Easy Newtonian Gravitation viewed from a rusty perspective.

In summary, this conversation discusses a problem involving the relationship between the forces between the sun, Earth, and moon, and the acceleration of the moon around the sun. By making assumptions about the distances and using the relationship between acceleration and velocity, it is possible to solve for the force between the sun and moon.
  • #1
zaarin_2003
3
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
 
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  • #2
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
 
  • #3
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.
 
  • #4
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.
 
  • #5
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!
 

Related to Easy Newtonian Gravitation viewed from a rusty perspective.

1. What is Newtonian gravitation?

Newtonian gravitation is a theory that describes the force of gravity between two objects as being proportional to their masses and inversely proportional to the square of the distance between them. It was first proposed by Sir Isaac Newton in the late 1600s and is still widely used today to explain the motion of objects in our solar system.

2. How does Newtonian gravitation differ from Einstein's theory of relativity?

While Newtonian gravitation is based on the idea that gravity is a force acting between two objects, Einstein's theory of relativity views gravity as a result of the curvature of space-time caused by the presence of massive objects. This allows for a more accurate description of gravity in extreme conditions, such as near black holes.

3. What does it mean to view Newtonian gravitation from a "rusty perspective"?

Viewing Newtonian gravitation from a "rusty perspective" refers to looking at it from a more simplistic or outdated point of view. This could mean using outdated equations or not taking into account newer discoveries or theories, such as Einstein's theory of relativity.

4. Is Newtonian gravitation still relevant in modern science?

Yes, Newtonian gravitation is still widely used and taught in modern science. While it may not be as accurate as Einstein's theory of relativity in certain situations, it still provides a good approximation for many real-world scenarios and is easier to understand and calculate.

5. How does Newtonian gravitation affect the motion of objects in our everyday lives?

Newtonian gravitation affects the motion of objects in our everyday lives by keeping objects, such as the planets in our solar system, in their orbits around the sun. It also causes objects to accelerate towards the Earth, giving us the sensation of weight and allowing us to walk and move around. Without the force of gravity, our everyday lives would be drastically different.

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