:52:09
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Given the radius of earth is 6400km, show the orbital speed of the ISS around the centre of the earth? The ISS orbits the earth at 410km and it takes 93 mins to orbit the earth once.
Google says the answer is 7.7 km s-1 but Id like to be able to work out how to get to this answer. Can anyone help please?
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Hi Phil
The equation and relevant information you require to calculate it is in the link provided
https://www.physicsclassroom.com/class/circles/Lesson-4/Mathematics-of-Satellite-Motion
🤔
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Using the numbers provided, and assuming a circular orbit (it's not quite circular, but close enough for this level of analysis), an orbit of once every 93 minutes (1.55 hours) means that the ISS travels the a distance of one circumference every 1.55 hours. The circumference of a circle is equal to 2 π times the radius of the circle, and in this case the radius of the orbit is equal to the radius of the Earth plus the altitude of the orbit.
so a distance of 2 times π times 6810 km ≈ 42790 km
if it can cover 42790 km in 1.55 hours it must be going 42790/1.55 km/hour = 27600 km/hour
and there are 3600 seconds in an hour (60 second per minute times 60 minutes per hour)
so 27600/3600 = 7.67 km/s
The difference could be due to rounding or deviation from circular orbit (also because the orbit isn't exactly circular, the speed isn't exactly fixed either)
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If you want to work it out via orbital mechanics, the equation is v^2 = u(2/r-1/a), also known as the vis-viva equation.
u is the gravitational parameter or GM (with M being the mass of the planet). For the Earth, it is 3.987e14 m^3/s^2
r is the present distance of the orbiting object from the center of the Earth.
a is the semi-major axis of the orbit, or average orbital distance from the center of the Earth (this allows you to account for elliptical, non-circular orbits.)
For a circular orbit, r=a, and the equation reduces to V^2 = u/r
With the numbers given in the OP, and assuming a circular orbit, you get 7.652 km/sec.
However, the actual radius of the Earth (at the Equator) is 6378 km. making this slight correction gives 7.664 km/sec, which when rounded up to two significant digits, gives 7.7 km/sec
The Wiki article on the ISS lists its orbital speed as 7.66 km/sec, which either of the two above answers would round out to.
Also, using the smaller, more accurate radius of 6378 km, and using chiralSPO's method, we get 7.64 km/sec, much closer to to the given value.
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This website gives the current orbital position and speed of the ISS:
https://spotthestation.nasa.gov/tracking_map.cfm
When I read this website, the ISS altitude was 419km, and speed was 27,586 km/h.
- The altitude is always decaying because of friction with the outer atmosphere.
- So they have to give it a periodic rocket boost to prevent it burning up in the denser atmosphere
- We tend to think of this as "the atmosphere slowing it down", but surprisingly, as the altitude drops, the speed increases.
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Thank you everyone. I understand it a lot better now
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you're welcome!
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But if we delivery a 10 ton payload in a 10 ton transporter parked alongside will the ISS slow down and by how much?? We can assume the ISS weight 400 tons so it increases in combined weight by 5%
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But if we delivery a 10 ton payload in a 10 ton transporter parked alongside will the ISS slow down and by how much?? We can assume the ISS weight 400 tons so it increases in combined weight by 5%
We don't throw stuff at the ISS.
We carefully match its speed and location to the ship- the process is called docking.
So, when the supply ship attaches to the ISS they are travelling at very nearly identical speeds and trajectories and so they don't affect one another significantly.
Did you not realise this?
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Hi all,
So acsinuk
But if we delivery a 10 ton payload in a 10 ton transporter parked alongside will the ISS slow down and by how much?? We can assume the ISS weight 400 tons so it increases in combined weight by 5%
As BC has already stated "They" try to match velocities when docking.
And then Its down to the principle postulated by Galileo of acceleration/velocity of falling bodies is independent of Mass.
https://en.wikipedia.org/wiki/Galileo_Galilei#Falling_bodies
https://www.google.com/search?q=hammer+and+feather+on+the+moon&rlz=1CATTSD_enGB890&oq=hammmer+amd+&aqs=chrome.2.69i57j0i13i433j0i13j46i13i175i199j0i13l2j46i13i175i199j0i13l3.7378j0j15&sourceid=chrome&ie=UTF-8
:)