Circular Motion: Planet Earth & Moon Orbiting Sun

In summary, the moon orbits the Earth with the same angular frequency w as the Earth orbits the sun. The moon and Earth are orbiting together as well. The position vector of the moon is a function of time in the sun's frame.
  • #1
Parallel
42
0
hello

planet Earth is orbiting the sun with angular frequency w,and radius r1.
the moon,which is orbiting the Earth with the same angular frequency w(in planet Earth's frame) and radius r2.
and together they are orbiting the sun.

now I need to describe the position vector of the moon as a function of time,in sun's frame.

here's my thoughts about this:

I assume that at time t=0 the angle between all the planets is zero.
the speed of the moon in the sun's frame is 2w.
now after a time t,the angle between the moon and the Earth is: wt.
and the angle between the Earth and the sun is also wt.

now I can't seem to connect all these things.

thanks for your help
 
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  • #2
A position vector from the sun to the moon is the sum of the vector from the sun to the Earth plus the vector fom the Earth to the moon. I'm guessing the problem wants you to assume everything is moving in one plane.
 
  • #3
you're right,they are moving in one plane.

my problem is,when I wanted to sum those vectors(by first disolving them into components),it looked pretty weird,that I use the "same" w,to describe the angle(because the problem states that w is relative,in the sun's frame and in the Earth's frame).

for example the x component of the vector from the Earth to the moon is:
r2sin(wt)

and the x component of the vector from the sun to the Earth is:
r1sin(wt).
 
  • #4
Parallel said:
you're right,they are moving in one plane.

my problem is,when I wanted to sum those vectors(by first disolving them into components),it looked pretty weird,that I use the "same" w,to describe the angle(because the problem states that w is relative,in the sun's frame and in the Earth's frame).

for example the x component of the vector from the Earth to the moon is:
r2sin(wt)

and the x component of the vector from the sun to the Earth is:
r1sin(wt).
So you are assuming that at time zero the x components are both zero, which estblishes a starting point (any starting point you want to choose is fine). At this time zero, there are corresponding y cmponents for r1 and r2 that I am guessing you have written with cosines. And your problem is that you do not believe that r_x (r = sun to moon) = r1_x + r2_x = (r1+r2)sin(ωt)?
 
  • #5
yea I've written the y's with cosines.

actually my problem is with the angle,I don't understand,how can I sum the "sines"[actually I'm saying that in my opinion,it was supposed to be r1sin(w1t)+r2sin(w2t)],because the angular frequency,is different in each frame.

or maybe it's o.k,because I'm only looking for the vector sum,and I don't care,what are the vectors in each frame.

I hope this doesn't sound dumb.
 
  • #6
Parallel said:
yea I've written the y's with cosines.

actually my problem is with the angle,I don't understand,how can I sum the "sines"[actually I'm saying that in my opinion,it was supposed to be r1sin(w1t)+r2sin(w2t)],because the angular frequency,is different in each frame.

or maybe it's o.k,because I'm only looking for the vector sum,and I don't care,what are the vectors in each frame.

I hope this doesn't sound dumb.
For the real Earth and moon, the frequencies certainly are different. The way this problem was stated it sounds like you are supposed to use the same frequency for both. If they were different you would leave the expression the way you have it.

It might be interesting for you to graph the sum on your calculator or at a site like the one I linked below and play with some values for the r and ω to see what it looks like. If you have a TI graphing calculator (or equivalent) you can put it in paranmetric mode and plot the position of the moon relative to the sun for any combination of r1, r2, ω1 and ω2 you want, limited only by the resolution of the display.

http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=graphs&s2=equations&s3=advanced
 
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Related to Circular Motion: Planet Earth & Moon Orbiting Sun

1. How does the gravitational force between the Earth and the Sun affect their circular motion?

The gravitational force between the Earth and the Sun is what keeps them in orbit around each other. The Sun's mass creates a gravitational pull that keeps the Earth in a circular path around it. This force is balanced by the Earth's orbital velocity, creating a stable circular motion.

2. How long does it take for the Earth to complete one orbit around the Sun?

It takes the Earth approximately 365.24 days, or one year, to complete one orbit around the Sun. This is why we have leap years every four years to account for the extra quarter of a day.

3. Why does the Moon orbit around the Earth and not the Sun?

The Moon orbits around the Earth because of the Earth's stronger gravitational force. While the Sun's gravitational force is much stronger than the Earth's, the Moon is much closer to Earth and therefore experiences a stronger pull towards it.

4. What causes the Earth's orbit around the Sun to be slightly elliptical rather than a perfect circle?

The Earth's orbit around the Sun is not a perfect circle due to the gravitational pull of other planets and objects in our solar system. These forces can slightly alter the Earth's orbit, causing it to be slightly elliptical.

5. Is the Earth's orbit around the Sun constantly changing?

Yes, the Earth's orbit is constantly changing due to the gravitational pull of other celestial bodies. This is known as orbital perturbation. However, these changes are gradual and do not significantly affect the Earth's orbital path around the Sun.

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