# How to solve this exercise in orbital mechanics; the moon stops in its orbit, what happens next?

As an exercise in orbital mechanics, suppose the moon were to suddenly stop orbiting Earth, and instead at time $$t=0$$ begin moving in the same direction as Earth so that their relative velocity were zero.

How could one go about calculating the answer to the following questions?

1. Would the Moon hit the Earth or not?
2. If it did, how long would it take?
3. If it didn't, what would its new orbit be like?
4. Does the answer depend on where the Moon is in its orbit?

Do we have to do detailed numerical integration to find the answer, or can we use some simple equations that involve energy and/or eccentricity to get an answer?

• I'm voting to close this question as off-topic because it doesn't appear to be about space exploration. – SE - stop firing the good guys Feb 5 '20 at 19:54
• "Stopping orbiting the Earth" does not necessarily imply a collision. – Starfish Prime Feb 5 '20 at 21:13
• Well, that's more like it. – SE - stop firing the good guys Feb 6 '20 at 1:28
• That's very odd, it is a question about space, what other category can it be in ? ..it was raised in my science class, my teacher explained how the moon doesn't fall to earth because of its orbit, which I understood, but he couldn't explain the maths for a decline in orbit. If no one here is able to answer it perhaps you can suggest where I can pose it that someone might be able to solve it for me - Thank you. – Justin Feb 6 '20 at 5:38
• @neph "better-asked on..." is never a close reason in Stack Exchange. OP decides where they want to ask it, and as long as its not off-topic we shouldn't cast a close vote. – uhoh Feb 6 '20 at 7:55

This is a supplementary answer with calculations (for 1) two bodies and 2) three bodies) problem which confirms that @MatthewWells answer is correct.

Would the Moon hit the Earth or not?

Yes, it would hit the Earth.

Do we have to do detailed numerical integration to find the answer, or can we use some simple equations that involve energy and/or eccentricity to get an answer?

To perform calculations for this answer I applied combination of the use of simple equations (force, acceleration, energy/eccentricity/velocity) and simple iterative approach (spreadsheet in OpenOffice Calc), i.e. no numerical integration of complex equations.

tl;dr

1. The simplest case of two initially static bodies: the Earth and the Moon (assuming there's no Sun, no other planets, and no galaxy):

Gravitational force that acts on each of the bodies can be found from Newton's law of universal gravitation:

Fem=Fme=G*(Mm*Me)/(r^2),

where: Fem: Gravitational force acting on Earth, directed towards Moon; Fme: Gravitational force acting on Moon, directed towards Earth; G: universal gravitational constant; Mm: mass of the Moon; Me: mass of the Earth; r: distance between centers of the Moon and the Earth.

Fem=Fme= 1.982*10^20 N

Acceleration of each body can be calculated using second Newton's law of motion: F=Ma , where F: force; M: mass; a: acceleration.

Earth's acceleration: ae=Fem/Me= 3.31*10^-5 m/s^2

Moon's acceleration: am=Fme/Mm= 2.79*10^-3 m/s^2

Then I applied a simple iteration process: The above accelerations are calculated for time t=0. Velocity of both bodies Ve0, Vm0 and distance they traveled towards each other De0, Dm0 are equal to zero.

For the next step, after a time step dt: t1=0+dt, Earth's velocity grows from 0 to Ve1=0+ae0*dt, travelled distance is De1=0+Ve1*dt.

Moon's velocity grows from 0 to Vm1=0+am0*dt, travelled distance is Dm1=0+Vm1*dt.

Distance between the objects decreases: r1=r0-De1-Dm1

Then we obtain "step-1" gravitational force Fem1=Fme1 as function of new r1 and new accelerations ae1 and am1 from the two formulas mentioned above.

For the next step t2=t1+dt, new velocities are: Ve2=Ve1+ae1*dt, Vm2=Vm1+am1*dt; travelled distance is De2=De1+Ve2*dt, Dm2=Dm1+Vm2*dt.

Distance between the bodies: r2=r1-De2-Dm2

and so on, until the distance between the centers of the bodies becomes equal to the sum of Earth and Moon radii.

how long would it take?

It would take 6898 minutes = 114.96 hours = 4.79 days for the Moon to hit the Earth.

At impact, Earth would have traveled 4573km km (or 72% of its radius) and would have velocity of 117 m/s; the Moon would have traveled 371,721 km and would have impact velocity of 9,544 m/s.

The Energy of the impact can be calculated from the formula for kinetic energy:

E=0.5*Me*(Ve)^2=0.5*Mm*(Vm)^2.

So, the energy of the impact would be E=3.506*10^26 J, or an equivalent of 1.67 billion Tsar Bomba's.

2. Including the Sun (and the fact that both Earth and Moon are orbiting the Sun) is a bit more tricky, but with some simplifying assumptions can be coarsely approximated by applying balance of the forces that act on the two orbiting bodies and simple iteration process as described above.

Let's assume Earth's trajectory around the Sun is perfect circle.

In absence of the Moon, and considering the local Earth-centered reference frame, there are two forces acting on Earth: gravitational (from the Sun, calculated as per formula above), directed from Earth towards the Sun and centrifugal force due to orbiting motion (directed radially, i.e. perpendicular to orbital velocity vector, outwards from the Earth and Sun). These two forces are equal in magnitude but opposite in direction, therefore cancelling each other, hence the Earth's radius vector is constant.

Although, apparently, the centrifugal force is a fictitious one for this case, let's assume it would be numerically equal to centripetal force:

Fce=(Me*(Vte)^2)/Re,

where Vte is tangential (orbital) velocity, Re is distance between Earth and Sun centers.

Vte can be calculated from vis-viva equation with the special case of c=Re (I have substituted "a" in the formula from wikipedia with "c" to not confuse it with acceleration)

(Vte)^2=GMs(2/Re-1/c),

where Ms: mass of the Sun, c: semi-major axis of the orbital trajectory ellipse.

Now when we include the Moon, the third force would act on the Earth: gravitational force between the Earth and the Moon, directed outwards from Earth, towards the Moon.

It would depend where exactly around the Earth it has stopped. I've looked into two special cases of Sun, Earth and Moon aligned in one line, where the Moon is either "above" the Earth (i.e. on larger radius than Earth), or between the Earth and the Sun (i.e. on smaller radius then Earth).

2a) Moon "above" the Earth.

The same three forces would act on the Moon as described above for the Earth. The difference would be in the magnitude of centrifugal (calculated as centripetal) force: by constraints of the question, the Moon moves with the same speed as the Earth, but in a higher orbit (larger radius vector), which means it's going to orbit in ellipse and at t=0 it's in its periapsis.

Using the vis-viva equation,

(Vtm)^2=GMs(2/Rm-1/cm)

we can find semi-major axis cm, taking tangential (orbital) velocity Vtm equal to Vte (explained above to be calculated from circular case at Earth orbital radius) by definition of the question. Rm is the Moon radius vector, i.e. the distance from the Moon center to the Sun center: Rm=Re+r.

By applying the iterative process as described above for the case 1) (i.e. the vector sum of the three acting forces is used to calculate the acceleration, then calculating traveled distance iteratively) the calculation continues until the distance between the Earth and the Moon centers becomes equal to the sum of their radii.

The result of the calculation is that collision will happen in this case after 6914 min=115.23 hrs, which is 15.8 minutes or 0.229% longer than calculated in the (two-body) case 1).

2b) Moon between the Earth and the Sun.

Everything else, described above for the case 2a) applies here, except the Moon is now in lower orbit with the orbital speed of the Earth, which means it would orbit the Sun in ellipse, and at t=0 it's in its apoapsis.

The time to collision in this case is 6914 min=115.23 hrs, which is 15.9 minutes or 0.23% longer than calculated in the (two-body) case 1) and practically identical to the same as in case 2a). The Moon and the Earth velocities at impact are almost identical to case 1). Energy of collision is 3.505*10^26 J, a 0.04% deviation from that of case 1).

See below the plots of the {radius vector - 1au} (in meters) vs. time in minutes (my apologies for the quality): Bonus calculation:

3) Let's assume the Earth and the Moon stopped altogether in the Solar system, and are aligned in one line, as described in cases 2a) and 2b).

By zero'ing orbiting velocity and centrifugal force in the calculations for the cases 2a) and 2b), we obtain the following plots (with comparison to the case 1)): For the case 3a) (Moon is on larger radius than Earth), the time to collision is 6925 min=115.41 hrs, which is 26.9 minutes or 0.39% longer than calculated in the (two-body) case 1), but velocities at impact and impact energy differ significanly: Earth velocity 2352 m/s (20 times higher), Moon velocity 12030 m/s (26% higher); impact energy: 4.420*10^26 J, a 26.1% increase compared to case 1)

For the case 3b) (the Moon is on the lower radius than the Earth), the time to collision is 6925 min=115 41 hrs, which is 27.0 minutes or 0.39% longer than calculated in the (two-body) case 1) and pretty much the same as for the case 3a), but velocities at impact and energy are significantly different again: Earth velocity 2587 m/s (22 times higher), Moon velocity 7093 m/s (26% lower). Energy of impact is 2.606*10^26 J, 25.7% lower than in case 1) and 1.7 times lower than case 3a).

So, to sum up, in the case of stopped Moon, the mutual gravitational force between the Moon and the Earth at the given distance between them prevails in comparison with the gravitational force of the Sun acting on these two bodies, resulting in relatively short time between the Moon stop and the impact of the two bodies, unless we consider stopping the Earth as well.

Does the answer depend on where the Moon is in its orbit?

It does very slightly and the Moon position can be neglected, unless we consider stopping the Earth as well, in which case, positioning of the moon around the Earth gives large effect on energy of the impact (but time to the impact stays very similar).

A numerical integration is the best option for me.
Using the JPL’s DE/LE438 ephemerides and the NAIF’s SPICE library, I found the dates when the angle between the Earth-Sun and the Earth-Moon vectors is about 180, 0 and 90 degrees and the Moon/Earth distance is about the same:

   Date      Angle  Distance
2000-11-12  175.54   371281
2008-03-08    4.25   371203
2002-12-27   90.29   371208 (the Moon leads the Earth)


I read the state of the bodies from the JPL’s files, then I put Vmoon = Vearth and I start the simulation.
The simulation includes the Newtonian and the relativistic accelerations of all the planets, Sun and Moon. The Earth's gravity field includes the J2 and J3 perturbations.
Here’s the result: The Earth/moon distance is the distance between the center of the bodies (the Moon hits the Earth when that distance is about 8115 km).
The label D_xxx means that the plot is for the distance when the starting Sun-Earth-Moon angle is xxx degrees.

The graph shows that the Moon always hits the Earth after about 4.6 days (the impact speed is about 10 km/s).

While the D_175.5 and D_4.3 plots almost overlap, we see a small difference for the D_90.3 plot; let’s see a detailed view: the graph shows that the impact for 90.3 deg happens about 35 minutes earlier.

• Beautiful plots! I like the way you chose different Sun-Earth-Moon angles but kept the initial separations almost the same. – uhoh Feb 9 '20 at 16:17
• @Cristiano Thank you for your effort. This allows validarion of my simplified model. When I plug in the initial separation that you used (I had slightly larger one), the 90deg case, which is most similar to two-bodies model, gives deviation for time to impact (compared to your full model) of -0.35%. For 0 and 180deg the simplified 3-bodies model gives deviation of -0.69% and compared to 90deg case, qualitatively gives the increase of the time to impact, which is the same direction (i.e an increase of the time) as the full model you used. – Sergiy Lenzion Feb 9 '20 at 21:24

This is easier than it sounds, provided we choose a suitable reference frame. In this case, we'll to consider the problem from the (doomed) perspective of someone standing on Earth's surface directly below the Moon. Since the question specifies that relative velocity between the bodies is zero, we can construct a free-body diagram, treating the gravity of other celestial objects as negligible. What remains is then figuring out the change in gravitational force over time, which can be derived from experimental data and some low-level calculus.

• Given this seems to be in regards to some academic problem, and the question mentions only methods, I believe it would be inappropriate to go into any further detail. – Matthew Wells Feb 6 '20 at 9:15
• It is not obvious, at least immediately that the gravity of the Sun can be neglected. It seems reasonable to at least check if there is any chance of the difference in gravitational attraction from the Sun on the Earth and the Moon is enough to change the trajectory significantly. – Steve Linton Feb 6 '20 at 15:42
• I agree with @SteveLinton. Also the trajectory would depend where exactly around Earth the Moon has stopped: every different position around the half of symmetrical (across Earth-Sun line) circumference of the moon ex-trajectory around the Earth would have its unique trajectory when including Earth and Sun gravitational influence. I.e. in the the two special cases of all three bodies aligning in one line: if Moon "behind" Earth, both Earth and Sun will pull it towards Earth, if Moon between Earth and Sun, Earth and Sun pull it in different directions, and it may matter how much Sun influences. – Sergiy Lenzion Feb 6 '20 at 19:52
• Crunching some numbers, the relative force of Earth should be at least twice that of the sun, and the moon roughly retains it's momentum relative to the sun, so solar gravity should not be the deciding factor; it may prove confounding in finding the precise time, as may other orbital factors, but that would be going into the detail I said I wouldn't. – Matthew Wells Feb 7 '20 at 3:35