4 ms·

In a uniform gravitational field, it isn't. On the earth's surface, acceleration due to gravity is 9.8 m/s^2, independent of the mass of the falling object. S

In a uniform gravitational field, it isn't. On the earth's surface, acceleration due to gravity is 9.8 m/s^2, independent of the mass of the falling object. See Galileo's Leaning Tower of Pisa experiment. https://en.wikipedia.org/wiki/Galileo%27s_Leaning_Tower_of_Pisa_experiment https://en.wikipedia.org/wiki/Galileo%27s_Leaning_Tower_of_P...

CoastalCoder 3y ago
> Hopefully I'm not embarrassing myself with this question, but: Yup, I'm kind of embarrassed :) I forgot that maintaining orbit is just a matter of falling at the same pace that the earth is falling away from you.
- flainne 3y ago
  > maintaining orbit is just a matter of falling at the same pace that the earth is falling away from you Perhaps a better visualization: moving sideways fast enough that you miss the earth? :)
- mmh0000 3y ago
  If you've never tried it, I highly recommend playing Kerbel Space Program[1] (it works on Linux, Mac, and Windows!). That game taught me so much about orbital mechanics, which led to rabbit holes of textbooks and videos[2]. The first big lesson KSP taught me was: why, when launching a rocket, you don't just go straight up but, instead, have to lean over pretty aggressively. [1] https://store.steampowered.com/app/220200/Kerbal_Space_Program/ https://store.steampowered.com/app/220200/Kerbal_Space_Progr... [2] https://www.youtube.com/watch?v=dhYqflvJMXc https://www.youtube.com/watch?v=dhYqflvJMXc
deleted 3y ago
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staunton 3y ago
The field is not uniform though. So in theory, if you know the orbit and firld exactly, you can calculate it. In the present case, I guess the precision with which one knows the orbit and other stuff (like the exact gravitational fiel of the earth) doesn't work out.
- martincmartin 3y ago
  True. It turns out, it also applies for the two body problem, as long as one body is much more massive than the other. https://en.wikipedia.org/wiki/Orbital_period#Small_body_orbiting_a_central_body https://en.wikipedia.org/wiki/Orbital_period#Small_body_orbi...

CoastalCoder 3y ago

> Hopefully I'm not embarrassing myself with this question, but: Yup, I'm kind of embarrassed :) I forgot that maintaining orbit is just a matter of falling at the same pace that the earth is falling away from you.

flainne 3y ago
> maintaining orbit is just a matter of falling at the same pace that the earth is falling away from you Perhaps a better visualization: moving sideways fast enough that you miss the earth? :)
mmh0000 3y ago
If you've never tried it, I highly recommend playing Kerbel Space Program[1] (it works on Linux, Mac, and Windows!). That game taught me so much about orbital mechanics, which led to rabbit holes of textbooks and videos[2]. The first big lesson KSP taught me was: why, when launching a rocket, you don't just go straight up but, instead, have to lean over pretty aggressively. [1] https://store.steampowered.com/app/220200/Kerbal_Space_Program/ https://store.steampowered.com/app/220200/Kerbal_Space_Progr... [2] https://www.youtube.com/watch?v=dhYqflvJMXc https://www.youtube.com/watch?v=dhYqflvJMXc

deleted 3y ago

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staunton 3y ago

The field is not uniform though. So in theory, if you know the orbit and firld exactly, you can calculate it. In the present case, I guess the precision with which one knows the orbit and other stuff (like the exact gravitational fiel of the earth) doesn't work out.

martincmartin 3y ago
True. It turns out, it also applies for the two body problem, as long as one body is much more massive than the other. https://en.wikipedia.org/wiki/Orbital_period#Small_body_orbiting_a_central_body https://en.wikipedia.org/wiki/Orbital_period#Small_body_orbi...