I Would the rogue planet in this picture slow Earths spin?

[Edward Barrow]

The following picture shows a rogue planet (in purple) approaching Earth (in blue). It is spinning prograde like Earth. We're talking about a close approach. For example a planet the size of the moon approaching within 12,000km of Earth. That would be just outside the roche limit.

I presume the interaction would result in a slight change in Earth orbit around the sun, pulling it outwards. This is why I have drawn a new line for Earths orbit after the interaction.

This would make the length of a year on Earth longer I presume?

Would the interaction also slow down Earths spin? ...resulting in a day becoming longer? For example, resulting in there being maybe 28 hours in a day instead of 24?

Any insight would be greatly welcomed.

 

[mfb]

This would make the length of a year on Earth longer I presume?

Depends on the details of close approach.

The influence on spin would be negligible. To a good approximation, Earth is a sphere - there is no torque acting on it. The interaction with the equatorial bulge would influence the orientation of the spin axis a bit, the induced tides would change the length of a day a bit - but we are talking about microseconds per day or something like that.

 

[Edward Barrow]

Depends on the details of close approach.

The influence on spin would be negligible. To a good approximation, Earth is a sphere - there is no torque acting on it. The interaction with the equatorial bulge would influence the orientation of the spin axis a bit, the induced tides would change the length of a day a bit - but we are talking about microseconds per day or something like that.

We're talking about a close approach. For example a planet the size of the moon approaching within 12,000km of Earth. That would be just outside the roche limit. Surely such a close approach would alter Earths spin (length of a day)?

 

[mfb]

Why?

Rough estimate: Tidal effects scale with mass*time/distance³. Let's give the planet one Earth mass. Our close approach would take something like an hour, and let's say the planets are nearly touching. That will lead to massive earthquakes, but does it lead to notable tidal effects? The scaling suggests it has the same influence on the length of a day as the Moon over 200 years.
The estimate is certainly not very accurate, but even if it is wrong by 4 orders of magnitude, the effect on the length of a day is not notable in everyday life.

 

[lifeonmercury]

What if a Jupiter-sized planet came that close?

 

[mfb]

Then Earth is ripped into pieces, and the length of a day becomes ill-defined.

 

[Edward Barrow]

Why?

Rough estimate: Tidal effects scale with mass*time/distance³. Let's give the planet one Earth mass. Our close approach would take something like an hour, and let's say the planets are nearly touching. That will lead to massive earthquakes, but does it lead to notable tidal effects? The scaling suggests it has the same influence on the length of a day as the Moon over 200 years.
The estimate is certainly not very accurate, but even if it is wrong by 4 orders of magnitude, the effect on the length of a day is not notable in everyday life.

I think a close approach of such a large body to Earth would create far more significant forces on the Earth than that. For example, if the rogue planet passed within 8000km of Earth (the roche limit for a rogue planet the size of the moon), the forces on it would be great enough to cause the rogue planet to disintegrate. It only follows logically then that the forces enacted back on Earth must be quiet big ...and well capable of significantly changing the Earths spin.

One would assume Earths spin would slow down as the rogue planet approached, spin at its slowest rate when the rogue planet is right beside it, and then spin faster again once the rogue planet leaves the vicinity of Earth. The motion of the rogue planet leaving Earth would be like setting a spinning top spinning faster. So, overall, the Earths spin would be slowed down and then sped up again. But would the rate of slowing down equal the rate of speeding up? ...in which case both interactions would cancel each other out and the length of a day on Earth would be the same before the interaction as after.

Of course if the rogue planet disintegrated (by coming within the roche limit), the Earth would slow down as the rogue planet approached but would have nothing to speed it up again as the rogue planet would disintegrate into pieces once it passed by Earth. This could result in the significant shortening of a day. It could shorten it from 24 hours to 30 hours or more just by having one close encounter with a rogue planet like this. This is my theory anyway, but i could be wrong.

 

[mfb]

It only follows logically then that the forces enacted back on Earth must be quiet big ...and well capable of significantly changing the Earths spin.

Is that claim backed by calculations?
Where do you expect a large torque to come from?

Of course if the rogue planet disintegrated (by coming within the roche limit), the Earth would slow down as the rogue planet approached but would have nothing to speed it up again as the rogue planet would disintegrate into pieces once it passed by Earth.

The pieces still have the same total mass as the planet before.

This could result in the significant shortening of a day. It could shorten it from 24 hours to 30 hours or more just by having one close encounter with a rogue planet like this.

30 hours is longer, not shorter.

This is my theory anyway, but i could be wrong.

It is not a theory. It is speculation without any numbers backing it, and in huge disagreement with calculations done already.

 

[Edward Barrow]

If the rogue planet passed above planet Earth in its trajectory, this would cause the top hemisphere to rotate at a different speed than the bottom hemisphere as more gravitational force is being applied to the northern hemisphere than the bottom.

The question is - would this cause the top hemisphere to rotate faster or slower in relation to the bottom hemisphere? ...overall speeding up or slowing down planet Earth?

If the rogue planet flew by in just an hour or if it flew by the Earth over the course of a week, this would create different torque values on planet Earth.

 

[mfb]

If the rogue planet passed above planet Earth in its trajectory, this would cause the top hemisphere to rotate at a different speed than the bottom hemisphere as more gravitational force is being applied to the northern hemisphere than the bottom.

No it would not. The force on a spherical symmetric object, as you draw it, is exactly spherical symmetric. Only deviations from the spherical shapes matter. And in terms of the length of the day, not even the equatorial bulge helps.

If the rogue planet flew by in just an hour or if it flew by the Earth over the course of a week, this would create different torque values on planet Earth.

The planet cannot be close to Earth for more than 1-3 hours (depending on what we call "close" - but certainly not more).

 

[Bandersnatch]

One would assume Earths spin would slow down as the rogue planet approached, spin at its slowest rate when the rogue planet is right beside it, and then spin faster again once the rogue planet leaves the vicinity of Earth.

No, one wouldn't, or at least shouldn't, because that's not how it works.
Look at the picture below:

The top row depicts a static situation, where the smaller body raises a tidal bulge on the larger one (the opposite case is of little interest to us here).
The lower rows have the larger body rotating at angular velocity ω1, while the smaller body flies by at closest approach R, with instantaneous tangential velocity V, which translates to angular velocity ω2=V/R.

If the angular velocity of the planet's rotation is larger than the angular velocity of the rogue moon, then the tidal bulge is displaced forward, giving rise to a torque that is retarding the rotation.

If it's the angular velocity of the rogue moon that is larger, then the tidal bulge is displaced backward, and the moon's gravity produces torque on the bulge resulting in acceleration of the rotation.

Let's find out how far the rogue moon needs to pass by for one effect to win over the other.

If it's a rogue body, then it must have at least the escape velocity w/r to the Sun, which at Earth's orbital radius is approx 42 km/s. Earth's orbital velocity is 30 km/s, and the rogue body will have gained up to ~11 km/s falling into Earth's gravity well, so depending on the direction of approach, the minimum relative velocity is between ~80 and 20 km/s.

We are going to assume that the direction of approach w/r to the Earth's rotation is as shown in the first picture, and that the encounter happens in the plane of Earth's rotation.
Taking the lowest value, for ω2 to be larger than ω1 the body would have to pass Earth closer than R=V/ω1, where ω1 = 2π/(24*3600) rad/s. This gives approx 275 000 km, or a distance comparable to Lunar orbit.

This means, that the tidal interaction will have retarding effect on the rotation only when the rogue body is further than that. While it'll spend most of the time at such larger distances, its tidal effects on Earth will match that of the Moon only for a short time.
The reminder of the approach, where tidal effects are the strongest (remember, tidal acceleration scales with third power of distance), will accelerate Earth's rotation.
You could get deceleration (and only deceleration) if the body passed from the other side of Earth, so that their angular velocities have opposite directions.

Having said that, and reinforcing what mfb stressed, even at the Roche limit, a single passage of a Moon-like body would not produce enough torque to cause noticeable changes in day length of the magnitude you're envisioning. We're talking millions of years of constant tidal interactions here, not a few hours.

I presume the interaction would result in a slight change in Earth orbit around the sun, pulling it outwards. This is why I have drawn a new line for Earths orbit after the interaction.

This would make the length of a year on Earth longer I presume?

The way it's drawn, the interaction is a slingshot manoeuvre, where the smaller body 'steals' orbital momentum of the other body. It'd result in a boost for the rogue body, and a decrease in instantaneous orbital velocity of Earth. This in turn means that Earth's orbit would become more eccentric, with its perihelion distance going down and its orbital period reduced = shorter year. The deflection of the velocity direction is less important.

 

[Janus]

To help illustrate the comment Bandersnatch made on the resultant effect on the Earth's orbit, we can use this diagram:

The green arrow represents the vector of the Earth's orbital velocity prior to the flyby. The wider white arrow is the vector of the net velocity deflection of the rogue planet by its passage. The yellow arrow is the deflection vector of the Earth due to the passage. If we add this to the initial velocity vector of the Earth, we get the resultant orange arrow which is the final velocity of the Earth after the passage. Note that it is shorter than the green arrow, meaning a smaller magnitude velocity.
A smaller velocity magnitude means a lower total orbital energy, which results in a smaller semi-major axis, which, in turn, means a shorter orbital period.
The direction of the orange vector will determine the eccentricity of the new orbit.

 

[Edward Barrow]

The force on a spherical symmetric object, as you draw it, is exactly spherical symmetric. Only deviations from the spherical shapes matter. And in terms of the length of the day, not even the equatorial bulge helps.

If the rogue planet was not a planet but a giant irregular shaped asteroid. By "giant", i mean something the size of Mars or Venus. Would this irregular shape help create torque on the Earth slowing its rotational spin on its axis? And thus slowing the length of a day from 24 hours in a day to say 30 or 40 hours in a day?

You could get deceleration (and only deceleration) if the body passed from the other side of Earth, so that their angular velocities have opposite directions.

Ok, so Earths rotational spin on its axis would slow in the following diagram?

So in this scenario, Earths rotational spin would slow down. Meaning a day would become longer.

In addition, Earths velocity would speed up meaning it would push out further from the sun.

 

[Janus]

If the rogue planet was not a planet but a giant irregular shaped asteroid. By "giant", i mean something the size of Mars or Venus. Would this irregular shape help create torque on the Earth slowing its rotational spin on its axis? And thus slowing the length of a day from 24 hours in a day to say 30 or 40 hours in a day?
Ok, so Earths rotational spin on its axis would slow in the following diagram?

So in this scenario, Earths rotational spin would slow down. Meaning a day would become longer.

In addition, Earths velocity would speed up meaning it would push out further from the sun.

Something as large as Mars or Venus is too big to hold an irregular shape. Gravity will smooth it out to a near sphere.

Even if a Venus sized planet passed ~2.5 Earth radii away, it would have only a very small effect on the Earth's rotation. Something that massive and that close would have ~1.2 million times as much tidal effect on the Earth as the Moon does. However since the Moon's influence only adds 2 milliseconds per century to the Earth's rotational period, it would still require a century of orbiting close to the Earth to increase the rotational period by even 37 min. Since this would be a quick fly-by, the planet is just not going to be close enough for long enough to cause any large effect on the Earth's rotation.

 

[mfb]

You can alter the spin of an irregularly shaped asteroid notably (but it has to be smaller to be irregularly shaped), but you cannot alter the spin of a spherical object notably.
How often do we have to repeat "no" until you stop asking the same question over and over again? Do you expect the answer to change after the 10th time?