A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
@experimentX
anonymous
 3 years ago
@experimentX

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Let's start with Keplers laws

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Let's see what I remember without looking at my cheat sheet

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Something about \[T^2=R^3\]

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1seems like I forgot Kepler laws ... let's see

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The period of an object orbital around the sun is proportional to the radius?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0whose radius? or perhaps some distance? The distance between that object and the sun?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1http://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion ellipse ... semi major axis.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0draw it with medw:1361862577312:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1361862616815:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Let's see what I remember about the semi major axis.......

DLS
 3 years ago
Best ResponseYou've already chosen the best response.0\[\LARGE (\frac{T_1}{T_2})^2=(\frac{R_1}{R_2})^3\]

DLS
 3 years ago
Best ResponseYou've already chosen the best response.0\[\LARGE \frac{dA}{dT}=\frac{L}{2M}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0why do we have two semi major axis?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1it doesn't matter which side you take ... lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1361862832161:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1361862878591:dw Where is \(R_2\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wiki didn't explain the ratio though

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0LOL I'm tired. ok I get it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Let's talk about escape speed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1361863110279:dw sweetheart, I have all the formulas staring at me from my notebook. I'm trying to have a discussion about those wonderful formulas

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0something about when the kinetic energy reaches \(\frac{GMm}{r^2}\)?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1361863195805:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0THanks! When do we know we have reached escape speed? \[U_f+K_f=U_i+K_i\] Let's derive escape speed. we don't have a final kinetic energy when we've reached escape speed correct?

DLS
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{GMm}{R}+\frac{mv^2}{2}=0\]

DLS
 3 years ago
Best ResponseYou've already chosen the best response.0put the total energy=0 find V

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Why what's the logic behind it? WHy is the total energy zero?

DLS
 3 years ago
Best ResponseYou've already chosen the best response.0If a body's total net mec. energy=0,it will escape from the earth's gravitational field

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1find the total work done when bringing object from infinity to position 'r'

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh ok, so when the kinetic energy equals the potential energy?

DLS
 3 years ago
Best ResponseYou've already chosen the best response.0have u heard of binding energy

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Let's see if I remember. When E<0 or =0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0parabolic and hyperbolic orbits?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that's when they're unbound correct?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0nope when E>0 is unbound

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0when E is less than zero is the only time when it's bound

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So when the potential is greater than the kinetic energy the energy is bound?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0gotta sleep =) Thanks for the discussion everyone. I look forward to hear more about bounded and unbounded Energy when I wake up. See ya :)

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1sorry ... was kinda busy not paying attention

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1this way you can do it ... for escape velocity. dw:1361863974972:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1You can equate those two, and hence get the result ...

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1I think ... if the velocity is less than esc velocity, the orbit will be elliptical or circular at escape velocity, the orbit is parabolic, and beyond that .... it's hyperbolic.
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.