Easy Newtonian Gravitation viewed from a rusty perspective.

AI Thread Summary
The discussion revolves around solving a physics problem related to Newtonian gravitation involving the forces between the Earth, Moon, and Sun. The key point is the assumption that the distances between the Earth and Sun, and the Moon and Sun, are approximately equal, allowing for the simplification of accelerations. By substituting the relationship between forces and accelerations, the equation simplifies to show that the force between the Earth and Moon can be expressed as a function of the Moon's mass and its acceleration. Participants acknowledge the challenge of the problem while emphasizing the importance of understanding the underlying concepts rather than just the calculations. The discussion concludes with a recognition that revisiting foundational physics can provide clarity and ease in problem-solving.
zaarin_2003
Messages
3
Reaction score
0

Homework Statement


Given

if:
(1) Fsm + Fem = Mm x Ams
(2) Ams = Ame + Aes
(3) Res ~= Rms

Show that (4) Fem ~= Mm x Ame

Where:
'Fsm' is the force between the sun and moon, 'Fem' the force between Earth and moon, etc.
'Mm' is the mass of the moon, 'Ams' the accelleration of the moon round the sun, 'Aes' the accelleration of the Earth in the frame of the sun, etc.
'Res' is the distance between the Earth and sun, etc.
And '~=' is a rubbish looking 'approximately equal to' sign.



Homework Equations


Oh... see above.


The Attempt at a Solution


Right, well, trust me, I'm not trying to cheat, I do have a physics degree gained in 2001 (a mere 2.1), but am finding myself more rusty/stupid than I realized. I'm going through my undergraduate physics book slowly, refreshing my memory and destroying my ego simultaneously.

I'm not going to detail my working (although written from my future self's proof reading perspective, it would have been easier), but effectively my attempt to solve this has centred around the given statement (3) that the distances between the Earth and sun, and the moon and sun, can be assumed to be approximately equal. The only relevance I can see for this statement is to enable you to make the assumption that Aes and Ams are also approximately equal (the mass of the Earth and moon being irrelevant obviously and the only variable being R). However, once I've made this assumption I'm stuck with a problem. Looking at (2), Ame now may as well ~=0. Which is completely the opposite of what I want. Clearly, by assuming Res and Rms are equal, I'm supposed to take Ams and Aes out of the picture. But I don't know how!

I've tried using (1), which is an addition of the forces around the moon, resulting in it's circular path around the Sun, constructing a similar expression for the Earth and rearranging... but no luck.

Any help will be appreciated. My book only gives the answers to problems with a numerical answer, so I'm never going to find out otherwise!
Thanks

Matt
 
Physics news on Phys.org
Welcome to PF.

First of all have a free " ≅ ".

I think it is a fairly simple vector problem.

Plug the identity of 2) into 1).

What they are saying is that using Res ≅ Rem,

I think you can infer that using the relationship A = V2/R

that Fsm ≅ MmAes
 
So, plugging 2) into 1) gives:

Fsm + Fem = (Mm x Ame) + (Mm x Aes)

Then eliminating Fem = Mm x Ame

gives Fsm = Mm x Aes

If you assume Aes ≅ Ams (given A = V2/R and that Res ≅ Rem)

All good? I feel like that was too easy.
 
Sometimes the math is easy, but the understanding harder to grasp.

(This makes for a lot of sloped foreheads.)

Given that they give you Res ≅ Rem... it looks to me any way like it just falls out the bottom after cranking not even half a turn.
 
Awesome. Thanks for that! I truly am rusty. However, leaving it a couple of weeks and looking at some electromag for a while made it a lot easier with fresh eyes. Your help was much appreciated!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top