anonymous
  • anonymous
Gravity and space-time
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1434420110948:dw|
anonymous
  • anonymous
|dw:1434420315626:dw|
anonymous
  • anonymous
And \((0,2,0,1)\) would appear be say \(0.9\) units away from the body, as was \((0,1,0,0)\) at an earlier time.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
This is an attempt to put it in 2D, but we can be even more simpler in 1D, I think.
Empty
  • Empty
Ahhh I am thinking more in terms of trajectories, which are lines perpendicular to yours. So to the object in freefall's view they never change speed, their speed is constant and their path is straight since they have momentum in one direction. It's just that the closer they travel to the earth, the smaller distances become, they just don't realize it.
Empty
  • Empty
I guess I'm trying to hold onto the idea that while you're in free fall with your eyes closed you feel no gravity, you're just weightless feeling.
Empty
  • Empty
I don't know if what I'm saying is mutually exclusive from what you're saying however, I'm just saying this is the angle from which I was thinking, while you have sorta like an equipotential gravity field lines slant I suppose
anonymous
  • anonymous
|dw:1434420951568:dw| The idea here is that the point \((1,0)\) is sucked into be the effective point of \((0.2,0.2)\). And this suction is always happening, which is different than a one-time bend. A one time bend would mean that if you went all the way around, you would be back at point \((0,0)\), but in reality, whatever was at \((0,5)\) would probably now occupy \((0,0)\).
anonymous
  • anonymous
Ummm, I should put \((0.2,-0.2)\) I guess, it's just a made up number.

Looking for something else?

Not the answer you are looking for? Search for more explanations.