# Period of the moon

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1. Nov 18, 2014

### Boe Nancy

1. The problem statement, all variables and given/known data
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

2. Relevant equations
Newtonian law of gravity

3. 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 ???

2. Nov 18, 2014

### SteamKing

Staff Emeritus
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. Nov 18, 2014

### Boe Nancy

4. Nov 18, 2014

### Boe Nancy

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. Nov 18, 2014

### SteamKing

Staff Emeritus
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. Nov 18, 2014

### Boe Nancy

Hey are you going to answer the question yes or no its one set of small talk you pulling ?

7. Nov 18, 2014

### Staff: Mentor

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. Nov 18, 2014

### lep11

9. Nov 18, 2014

### Staff: Mentor

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. Nov 18, 2014

### lep11

Your formula for v is not correct so maybe that is why you didn't get the right answer.

11. Nov 18, 2014

### Boe Nancy

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. Nov 18, 2014

### Boe Nancy

Yeah sorry i know just lost in a question an some guy was giving me a hard time

13. Nov 18, 2014

### Boe Nancy

this is what I did and got a period of 0.457 days, in the 2 attachment

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14. Nov 18, 2014

### Staff: Mentor

$$G \frac{M_E m_M}{r^2} = m_M g$$
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...

15. Nov 18, 2014

### Boe Nancy

Oh lawd this question is head ache could you give me a little more detail please

16. Nov 18, 2014

### Boe Nancy

So do i have to use the value of the moon gravity

17. Nov 18, 2014

### Ritzycat

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).

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. Nov 19, 2014

### lep11

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: Nov 19, 2014
19. Nov 19, 2014

### Staff: Mentor

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).