Gravitation between the Moon and the Earth: physics project

  • #1

Homework Statement


projectphysics.PNG


Homework Equations


f = ma
m1 = mass of moon
m2 = mass of earth

The Attempt at a Solution


Ok this is crunch time here and i am NOT Kobe Bryant

I have chosen gravitation between the moon and earth for this project. I will start with the net force on the moon as
∑FSystem = (Gm1m2)/(R2) = ma

i will only be focusing on the moving body (the moon) so to find the acceleration i will use

(Gm1m2)/(R2*m1) = a

ok. Now this project is in excel. So i will use rienmann sums to derive my velocity and position graphs.

This is where i am stuck, i will integrate the acceleration which is
(Gm1m2)/(R2*m1) = a
now the radius is changing with respect to time between the earth and the moon, now i need to figure out how to find the radius with respect to time as well as figure out my constants of integration (which i think will be the previous velocity in the row on top in excel. i will start with some random velocity and position and integrate them.
 

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  • #2
tnich
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Homework Statement


View attachment 225233

Homework Equations


f = ma
m1 = mass of moon
m2 = mass of earth

The Attempt at a Solution


Ok this is crunch time here and i am NOT Kobe Bryant

I have chosen gravitation between the moon and earth for this project. I will start with the net force on the moon as
∑FSystem = (Gm1m2)/(R2) = ma

i will only be focusing on the moving body (the moon) so to find the acceleration i will use

(Gm1m2)/(R2*m1) = a

ok. Now this project is in excel. So i will use rienmann sums to derive my velocity and position graphs.

This is where i am stuck, i will integrate the acceleration which is
(Gm1m2)/(R2*m1) = a
now the radius is changing with respect to time between the earth and the moon, now i need to figure out how to find the radius with respect to time as well as figure out my constants of integration (which i think will be the previous velocity in the row on top in excel. i will start with some random velocity and position and integrate them.
I think the assumption that only the moon is moving will turn out to be an issue. That would violate conservation of momentum. The problem statement says that you can fix one of the objects (at a constant position). Be aware that the earth and the moon revolve about a common point. If you consider the earth to be fixed, you will probably have to justify your assumption and explain why momentum in not conserved in your answer for goal #5.

I think it would help you to think of the the force equation for your chosen problem in terms of the vector ##\hat r## between the Earth and the Moon. Remember that the unit vector associated with that vector is ##\frac {\hat r} {|\hat r|}##.
 
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  • #3
tnich
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I think the assumption that only the moon is moving will turn out to be an issue. That would violate conservation of momentum. The problem statement says that you can fix one of the objects (at a constant position). Be aware that the earth and the moon revolve about a common point. If you consider the earth to be fixed, you will probably have to justify your assumption and explain why momentum in not conserved in your answer for goal #5.

I think it would help you to think of the the force equation for your chosen problem in terms of the vector ##\hat r## between the Earth and the Moon. Remember that the unit vector associated with that vector is ##\frac {\hat r} {|\hat r|}##.
You might avoid that problem altogether by looking at the orbit of a spacecraft around the earth instead. In that case, the difference in the masses of the two bodies is so large that you could reasonable assume that the earth is fixed.
 
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  • #4
BvU
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this is crunch time
As my favorite IT guru says: Lack of planning on your part does not constitute an emergency on my part !
I will start with some random velocity and position
Not a good idea. The moon is known for a long time as approximately at the same distance from the earth and certainly not heading straight toward or away from it.
1. Homework Statement
You left out part 1. For a reason ?
Re 3. Choices: You pick earth and moon. OK, fine. You pick earth as not moving ? Not so good ! Why do you think we have high tide twice a day ?
 
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  • #5
You might avoid that problem altogether by looking at the orbit of a spacecraft around the earth instead. In that case, the difference in the masses of the two bodies is so large that you could reasonable assume that the earth is fixed.
I have already told my professor that i will be doing the moon and earth so it is to late to change. I will fix the earth at a constant position.
So the acceleration in the x direction will be
(Gm1m2)/(R2*m1) * (Xposition/r)= a
the acceleration in the y direction will be
(Gm1m2)/(R2*m1) * (Yposition/r)= a

sin = opp/hyp
cos = adj/ hyp

rather than trig angles i will use the coordinates

i need to get the R value with respect to time, so at any given time T i need to find the radius R, i don't know if i should use the eqation of a circle around the fixed point or not because the moons orbit is an ellipse.
 
  • #6
As my favorite IT guru says: Lack of planning on your part does not constitute an emergency on my part ! Not a good idea. The moon is known for a long time as approximately at the same distance from the earth and certainly not heading straight toward or away from it.
You left out part 1. For a reason ?
Re 3. Choices: You pick earth and moon. OK, fine. You pick earth as not moving ? Not so good ! Why do you think we have high tide twice a day ?
part1.PNG


well i need to start with some velocity because i think i need that as my constant of integration
 

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  • #7
kuruman
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View attachment 225234

well i need to start with some velocity because i think i need that as my constant of integration
Yes you do, but what is that velocity going to be? If it's not right, your integrated trajectory may predict that the Moon will leave its orbit or (worse for us) collide with the Earth.
 
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  • #8
Yes you do, but what is that velocity going to be? If it's not right, your integrated trajectory may predict that the Moon will leave its orbit or (worse for us) collide with the Earth.
yes i will use the real velocity 3683 kilometers per hour
 
  • #9
Ok my partner has went AWOL and has shut off phone I'm all on my own for this doing it right now
 
  • #10
Ok i have my dt values which will be every 600 seconds, now heres the thing, i need to integrate the acceleration which is
(Gm1m2)/(R2*m1) = a

but the only thing changing in here is the R value with respect to time. Is there some kind of rotational formula that i should use for the radius of moon around the earth?

So R = sqrt((x2 - x1)2 + ((y2 - y1)2))
My thinking right now is i need some dx and dy values for the moon, put that formula in column F and change the dx and dy with respect to time then integrate the
(Gm1m2)/(R2*m1) = a

projectphysics.PNG
 

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  • #11
BvU
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As they say: dang happens.
From your posting of part 1 it is clear the exercise doesn't want you to have only one body moving. (Good thing I remarked on that)
Earth/moon sytem is well known, so start with (looking up or calcualting) reasonable starting values: radial positions and tangential velocities.
You need to keep track of 2 positions and 2 velocities. Force is radial, so angular momentum is conserved.
 
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  • #12
As they say: **** happens.
From your posting of part 1 it is clear the exercise doesn't want you to have only one body moving. (Good thing I remarked on that)
Earth/moon sytem is well known, so start with (looking up or calcualting) reasonable starting values: radial positions and tangential velocities.
You need to keep track of 2 positions and 2 velocities. Force is radial, so angular momentum is conserved.
bro please see post above. i am keeping earth at a fixed position which was an option
it will be more complicated if both are moving
 
  • #13
omg i have to use angular momentum? i just read about that yesterday i hardly understand it

found video
 
  • #14
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Can make do without it for quite a while. If you do everything well, it is a result of your calculations.
 
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  • #15
Can make do without it for quite a while. If you do everything well, it is a result of your calculations.
ok, do you agree with my post 10 thinking?
i need to keep earth in a fixed position or it will get complicated.
i am thinking i do what i say in post 10 and then integrate acceleration with Riemann sum formula.
after i integrate with Riemann sum i will tack on the constant of integration being vvelocity of row above
 
  • #17
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earth at a fixed position which was an option
I can understand you don't go for the bonus. So: a small moon (dutch word for satellite is 'artificialmoon' when transposed litterally) it is. Only one position (x,y or r, theta) and one velocity (also two components). The motion is in a plane (no forces perpendicular to the ##\vec r, \vec v## plane).

Post 10 is ok, but r(t) alone doesn't cut it.
Think circular motion: orbit is circular because of a constant centripetal force (your gMm/r^2)

Funny how these four stars appear automatically :rolleyes:
 
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  • #18
I can understand you don't go for the bonus. So: a small moon (dutch word for satellite is 'artificialmoon' when transposed litterally) it is. Only one position (x,y or r, theta) and one velocity (also two components). The motion is in a plane (no forces perpendicular to the ##\vec r, \vec v## plane).

Post 10 is ok, but r(t) alone doesn't cut it.
Think circular motion: orbit is circular because of a constant centripetal force (your gMm/r^2)

Funny how these four stars appear automatically :rolleyes:
ok so you are saying acentripical = (Gm1m2)/(R2*m1)
a centripetal = v^2/r

i could've sworn my prof said not to use a centripetal but chances are i miss undersood him because it makes sense.

dang has hit the fan but my mentality right now is Kobe Bryant 2009 world cup
 
  • #19
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Prof is right: centripetal is a crutch, a result force (in this case from gravity, the only real force present, that you are dealing with by integrating) -- but it's very useful for reverse engineering to get initial conditions that come close to the actual motion ...
 
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  • #20
Prof is right: centripetal is a crutch, a result force (in this case from gravity, the only real force present, that you are dealing with by integrating) -- but it's very useful for reverse engineering to get initial conditions that come close to the actual motion ...
(Gm1m2)/(R2*m1) = acentripital
(Gm1m2)/(R2*m1) = v2/R
(Gm1m2)/(R1*m1) = v2initial
sqrt((Gm1m2)/(R1*m1)) = vinitial

= 32202 km/h​
 
  • #21
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Yes. -- provided you use appropriate units for all variables :wink:
 
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  • #22
Yes. -- provided you use appropriate units for all variables :wink:
Ok i will recheck units later, i am worried about the rienmann sums integration right now, this is my first time using excel
Now i will need to determine my dx and dy values for the moon rotating around the fixed point 0,0
i have my velocity as roughly 32000km/h
i cannot use trig functions here so i think i may have to start with an arbitrary x and y coordinate for the moon. rather than trig function i can use
dx = 32000*(x/R)
dy = 32000*(y/R)
my time intervals are every ten minutes
 
  • #24
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As I said, R and v alone are not enough.
i have my velocity as roughly 32000km/h
From $${GMm\over R^2} = {\ mv^2\over R}\quad ???$$
You can check it by calculating how long one month is

Now your integration:
Start in the xy plane with moon at ##(R, 0)## velocity ##(0, v_y)##
(with ##v_y## a better value that what you have now :rolleyes:)​

upload_2018-5-6_23-51-32.png


(the pink ones you stil have to calculate)

and at t = 600:
the new x ##\ ## is the x ##\ ## just above + vx * dt
the new y ##\ ## is the y ##\ ## just above + vy * dt
the new vx is the vx just above + ax * dt
the new vy is the vy just above + ay * dt
and the other ones you calculate​

this is called Euler integration. See where you end up after a month....
 

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  • #25
As I said, R and v alone are not enough.
From $${GMm\over R^2} = {\ mv^2\over R}\quad ???$$
You can check it by calculating how long one month is

Now your integration:
Start in the xy plane with moon at ##(R, 0)## velocity ##(0, v_y)##
(with ##v_y## a better value that what you have now :rolleyes:)​

View attachment 225305

(the pink ones you stil have to calculate)

and at t = 600:
the new x ##\ ## is the x ##\ ## just above + vx * dt
the new y ##\ ## is the y ##\ ## just above + vy * dt
the new vx is the vx just above + ax * dt
the new vy is the vy just above + ay * dt
and the other ones you calculate​

this is called Euler integration. See where you end up after a month....
ok let me try again thank you for coming back be right back
my prof said i cannot use angles i must use x/r y/r which is what i think you mean
 

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