The moons gravitational influence on the earth causes the tides, which dissipate

Question

The moons gravitational influence on the earth causes the tides, which dissipate the kinetic energy of the earth-moon system. As a consequence the earth's rotation is slowing and the moon is moving away from the earth at a rate of 3.84 cm each year (same speed your fingernails grow at). At some point in the earths future it is possible that the moons orbit will correspond to the time it takes the earth to rotate – in other words a day will equal a lunar month – and the moon will be in geostationary orbit (the moon will only be visible from one side of the earth). By conserving angular momentum find how far into the future (in years) that this will occur (assuming the sun doesn't go supernovae before then). The moment of inertia of the earth is 8.008 x 1037 kg m2, the mass of the earth is 5.972 x 1024 kg, the mass of the moon is 7.347 x 1022 kg, and the angular momentum of the earth and moon combined is 3.468 x 1034 kg m2 s-1. The distance from the earth to the moon is currently 3.84 x 108 m, although this will increase with time. Assume a circular orbit throughout.

Explanation / Answer

Hi,

In this problem there are many considerations to make:

1. The Moon and the Earth formed an insulated system, so the angular momentum of the system remains the same.

2. The Moon can be treated as a particle in this problem as it is much smaller than the Earth.

3. The movement of the Moon is similar to the one of a particle in an uniform circular movement.

4. The axis of the rotational movement passes right thorugh Earth's center.

The angular moment of this system can be found as follows:

L = IE w + IM w' (1); where L is the angular momentum, IE is the moment of inertia of the Earth while IM is the moment of inertia of the Moon and w and w' are the angular velocities of the Earth and the Moon respectively.

If we use some equations from dynamics and the relations from uniform circular movement then we can express the angular velocity of the moon in other terms:

w' = ( GM/ r3 )1/2 (2); where M is the mass of the Earth, r is the distance between the Moon and the Earth and G is the constant of gravity which value is 6.673*10-11 Nm2/kg2

If the equation (1) is applied millions of years in the future, in a time when the Moon and the Earth rotate at the same speed we will have the following:

L = w' (IE + IM) = ( GM/ r3 )1/2 (IE + IM)

If we remember the equations of moments of inertia, we have the following:

IM = r2Mm ; where Mm is the mass of the Moon and r is the distance from the Earth. This is right as long as we consider that, compared to the Earth, the Moon is a particle.

L = ( GM/ r3 )1/2 ( IE + r2Mn) ; this equation has as sole unknown the value of r, however, it is difficult to solve right away. To find the value of r we can use and iterative process, using the actual value of r (3.84*108 m) as a seed value, then we have the following:

r = 5.56*108 m

Note: you can use solver in Excel to solve this part

Now that we know the value of r when the Earth and the Moon have the same angular speed, we can find the time using the relation (dr/dt) they gave us:

ro = r - ro = ( 5.56 - 3.84)*108 m = 1.72*108 m

y = ro / [dr/dt] =  1.72*108 m / [ 3.84*10-2 m/years ] = 4.48*109 years

It is a really long time, as expected

I hope it helps.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.