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Originally Posted by Tassel
Continuing to assume for the sake of example that G = 5G at Jupiter (or G=10G...whatever you want), and assuming a radius of approximately 68,000km for Jupiter, can you show how you would calculate the acceleration due to gravity at 68,000km from the center of Jupiter (ie: surface gravity)? I'm interested in the steps you would take to measure the mass of Jupiter (and what value you think that should be), and then the steps you would take to use that mass to determine the acceleration due to gravity.
Perhaps working through an example would help us understand what you are trying to say. |
Comparing apples to oranges. Earth's surface gravity is 9.8 m/s^2, while Jupiter's "surface" gravity is 23 m/s^2, but the two are not comparable directly, since we don't know Jupiter's real surface (use top of atmosphere, not same on Earth where we use planet's crust surface).
Per the above listed examples, at Jupiter (hypothetical) 5 G, the force would not affect surface (whatever that is) acceleration since the mass of the planet is adjusted for G (in local terms): i.e., F = 5(0.2 m)a
[I think this is what
papageno is referring to in his:
"In your reasoning you adjusted G and m(gravitational) so that the force would not change."
This is how I see it, that the
force does not change, nor how the planet interacts gravitationally per (G*m) per orbital equation, even if G and m are different (adjusted separately) so product remains same.]
Jupiter has not changed, same planet, so its gravity acceleration force has not changed. The mass of Jupiter, hypothetically, is adjusted for local G only, but it is still the same planet we figured in Earth's 1 G. It's just that Jupiter's "kilograms" are different from Earth derived kilograms. This has been my point for all of the above posts, though I am sure there would be some (all?) who would disagree.
[Edited for error in prior, viz. 5F = (0.2m)a, which was wrong. I sometimes think too fast for my fingers.]
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