# Why doesn't the moon fall on Earth?

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It's actually *moving away* from Earth - by about an inch and a third every year... Tidal effects cause this; the moon raises tides on the Earth.  The Earth's rotation carries that tidal bulge *ahead* of the moon.  The moon, in turn is attracted to the bulge through gravity - and this accelerates the moon, boosting it's orbit.

But, the energy going into boosting the moon's orbit isn't free - it's coming from our rotational energy; Earth's day is slowly getting longer, as the Earth slows in it's rotation due to this tidal effect.

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• The moon doesn't fall on Earth because it has enough transverse velocity to remain in an orbit around Earth, and because angular momentum is conserved as a natural law.

Now, if you removed all of the moon's transverse velocity when the moon reached its aphelion, leaving the moon momentarily stationary with respect to Earth, then the moon and Earth would indeed fall together. How long would it take?

Two bodies having a total mass M (i.e., M=M₁+M₂) are initially at rest, separated by a distance d, in vacuum, and isolated from all forces except their mutual gravitational attraction. Find the time elapsed from the initial moment to the moment at which the separation is r, such that 0<r<d.

The problem involves the conservation of energy in a gravity dominated physical system, so we begin with the Vis Viva equation, solved for the speed in orbit:

v = √[GM(2/r − 1/a)]

where G = 6.6743e-11 m³ kg⁻¹ sec⁻²

The eccentricity of a plunge orbit is one, just as it is for a parabolic orbit. Why? Because for an elliptical orbit,

e = √(1 − b² /a² )

where b is the length of the semiminor axis, and a is the length of the semimajor axis. As b→0, the elliptical orbit becomes a plunge orbit, and e→0, also.

The apoapsis separation of M₁ and M₂ is therefore twice the semimajor axis.

e = 1

d = a(1+e)

d = 2a

a = d/2

And, since the angular momentum in a plunge orbit is zero, all of the motion therein is radial, and so

v = ∂r/∂t

Therefore, the Vis Viva equation can be rewritten as

∂r/∂t = √[2GM(1/r − 1/d)]

which is a non-linear differential equation with variables separable. We shall integrate both sides over the entire course of the fall, from the apoapsis of the plunge orbit to contact, assuming that both M₁ and M₂ are point masses.

∫(t₀,tᵢ) ∂t = − ∫(rᵢ,d)  ∂r / √[2GM(1/r − 1/d)]

The minus sign appears because the actual motion in the plunge orbit begins at d and ends at rᵢ, and I flipped those limits. However, hereafter I'll omit the limits and treat the indefinite integral.

tᵢ−t₀ = −√[d/(2GM)] ∫ ∂r/√(d/r−1)

We make this substitution:

u = √(d/r−1)

∂u/∂r = −½ d r⁻²/√(d/r−1)

r = d/(u²+1)

∂r = (−2d) u [∂u/(u²+1)²]

Plugging them in, we get

t−t₀ = 2d √[d/(2GM)] ∫ ∂u/(u²+1)²

Integrating by parts,

t−t₀ = 2d √[d/(2GM)] { ½ ∫ [∂u/(u²+1)] + ½ u/(u²+1) }

t−t₀ = d √[d/(2GM)] { u/(u²+1) + ∫ [∂u/(u²+1)] }

We notice an integral-trigonometric identity.

∫ ∂u/(u²+1) = arctan u

So,

t−t₀ = d √[d/(2GM)] { u/(u²+1) + arctan u }

Reversing the substitution,

t−t₀ = d √[d/(2GM)] { √(d/r−1)/[(d/r−1)+1] + arctan √(d/r−1) }

After simplification, we get as the general solution for the time to fall in a plunge orbit:

tᵢ−t₀ = √[d/(2GM)] { √(rᵢd−rᵢ²) + d arctan √(d/rᵢ−1) }

In the case where the motion in a plunge orbit begins at the apoapsis and ends in contact (assuming point masses),

tᵢ−t₀ = π √[d³/(8GM)]

When determining the time to fall from r₁ to r₂, such that r₁>r₂>0,

t₂−t₁ = (t₂−t₀) − (t₁−t₀)

Now let's consider what happens when we let the moon drop from its apogee to first contact between the two bodies (allowing them to keep their actual size and shape).

d = 405,503,560 meters

r₁ = 8,108,400 meters

M = 6.0483e+24 kilograms

t₁−t₀ = 450871.423 seconds = 125.242062 hours

Now let's run the fall again, from apogee to center-to-center contact, assuming that Earth and moon are both point masses.

r₂ = 0

t₂−t₀ = 451416.430 seconds = 125.393453 hours

The difference is about 9 minutes and 5 seconds.

• Dump the liberals into Jupiter
Lv 6
1 month agoReport

Correction:
where b is the length of the semiminor axis, and a is the length of the semimajor axis. As b→0, the elliptical orbit becomes a plunge orbit, and e→1

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• Because your idiocy creates an antigravity forcefield that only works on specific materials. If you didn’t exist, life would have ended.

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• It's made of Emmental and floats.

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• It always is falling, just that its orbital velocity, it misses.

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• because the moon likes someone else.....

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• Mainly due to the effect of gravity, but it is "falling" and also moving away from the Earth.

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• Because it's in orbit.  It's falling all the time, but is also moving "sideways" fast enough that it keeps missing the Earth.  The force of Earth's gravity is balanced by the Moon's speed that would tend to make it fly off into space if the Earth wasn't there, so it keeps missing the Earth and just goes round and round it.  And that's an orbit.

Imagine throwing a ball.  Of course it falls down.  The harder you throw it, the further it will go before it falls down.  If you could throw it hard enough and nothing got in the way, it would never come down - you've thrown it into orbit.

And this is what happens with everything in an orbit.  But to make the two forces balance, you need the orbital speed to be exactly right.  Too fast and the orbiting thing will move away, too slow and it will move closer, until we get the balance.

Which is why the International Space Station also doesn't fall down.  It's going fast enough for the two forces to balance and it stays up.  In fact it needs a little boost up from time to time because of drag from the Earth's outer atmosphere, but other than that, it was put there by people who understand orbital mechanics and it stays there.

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• The moon is falling. But it is moving horizontally too. The combined motion is a circle. That's called an orbit.

The moon could not stay where it is without moving around the Earth.

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• Because she's falling round us and missing.

• Lv 5
2 months agoReport

Damn thing.

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• Because it is Orbiting at a fast enough Velocity to not fall towards Earth

In fact the Moon is retreating from Earth by about 3/4 Inch every year

Mostly because of Tidal Friction on both Bodies

Through time, the Moon and Earth will be Tidally locked

with one face of Earth facing the Same Face of the Moon

That will lock the Moon into its Orbit with the Centre of Gravity being Outside the Crust of the Earth

The Moon will have no outward Energy then and hold its Orbit

Just like Pluto and Charon

• Lv 5
2 months agoReport

nice. thank you :)

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