# Gravitation between the Moon and the Earth: physics project

• isukatphysics69
In summary, the conversation was about a student's approach to a project involving the calculation of gravitation between the moon and Earth. The student plans to use Riemann sums in Excel to derive velocity and position graphs, but is stuck on integrating the acceleration and finding the radius with respect to time. The issue of conservation of momentum was brought up and it was suggested to consider the orbit of a spacecraft around the Earth instead. The student also mentioned picking the Earth as a fixed object, which may not be accurate due to factors like high tide.
isukatphysics69

## 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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isukatphysics69 said:

## 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|}##.

isukatphysics69
tnich said:
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.

isukatphysics69
isukatphysics69 said:
this is crunch time
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.
isukatphysics69 said:
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 ?

isukatphysics69
tnich said:
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

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 equation of a circle around the fixed point or not because the moons orbit is an ellipse.

BvU said:
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 ?

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

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isukatphysics69 said:
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.

isukatphysics69
kuruman said:
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

Ok my partner has went AWOL and has shut off phone I'm all on my own for this doing it right now

Ok i have my dt values which will be every 600 seconds, now here's 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

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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 system 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.

isukatphysics69
BvU said:
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 system 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

omg i have to use angular momentum? i just read about that yesterday i hardly understand it

found video

Can make do without it for quite a while. If you do everything well, it is a result of your calculations.

isukatphysics69
BvU said:
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

**** has just hit the fan

isukatphysics69 said:
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

isukatphysics69
BvU said:
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
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

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 ...

isukatphysics69
BvU said:
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​

Yes. -- provided you use appropriate units for all variables

isukatphysics69
BvU said:
Yes. -- provided you use appropriate units for all variables
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

Currently stuck, thinking

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As I said, R and v alone are not enough.
isukatphysics69 said:
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

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 )​

(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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isukatphysics69
BvU said:
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

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 )​

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

I'm off to bed -- after midnight here and early rise ahead.

Your time steps have a problem too.
Tip: use a formula, preferably a defined name (Create from selection on the formulas tab)
defined names are a good thing in general. Your cell formulas become almost legible.

isukatphysics69
BvU said:
I'm off to bed -- after midnight here and early rise ahead.

Your time steps have a problem too.
Tip: use a formula, preferably a defined name (Create from selection on the formulas tab)
defined names are a good thing in general. Your cell formulas become almost legible.
ok man, i REALLY appreciate youre help. thank you

I am getting 1018 m/s now. i changed up the units from kilometers to meters for the distance

BvU said:
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

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 )​

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...
I am confused about the a column, i am getting an acceleration that doesn't make sense with an exponent of 10^26

So i have the acceleration formula now, do i have to integrate this formula here?

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1018 m/s is what I get too.
On earth, g = 9.81 m/s2 So out there it should be a lot less.
Can't read you formula, but it looks as if it calculates a force ?
Keep track of R too. If you see it change very rapidly, it's a bug.
Keep track of theta too. It should evolve approximately linearly

yes you should integrate ax to get vx and ay to get vy. So you need to keep track of those.
And integrate vx to get x and vy to get y according to the
BvU said:
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

isukatphysics69
BvU said:
1018 m/s is what I get too.
On earth, g = 9.81 m/s2 So out there it should be a lot less.
Can't read you formula, but it looks as if it calculates a force ?
Keep track of R too. If you see it change very rapidly, it's a bug.
Keep track of theta too. It should evolve approximately linearly

yes you should integrate ax to get vx and ay to get vy. So you need to keep track of those.
And integrate vx to get x and vy to get y according to the
hey we just made some serious progress 1 sec

BvU said:
1018 m/s is what I get too.
On earth, g = 9.81 m/s2 So out there it should be a lot less.
Can't read you formula, but it looks as if it calculates a force ?
Keep track of R too. If you see it change very rapidly, it's a bug.
Keep track of theta too. It should evolve approximately linearly

yes you should integrate ax to get vx and ay to get vy. So you need to keep track of those.
And integrate vx to get x and vy to get y according to the

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BvU said:
1018 m/s is what I get too.
On earth, g = 9.81 m/s2 So out there it should be a lot less.
Can't read you formula, but it looks as if it calculates a force ?
Keep track of R too. If you see it change very rapidly, it's a bug.
Keep track of theta too. It should evolve approximately linearly

yes you should integrate ax to get vx and ay to get vy. So you need to keep track of those.
And integrate vx to get x and vy to get y according to the
Now one thing we notice tho is that our x and y positions are increasing and not oscillating
any idea why?

Last edited:
Is the force attracting the moon or pushing it away ?

isukatphysics69

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