Mark44 said:1) v2 has to be >= 0, being the square of a real number.
2) u2 < 2MG/a, which means that u2 - 2MG/a < 0.
In order for the right side to be >= 0, 2MG/r has to be more positive than u2 - 2MG/a is negative. IOW, 2MG/r must be > -(u2 - 2MG/a).
Does that get you started?
Mark44 said:Part c is what you asked about, and that doesn't involve solving a differential equation. Since you asked about part c, I assumed you had already done parts a and b.
For what I said earlier, it doesn't matter whether v or r are constants or not.
They're not asking you to solve this differential equation. Did you do part b? If so, you should have ended up with the equation they show in part b.cloud360 said:because the thing am confused about is that v^2=r'^2
No, you can't assume that u2 = 2MG/a. They already told you that u2 < 2MG/a.cloud360 said:, and i was thinking i need to get r on one side, if so i need to solve the ODE?
if not, i am even more confused as my assumptions were dead wrong.
Would i get the correct answer if i assumed u^2=2MG/a then made the bracket =0, then solved to find r?
Reread what I said in post #2. I believe that's the direction you should take. I was able to get an upper bound on r from what I did.cloud360 said:i still can't start this :(
hope this is not to much to ask of you, but can you tell me what i am supposed to aim for in this question, do i have to get r on its own on one side?
Mark44 said:1) v2 has to be >= 0, being the square of a real number.
2) u2 < 2MG/a, which means that u2 - 2MG/a < 0.
In order for the right side to be >= 0, 2MG/r has to be more positive than u2 - 2MG/a is negative. IOW, 2MG/r must be > -(u2 - 2MG/a).
Does that get you started?
Solve this inequality for r.cloud360 said:so now should i solve
2MG/r > -(u2 - 2MG/a)?
still finding it difficult to start this question :(
See above.cloud360 said:i don't know what am supposed to solve or do to get max value of r.
cloud360 said:i am certian max values are find by solving r'=0, they will give criticial points which are either min or max, this is how i have been always thought to find max and min
Mark44 said:Solve this inequality for r.
See above.
That's incorrect. You have an incorrect sign somewhere, and you are supposed to show that r is less than some value.cloud360 said:1) ok, i got
r>-2MG/(u2 +2MG/a)
It's given that u2 - 2MG/a < 0. The quantity on the left side, 2MG/r, is positive, so saying that 2MG/r is larger than some negative number doesn't help you get an upper bound on r.cloud360 said:but then r'>0
what to do next
2) also why is it 2MG/r > -(u2 - 2MG/a).
why is it not 2MG/r > (u2 - 2MG/a),isnt that enough to solve r?
The answer is supposed to be an inequality, so you have to start with an inequality.cloud360 said:i understand that just being greater than a number i.e 2MG/r > (u2 - 2MG/a) won't make RHS positive it has to be 2MG/r > -(u - 2MG/a)
but why are you doing this i.e why are you making sure RHS is positive (i know that r'2>0, meaning RHS has to be positive and also >0), but why isn't 2MG/r > (u2 - 2MG/a), enough to solve r, its true isn't it?
3) also , why are you creating an inequality liek this while ignorign r'2. is it ok to ignore r'2
A better way to write this iscloud360 said:Ok i got (after multiplying top and bottom of RHS by a)
r<-(2aGM/(au^2-2Gm)
meaning r'<0, since a,u,G,M are constants?
but we don't need to find r' right? the answer is just r<-(2aGM/(au^2-2Gm) ?
(again i am so grateful to you, for your help)
Mark44 said:A better way to write this is
r < 2aMG/(2MG - u2), but what you have is also correct.
I don't know how you conclude that r' < 0.
And right, this part doesn't ask you to find r by integrating r'.
Yes, but you should write this as rmax = 2aMG/(2MG - au^2)cloud360 said:so is the answer just the maximum is r=-(2aGM/(au^2-2Gm)
cloud360 said:or do we have the differentate to get r', also are you saying r' is not <0, since if we differentiate RHS we should get 0, as all of RHS is constant?
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