Need assistance with orbital dynamics questions

In summary: The 54 cm is the theoretical amplitude, but in reality, the actual amplitude is much less due to various factors such as the shape of the ocean basins, landmasses, and other astronomical bodies. In general, tides can range from a few centimeters to several meters, depending on the location and other factors. The exact value for a specific location and time can be determined using tidal prediction software or tables.
  • #1
Kkelloww
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Hello,
I am currently writing a novel and there are quite a few situations that I would like to create that involve orbital dynamics. I would like to be as realistic as possible. I am seeking a knowledgeable person that can ballpark these problems. I do not think that it would take to much of your time and I would certainly provide an acknowledgment in the novel for your help! And a free copy!

Thanks,
Ken
 
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  • #2
What type of info are you after?
 
  • #3
The effects on Earth if the moon was in a highly eccentric orbit. perigee about 35,000 km which is outside the Roche limit I believe. Could such an orbit be sustainable. Also, what would the orbital period be and how far would apogee distance be?
How fast would it orbit Earth during perigee distance?

Specific effects

Tides?
Earth's orbital tilt
Increased volcanism
Affect on wind speed and patterns
Erratic seasonal weather.
Any possible effects on Earth's magnetic field?
Could tidal forces on the Earth and moon cause this orbit to decay?

Finally,
Is there any possible way in which the moon could orbit the Earth at the perigee distance for an extended period of time?

Crazy stuff I know. Thanks in advance!
 
  • #4
Kkelloww said:
The effects on Earth if the moon was in a highly eccentric orbit. perigee about 35,000 km which is outside the Roche limit I believe. Could such an orbit be sustainable. Also, what would the orbital period be and how far would apogee distance be?
How fast would it orbit Earth during perigee distance?
These things are pretty interdependent on each other. Just having the perigee isn't enough, as a perigee of 35,000 km can describe an infinite number of orbits. If you had the perigee and one of the other properties of the orbit, (period, apogee, orbital speed at apogee) you can work out the rest. So for example , if the apogee happens to be equal to the Moon's present distance, then you can work out that the orbital period would be ~11 days and the orbital speed at perigee would be ~4.6 km/sec
Specific effects

Tides?
Tidal effects vary by the inverse of the cube of distance. Since 35,000 km is almost 1/11 the present distance of the Moon, tidal effects would be over 1300 times greater.
Earth's orbital tilt
The major effect the Moon has on the Earth's axial tilt is precession, Which alters the direction of the tilt, but not its angle. when the Moon is at perigee this will be magnified. But he over all effect will depend on the overall orbit. The present effect causes the Earth to go through one full precession cycle every ~25,000 yrs. If the average distance of the new orbit is smaller, this will speed up. If it is larger, then it could well slow down. [/quote]
Increased volcanism
[/quote] Maybe. The increased flexing of the Earth's crust could lead to more crustal heating. It would be really hard to estimate to what degree this will effect volcanism.
Affect on wind speed and patterns
Probable. The Moon already produces an atmospheric tide, but it is much weaker than the effect caused by solar heating. Amping it up by a factor of 1,300 would likely turn it into a major factor, producing low pressure regions at the antipodes during perigee.
Erratic seasonal weather.
Weather is a very complex system, so it is very hard to predict exactly what kind of overall climatic changes can be produced by something like severe lunar atmospheric tides, but it is likely to upset the climate.
Any possible effects on Earth's magnetic field?
probably not
Could tidal forces on the Earth and moon cause this orbit to decay?
No. Not unless the orbit is such that the orbital period is less than the rotational period of the Earth. If this is not the case, then tidal effects will tend to raise ( and circularize the orbit.) But this effect will still be fairly slow.
Finally,
Is there any possible way in which the moon could orbit the Earth at the perigee distance for an extended period of time?
Not without a secondary effect from something. For instance, it would take something acting on the Moon to throw it into a new orbit with a 35,000 km perigee. It would take something else acting on the Moon in just the right way while it was at perigee to cause it to settle into an orbit at perigee distance. Otherwise it swings into perigee and then swing out to apogee. Depending on the eccentricity of the orbit, it could spend the vast majority of its time much further away, since it will be moving much faster near perigee than when near apogee and spends more of its time at a further than average distance than it does at a closer than average distance. (think about Comets that speed decades in the outer reaches of the solar system, and then come swooping in, spending just a few months in the inner system.)
Crazy stuff I know. Thanks in advance!

Hope this was of some help.
 
  • #5
Kkelloww said:
And a free copy!

Welcome to the PF.

We do not allow offers of payment or gifts for the help you receive here.

Sorry Janus! :smile:
 
  • #6
Helping out what little I can is enough for me. In fact, if you find anything useful here and fell the need to do an acknowledgment, just acknowledge PF as a whole.
 
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  • #7
Janus said:
Tidal effects vary by the inverse of the cube of distance. Since 35,000 km is almost 1/11 the present distance of the Moon, tidal effects would be over 1300 times greater.
(the cube law - get it)
How to translate this 1300 into tide levels?
wiki said:
The theoretical amplitude of oceanic tides caused by the moon is about 54 centimetres (21 in) at the highest point, which corresponds to the amplitude that would be reached if the ocean possessed a uniform depth, there were no landmasses, and the Earth were rotating in step with the moon's orbit.

Can it be taken at face value? With tidal change in the middle of ocean trying to reach of 0.7 km? And even greater results in some straits?
 
  • #8
Czcibor said:
(the cube law - get it)
How to translate this 1300 into tide levels?Can it be taken at face value? With tidal change in the middle of ocean trying to reach of 0.7 km? And even greater results in some straits?
That 54 cm height is for a ocean of uniform depth and no continents and the Earth tidally locked to the Moon. Since none of these conditions apply, Neither does the 54 cm height. The Earth rotating withj respect to the tide, the varying depths of the Ocean and the existence of continents all work to magnify the effects of the tides. If you look at a typical coastal tide chart, you will note that even neap tides can reach heights of +8 ft or 2.44 meters. That multiplied by 1300 would give you coastal tides of over 3 km.
You will basically have tides that will flood the vast majority of land masses every time the Moon approaches perigee.
 
  • #9
Janus said:
That 54 cm height is for a ocean of uniform depth and no continents and the Earth tidally locked to the Moon. Since none of these conditions apply, Neither does the 54 cm height. The Earth rotating withj respect to the tide, the varying depths of the Ocean and the existence of continents all work to magnify the effects of the tides. If you look at a typical coastal tide chart, you will note that even neap tides can reach heights of +8 ft or 2.44 meters. That multiplied by 1300 would give you coastal tides of over 3 km.
You will basically have tides that will flood the vast majority of land masses every time the Moon approaches perigee.

I was asking about possibility of just making simple multiplication because I wonder whether it wouldn't be a challenge to move all those water in a such a short time? (I mean in a few seas trying to move a few km would be a bit tricky, because they are only a few hundred meters deep, what makes them just dry shortly after starting a low tide)
 
  • #10
Czcibor said:
I was asking about possibility of just making simple multiplication because I wonder whether it wouldn't be a challenge to move all those water in a such a short time? (I mean in a few seas trying to move a few km would be a bit tricky, because they are only a few hundred meters deep, what makes them just dry shortly after starting a low tide)

Working out the actual tide heights would be a huge challenge. Even now, tide heights and times can vary along the same coastline due to topology. Coastal ranges will contain the rising tides, but then it will pour through passes. Much of this water will end up trapped on the other side when the tide recedes (like giant tidal pools). Where river valleys ( like the Columbia river basin) cut through such ranges, the tides will rush through like a torrent, flooding huge regions. In addition, the eccentricity of the Moon's orbit adds additional complexities. Consider the scenario I worked out earlier with a Moon apogee at its present orbital distance. In such a case, the moon would go from 90 degrees before perigee to 90 degrees after in ~44 hours or just under 2 days, it will spend the other 9 days in the othr "half" of its orbit.
During this time it will go from ~58,000 km to 35,000 km orbital distance. and the tidal force will go from 290 to over 1300 times its present value. During this same period the angular velocity of the Moon will go from just a little less than the angular velocity of the Earth to a bit under twice as fast. So the speed and direction of the tidal bulge movement with respect to the Earth will vary considerably during those 44 hours. As seen from the surface of the Earth during that 44 hrs it will be at first moving Westward quite a bit slower than it is now, slow to a stop, then start moving Eastward, picking up speed until moving a bit slower than it presently is, then it will slow to a stop again and then start moving Westward again picking up speed until it is moving that same speed it was when it started. During this time the tidal bulge will swell and shrink as noted above.
 
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  • #11
Thanks for the replies guys. Basically, we would be talking about the end of most life. How large would the moon appear at perigee? I guessing about 10 degrees? How fast would it traverse the visible sky? Would it go through phases from full to new like it does now? Would the variable gravitational effects destabilize the orbits of all satellites?
 
  • #12
Kkelloww said:
Thanks for the replies guys. Basically, we would be talking about the end of most life. How large would the moon appear at perigee? I guessing about 10 degrees?
between 5.7 and 7 degrees depending on whether or not the Moon is directly ahead or on the horizon. ( at perigee the Moon will be close enough that the difference in distance in viewing it from these points is enough to make a visible difference.
How fast would it traverse the visible sky?
that will vary from ~12.8 degrees/hour from West to East to ~14 degrees/hour from East to West, depending on which part of the orbit it is in. Most of the time its angular velocity will be less that of the Earth's and you will see it pass from East to West. But as it get closer to perigee it will increase to being greater than that of the Earth's rotation and you will see it go from West to East. In between these two, there will be moments where it will actually appear to slow to a stop in the sky as it passes between these two phases.( again, these numbers are based on my earlier assumed orbit. There are tons of orbits that have a 35,000 km perigee, and they would give different results. )
Would it go through phases from full to new like it does now?
Assuming that whatever changed its orbit didn't change its inclination by too large a degree, the phases would stay pretty much like they are now, the timing will be different due to the shorter period, and the phases would not be of equal length. ( for instance if the Full moon were to happen at perigee, then the time from first quarter through full moon to third quarter will be much shorter that the time from third quarter through new moon back the first quarter will be. Since the synodic month ( full moon to full moon will not be the same length as the anomalistic month ( perigee to perigee) These timings between quarters will also be constantly changing.
I'd like to point out something else here. It's about the Moons own rotation. At the present time it rotates once every 27 1/4 days. If this remains unchanged by what causes the orbit change, then we will no longer just see one side of the Moon. In fact, since it would take an amazing stroke of chance for whatever changes the Moon's orbit to change its rotation by just the right amount for its new rotational period to match that of its new orbit, you can pretty much assume that we would see all sides of the Moon. ( And actually, even if this highly unlikely chance were to occur, we would still see the vast majority of the Moon's surface due to libration.
Would the variable gravitational effects destabilize the orbits of all satellites?

There will be some perturbing effects, but most of the lower satellite will stay in orbit (including the GPS ones, however, their orbits would be perturbed enough to make them useless). The geostationary communication satellites won't fare as well as they orbit at greater than the 35,000 km perigee. Close approaches and out right collisions will sweep them clear.
 

Related to Need assistance with orbital dynamics questions

1. What are orbital dynamics?

Orbital dynamics is the study of the motion of objects in space, specifically how objects move and interact in orbit around a central body, such as a planet or star.

2. Why is understanding orbital dynamics important?

Understanding orbital dynamics is important for many applications, such as space exploration, satellite communication and navigation, and space mission planning. It also helps us better understand the behavior of celestial bodies in our solar system.

3. What factors affect orbital dynamics?

The main factors that affect orbital dynamics include the mass of the central body, the distance between the two objects, and the velocity and direction of the orbiting object. Other factors such as atmospheric drag and gravitational perturbations from other objects can also influence orbital dynamics.

4. How do you calculate orbital dynamics?

Orbital dynamics can be calculated using mathematical equations based on Newton's laws of motion and universal gravitation. These equations take into account the mass of the objects, their velocities, and the distance between them.

5. What are the different types of orbits?

There are several types of orbits, including circular, elliptical, parabolic, and hyperbolic orbits. These orbits differ based on the shape of the path and the velocity of the orbiting object. In addition, there are also different types of orbits such as geostationary, polar, and heliocentric orbits, which have specific applications and characteristics.

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