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Period of the moon

  • #1
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Homework Statement


The orbit of the moon is approximately a circle of radius 60 times the equatorial radius of the earth. Calculate the time taken for the Moon to complete one orbit, neglecting the rotation of the earth.

Equatorial radius of the earth = 6.4 *10^6m
1 day = 8.6*10^4s
Acceleration of free fall at the poles of earth = 9.8ms^-2

Homework Equations


Newtonian law of gravity

The Attempt at a Solution


GMm/r^2 = mv^2/r and solve for v ---> v = sqrt(2GM/r)

Now it takes a period of time T fort he mion to make one full orbit. Thus it travels a distance C

C = 2*pi*r and using C = v*T --> T = C/v = 2*pi*r/sqrt(2GM/r)

Now i am told that r = 60Re where Re = radius of earth.


were i go from here since I was not given the mass of the earth. And what do i use the value of pi as ???
 

Answers and Replies

  • #2
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
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Homework Statement


The orbit of the moon is approximately a circle of radius 60 times the equatorial radius of the earth. Calculate the time taken for the Moon to complete one orbit, neglecting the rotation of the earth.

Equatorial radius of the earth = 6.4 *10^6m
1 day = 8.6*10^4s
Acceleration of free fall at the poles of earth = 9.8ms^-2

Homework Equations


Newtonian law of gravity

The Attempt at a Solution


GMm/r^2 = mv^2/r and solve for v ---> v = sqrt(2GM/r)

Now it takes a period of time T fort he mion to make one full orbit. Thus it travels a distance C

C = 2*pi*r and using C = v*T --> T = C/v = 2*pi*r/sqrt(2GM/r)

Now i am told that r = 60Re where Re = radius of earth.


were i go from here since I was not given the mass of the earth. And what do i use the value of pi as ???
You can look up the mass of the earth on this thing known as the internet.

Seriously, you don't know the value of pi (π), you know, the ratio of the circumference of a circle to its diameter? It's like a constant.
 
  • #3
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You can look up the mass of the earth on this thing known as the internet.

Seriously, you don't know the value of pi (π), you know, the ratio of the circumference of a circle to its diameter? It's like a constant.
 
  • #4
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I wanted an alternative way to work it with the mass of the earth since it was not given. And when i use 3.14 for pi i not getting the proper answer. Do you have any thing good to add to this question if not go .........
 
  • #5
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
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I wanted an alternative way to work it with the mass of the earth since it was not given. And when i use 3.14 for pi i not getting the proper answer. Do you have any thing good to add to this question if not go .........
Thanks for your suggestion. Back at you.

If your answer isn't coming out correctly, I doubt it's because you're using an incorrect value of π.

Since you don't show any of your calculations, which is what the rules of the forum ask you to do, it's hard to tell where you might be making a mistake.
 
  • #6
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Thanks for your suggestion. Back at you.

If your answer isn't coming out correctly, I doubt it's because you're using an incorrect value of π.

Since you don't show any of your calculations, which is what the rules of the forum ask you to do, it's hard to tell where you might be making a mistake.
Hey are you going to answer the question yes or no its one set of small talk you pulling ?
 
  • #7
gneill
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Okay you two, please turn up the courtesy knob on your transmitters :)

Boe Nancy, consider that if you know the acceleration due to gravity at one Earth radius (g at the surface) then you can determine the local value of 'g' at any radius using a method of ratios. That will give you the centripetal acceleration...
 
  • #8
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Hey are you going to answer the question yes or no its one set of small talk you pulling ?
Hey if you are rude nobody will help you.
 
  • #9
gneill
Mentor
20,801
2,778
You can look up the mass of the earth on this thing known as the internet.

Seriously, you don't know the value of pi (π), you know, the ratio of the circumference of a circle to its diameter? It's like a constant.
I wanted an alternative way to work it with the mass of the earth since it was not given. And when i use 3.14 for pi i not getting the proper answer. Do you have any thing good to add to this question if not go .........
My interpretation is that the poster is wondering how many digits of accuracy to use for the pi constant. Since nearly every calculator these days has a ##\pi## key, just use that and all the digits provided. Other factors will determine the overall accuracy and precision of your calculated values.
 
  • #10
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Your formula for v is not correct so maybe that is why you didn't get the right answer.
 
  • #11
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Okay you two, please turn up the courtesy knob on your transmitters :)

Boe Nancy, consider that if you know the acceleration due to gravity at one Earth radius (g at the surface) then you can determine the local value of 'g' at any radius using a method of ratios. That will give you the centripetal acceleration...
LOL i will send a picture of what i did is just when I use the normal way of using the mass of the earth i am getting the answer just the way without using the mass i am seeing trouble with i dont know if is transposing or some simple mistake i will put up a picture shortly
 
  • #12
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Hey if you are rude nobody will help you.
Yeah sorry i know just lost in a question an some guy was giving me a hard time
 
  • #13
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this is what I did and got a period of 0.457 days, in the 2 attachment
 

Attachments

  • #14
gneill
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Unfortunately your formula:
$$G \frac{M_E m_M}{r^2} = m_M g$$
is not correct. F = Mg only applies to objects located near the surface of the Earth where the acceleration due to gravity is approximated by a constant g. If the Moon happened to orbit at the Earth's surface you would be fine, and your period of about half a day might even be credible. But we know that the period of the Moon is closer to a month (hence the name "month" derived from "Moon").

I gave you a hint about finding the acceleration due to gravity at the Moon's orbital radius given the known acceleration due to gravity near the Earth's surface...
 
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Likes Boe Nancy
  • #15
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Unfortunately your formula:
$$G \frac{M_E m_M}{r^2} = m_M g$$
is not correct. F = Mg only applies to objects located near the surface of the Earth where the acceleration due to gravity is approximated by a constant g. If the Moon happened to orbit at the Earth's surface you would be fine, and your period of about half a day might even be credible. But we know that the period of the Moon is closer to a month (hence the name "month" derived from "Moon").

I gave you a hint about finding the acceleration due to gravity at the Moon's orbital radius given the known acceleration due to gravity near the Earth's surface...
Oh lawd this question is head ache could you give me a little more detail please
 
  • #16
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Oh lawd this question is head ache could you give me a little more detail please
So do i have to use the value of the moon gravity
 
  • #17
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What he's trying to say is that g varies depending on your distance from Earth's core. For example, think if you are at Mount Everest. You are further away from Earth's core on Mount Everest then you are at sea level. So, your acceleration due to gravity is lower (albeit very small) at Mount Everest than at sea level. Now think of how far away the moon is from the Earth's core: (3.84*108 meters).

So do i have to use the value of the moon gravity
No. You still need to place the moon in relation to Earth. But, you should know that you can't use Earth's g on the moon - Earth doesn't exert a full 9.8m/s2 on the moon. It's only the acceleration due to gravity on Earth's surface.
 
  • #18
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GMm/r^2 = mv^2/r

v=ωr and T=(2π)/ω

M=5.97219 × 1024 kg

What is the problem? Is it not allowed to use the known value of M?
 
Last edited:
  • #19
gneill
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What is the problem? Is it not allowed to use the known value of M?
That is the problem, yes. It wasn't a given in the original problem statement. But it isn't required if you know how g varies with distance.

@Boe Nancy , what is the formula for the acceleration due to gravity? (You can use Mearth here, it will disappear from the equations shortly).
 

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