Space Missions and Lagrangian Points

https://isaacphysics.org/questions/lagrangian_points?board=21312db2-9047-4f1c-be16-62927169c326&stage=a_level

For part a of this question I dont really know where im going wrong (or if im even doing the right thing).
I watched the video and I believe im implementing what theyre saying? I get a binomial in the end but it doesnt match to the answer they want you to get at all.

If anyone could offer help, it would be greatly appreciated.

Reply 1

mosaurlodon

Working (the weird colours and blobs is editing to remove my scribbles/unnecessary info to try to make it look more readable - hopefully you can still see the gist of what im trying to do)

edit note: in the 4th line if youre confused I used (1+e)^3 = (1+3e) and multiplied both sides by e^2(1+3e)

(edited 11 months ago)

Reply 2

mqb2766

Original post by mosaurlodon

https://isaacphysics.org/questions/lagrangian_points?board=21312db2-9047-4f1c-be16-62927169c326&stage=a_level
For part a of this question I dont really know where im going wrong (or if im even doing the right thing).
I watched the video and I believe im implementing what theyre saying? I get a binomial in the end but it doesnt match to the answer they want you to get at all.
If anyone could offer help, it would be greatly appreciated.

Youve done (centripetal/gravitational) for the earth/sun, but you should get something like
Ms [(1 + e)^3 - 1] = Me (1 + 1/e)^2
which you can then do the series / approximation.

Edit - what youve done is similar / equivalent, so really you just need the "series" for (1+e)^(-2) and multiply by 3e^3 to get ...

(edited 11 months ago)

Reply 3

mosaurlodon

Oh so, 3e^3*(1+e)^(-2) = 3e^3*(1-2e) = 3e^3 - 6e^4
e^4 -> 0 so
= 3e^3
so thats how that works - thanks
I think I did smth similar initially, I just didnt realise that the e^4 -> 0 since its only "1" away from the answer's exponent (if that makes sense)

(edited 11 months ago)

Reply 4

mqb2766

Original post by mosaurlodon

The question asks for the dominant term, and as you say e^4 goes to zero quicker than e^3.

A small note in your original working, when you multiply through by e^2(1+e)^3 near the bottom, its worth putting a note in your working that youre doing a truncated expanion of the cubic term as you drop the e^2 and e^3 terms without any comment. Its correct for this question, but its often worth trying to do all the series truncations together so that the reasoning is consistent. Hence when youd eliminated G and w in the original working Id have done like in #2 and put the Me's on one side and the Ms's on the other (dont put over a common denominator) and got
alpha = ...
in a couple of lines and then reasoned about the series.

Reply 5

mosaurlodon

ok i added the note and will keep that in mind for next time.
my thought process for the big fraction (common denominator) was just hoping it would help simplify things but looks like I just used a method with more trouble.

Actually looking back at my working I condensed a lot of the steps at the end which makes it really confusing (and the messy working doesnt help either) im surprised you found out what I was doing.