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Moon Rotation


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#1 Anatanoshi

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Posted 20 March 2017 - 09:01 PM

Earth orbits itself as well as the sun due to its own gravity (if i remember correctly).

The moon has less gravity (i think its 1.6g) but it still has gravity.

So why doesn't the moon rotate like Earth but a bit slower?


Edited by Anatanoshi, 20 March 2017 - 09:01 PM.

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#2 Sensei

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Posted 20 March 2017 - 09:17 PM

Earth orbits itself as well as the sun due to its own gravity (if i remember correctly).
The moon has less gravity (i think its 1.6g) but it still has gravity.


1.6g would be stronger than Earth's gravitation.

"Surface gravity 1.62 m/s2 (0.1654 g)"
https://en.wikipedia.org/wiki/Moon
 

So why doesn't the moon rotate like Earth but a bit slower?


You're searching for tidal locking.
https://en.wikipedia...i/Tidal_locking

300px-Tidal_locking_of_the_Moon_with_the
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#3 Strange

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Posted 20 March 2017 - 09:32 PM

Earth orbits itself as well as the sun due to its own gravity (if i remember correctly).

 

 

I'm not sure what that means. When you say "orbits itself", do you just mean "it rotates"? This is not related to the Earth's gravity, as far as I know.

 

 

 

So why doesn't the moon rotate like Earth but a bit slower?

 

It does. It rotates once in 27 days. The same time it takes to orbit the Earth. (See Tidal Locking link above.)


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#4 Anatanoshi

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Posted 20 March 2017 - 09:52 PM

1.6g would be stronger than Earth's gravitation.

"Surface gravity 1.62 m/s2 (0.1654 g)"
https://en.wikipedia.org/wiki/Moon
 

What's the diffrence between m/s2 and g?

 

You're searching for tidal locking.
https://en.wikipedia...i/Tidal_locking
 

Thank you.


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#5 studiot

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Posted 20 March 2017 - 10:01 PM

 

 

So why doesn't the moon rotate like Earth but a bit slower?

 

You have been told that the moon does rotate, but not like the Earth due to a phenomenon called tidal locking.

 

So to complete the final part of this jigsaw.

 

First why does the Earth (or moon or other body rotate at all?)

 

Well during and after the formation of any substantial body in space other (smaller) bodies crash into it.

Whether they coalesce with the body or just bump into it they must hit it off centre which means that the collision exterts a moment about that central axis.

If all the collisions are from random directions they will tend to cancel out over time.

But if they are mostly from certain directions, as in the early solar system,

over time this builds up to generate a sizable rotational speed and there is not much in space to cause friction so the rotation will last a long time after the period of planet building has occurred.

 

So we have a body like the Earth, which is slowing down, rotating about its axis, but having substantial rotational momentum

 

The density of the Earth is 551 5510 compared to the Moon which is only 334  3340 kg/m3.  Edit see post7

 

Coupled with the huge size difference between the two bodies this leads to a huge difference in stored rotational momentum.

 

I will leave you to calculate the difference.

 

As strange noted, so far gravity is not involved, but introduce that into the situation and you enter the interaction between two spinning bodies via the tidal locking mechanism when the Moon became captured by the Earth's gravity.


Edited by studiot, 20 March 2017 - 10:40 PM.

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#6 Strange

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Posted 20 March 2017 - 10:08 PM

What's the diffrence between m/s2 and g?

 

 

g = 9.8 m/s2


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#7 Sensei

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Posted 20 March 2017 - 10:14 PM

The density of the Earth is 551 compared to the Moon which is only 334 kg/m3.

 

Small correction (sorry), multiply above by 10. Earth's density is 5514 kg/m^3 and Moon's density is 3344 kg/m^3


Edited by Sensei, 20 March 2017 - 10:14 PM.

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#8 studiot

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Posted 20 March 2017 - 10:40 PM

Thanks sensei +1


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#9 Sensei

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Posted 20 March 2017 - 11:06 PM

What's the diffrence between m/s2 and g?

 

If you divide acceleration on the surface of some cosmic object by Earth's surface gravity acceleration, you receive dimensionless fraction or multiply of g.

1.62 m/s^2 / 9.81 m/s^2 = 0.165

 

It tells how much weaker or stronger is gravitation relative to Earth.

 

Similar like if you divide distance from some planet by distance from Sun to Earth, ~ 150 mln km,

you receive distance in A.U. (Astronomical Units).

1 AU = ~ 150 mln km


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#10 swansont

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Posted 21 March 2017 - 09:58 AM

(i think its 1.6g) 

 

 

~0.16g


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#11 Anatanoshi

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Posted 21 March 2017 - 12:48 PM

 

If you divide acceleration on the surface of some cosmic object by Earth's surface gravity acceleration, you receive dimensionless fraction or multiply of g.

1.62 m/s^2 / 9.81 m/s^2 = 0.165

 

It tells how much weaker or stronger is gravitation relative to Earth.

 

Similar like if you divide distance from some planet by distance from Sun to Earth, ~ 150 mln km,

you receive distance in A.U. (Astronomical Units).

1 AU = ~ 150 mln km

Why are we squaring the m/s?

And the purpose of g is to check how much force acts on an object like in the formula for height energy? (Eh = m * g *h)


Edited by Anatanoshi, 21 March 2017 - 12:51 PM.

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#12 Argent

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Posted 21 March 2017 - 12:53 PM

Why are we squaring the m/s?

Acceleration is a rate of increase. So many metres per second increase in velocity for every second that passes. 9.81 metres per second per second. 9.81 m/s2


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#13 Kelly_c

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Posted 21 March 2017 - 02:25 PM

Really informative. Can you please guide me where should I find more GIF like this?


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#14 Sensei

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Posted 21 March 2017 - 07:30 PM

Why are we squaring the m/s?

And the purpose of g is to check how much force acts on an object like in the formula for height energy? (Eh = m * g *h)

 

Do you know how Newton's motion equations are derived?

 

If you have object at time t0 (we can simplify it's t=0) at location x0 (also we can simplify it's x0=0),

then it moves at time t1 to location x1,

then it moves at time t2 to location x2.

 

We can create equation describing this movement, to predict:

1) whether it's constant velocity (lack of acceleration, linear equation),

2) whether it's accelerating, or whether it's decelerating.

 

1) Movement with constant velocity equation will look like:

f(t)=x0+v*t

where v is in meters per second m/s:

v= (x1-x0)/(t1-t0) = (x2-x1)/(t2-t1) = dx/dt

 

2) Movement with acceleration/deceleration equation will look:

f(t)=x0 + (a*t^2)/2

 

a = (v1-v0)/(t1-t0) = ( (x2-x1)/dt - (x1-x0)/dt ) / dt = dv/dt

Acceleration/deceleration is change of velocity in unit of time. Velocity has already unit m/s. After further division by time in seconds, you get m/s^2.

If velocity remain constant. There is no acceleration or deceleration.


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#15 Anatanoshi

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Posted 21 March 2017 - 09:12 PM

 

Do you know how Newton's motion equations are derived?

 

If you have object at time t0 (we can simplify it's t=0) at location x0 (also we can simplify it's x0=0),

then it moves at time t1 to location x1,

then it moves at time t2 to location x2.

 

We can create equation describing this movement, to predict:

1) whether it's constant velocity (lack of acceleration, linear equation),

2) whether it's accelerating, or whether it's decelerating.

 

1) Movement with constant velocity equation will look like:

f(t)=x0+v*t

where v is in meters per second m/s:

v= (x1-x0)/(t1-t0) = (x2-x1)/(t2-t1) = dx/dt

 

2) Movement with acceleration/deceleration equation will look:

f(t)=x0 + (a*t^2)/2

 

a = (v1-v0)/(t1-t0) = ( (x2-x1)/dt - (x1-x0)/dt ) / dt = dv/dt

Acceleration/deceleration is change of velocity in unit of time. Velocity has already unit m/s. After further division by time in seconds, you get m/s^2.

If velocity remain constant. There is no acceleration or deceleration.

This was very helpful thank you!


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