A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
Gravity and spacetime
anonymous
 one year ago
Gravity and spacetime

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1434420110948:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1434420315626:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And \((0,2,0,1)\) would appear be say \(0.9\) units away from the body, as was \((0,1,0,0)\) at an earlier time.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This is an attempt to put it in 2D, but we can be even more simpler in 1D, I think.

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Ahhh 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
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0dw: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 onetime 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
 one year ago
Best ResponseYou've already chosen the best response.0Ummm, I should put \((0.2,0.2)\) I guess, it's just a made up number.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.