How to do both c and e is written in this thread so just read it. For d it's difficult for me to explain because I'm not sure myself. Thats why I asked here but don't know if I'm all that wiser on it anyways
For e you know that (G*M*m)/r^2 is the force and since F=m*a where a=v^2/r you can solve the problem by solving it for v and then setting in the formula for T instead. Then you will get the time for one lap. If you are a UiS student and that's why you're wondering about d I would suggest you ask...
I guess that makes sense.
So you feel that by answering the same as in b I've done everythin I need to? I mean, then I still haven't explained it as a function of θ. Why would he write that if it wasn't depending on the angle?
Awesome! You managed to see what i couldn't. But still, does it make sense to you that its the same anywhere on Earth because to me that's weird. If the centripital acceleration changes with the angle I would think that the amount of rotations would also change
Maby, but the bigger θ is the smaller the radius for which the centripital acceleration is conserned gets. in other words gcosθ gets smaller and thus the centripital acc. also gets smaller. Then the necessary acc. produced by g should get smaller.
Also from the way my teacher has formulated...
Thats what I'm concerned about, acording to haruspex earlier in this thread it doesn't matter if your on the equator or not, but to me that seems weird. if it was like sqrt(g*r*cosθ) i would actually have a function of θ which is what the task asks for. Honestly I don't know anymore. The task...
In c this is my calculation:
When I put this equal to gcosθ i get v = sqrt(g*r) and I've tried several times with same result. Still don't think its the correct answer though.
This is the two equations I then get. I've never devided two equations on each other so not quite sure how to do it but depending on whitch of them i divide on the other I will either get l/d + h or d/l + h on the right side. The left I don't know, it seems to me I get l^2/d^2 or the other way...