Velocity of Earth Around Sun: Recent Claims & Calculations

[monty37]

has the Earth been revolving around the sun with a constant velocity?if we
go about calculating the velocity ,wrt to we on Earth it will change ,when calculated
from another planet.does the Earth's rotation have a constant velocity,recent claims
have said that the Earth's velocity is slowing down.

 

[Cantab Morgan]

Be careful when you use the term "velocity," which has a different definition than "speed." You're really asking whether the speed is constant.

Kepler's laws of motion assure us that the speed of an orbiting planet is not constant. The rule is that a planet sweeps out equal areas (not equal arclengths) in equal times. There's a helpful picture here...

http://www.ancient-world-mysteries.com/kepler-equal-areas-law.html

I'm sorry I'm having trouble understanding the rest of your question. The velocity of an object with respect to itself is always zero, so I don't think that's quite what you mean.

 

[mgb_phys]

does the Earth's rotation have a constant velocity,recent claims
have said that the Earth's velocity is slowing down.

The Earth's rotation rate (ie the length of a day) is slowing down.

 

[Cantab Morgan]

The Earth's rotation rate (ie the length of a day) is slowing down.

That's really interesting! How can that be? Where is the momentum going?

 

[DaveC426913]

That's really interesting! How can that be? Where is the momentum going?

Into the Moon's orbit.

 

[mgb_phys]

Ultimately to the moon. Tidal friction between the Earth and the moon slows the Earth's rotation and increases the moon's orbital radius.

 

[Cantab Morgan]

Ultimately to the moon. Tidal friction between the Earth and the moon slows the Earth's rotation and increases the moon's orbital radius.

Whoa! That hurts my brain. So if I take mechanical energy out of the Earth-Moon system (presumably the tidal friction is creating heat), then the distance between them increases? That's really not what my intuition would have told me.

That must be why I'm finding this thread so interesting.

 

[DaveC426913]

Whoa! That hurts my brain. So if I take mechanical energy out of the Earth-Moon system (presumably the tidal friction is creating heat), then the distance between them increases?

What? No.

Not out the Earth-Moon system, just out of the Earth. It is being transferred to the Moon.

The Moon pulls on the various knobs, protrusions and water-bodies on Earth, slowing it down. (Eventually, Earth will become tidally-locked with the Moon, just like the Moon is with Earth. ) The knobs, protrusions and water-bodies conversely pull on the Moon. They accelerate it, speed it up. A faster orbit means a higher orbit. Thus, the Moon is slowly spinning away from us.

 

[Cantab Morgan]

DaveC426913, thanks for the kind explanation. Note that I wrote mechanical energy, not momentum. I'm trying to obey both the conservation of energy and the conservation of (angular) momentum. I had assumed that the tidal friction was converting mechanical energy to heat, so that the total mechanical energy of the Earth-Moon system is decreasing ever so slightly.

A faster orbit means a higher orbit.

Really? Mercury orbits more slowly than Mars?

 

[DaveC426913]

DaveC426913, thanks for the kind explanation. Note that I wrote mechanical energy, not momentum. I'm trying to obey both the conservation of energy and the conservation of (angular) momentum. I had assumed that the tidal friction was converting mechanical energy to heat, so that the total mechanical energy of the Earth-Moon system is decreasing ever so slightly.

Mayhap, but it is not significant to your question.

Really? Mercury orbits more slowly than Mars?

No you misunderstand. For a given body, if you accelerate its orbital speed, you will increase its orbital distance.

 

[Cantab Morgan]

Thanks for your patience. I am enjoying this exchange!

No you misunderstand. For a given body, if you accelerate its orbital speed, you will increase its orbital distance.

But if the Moon's centripetal acceleration is $$v^2/r$$ and the gravitational acceleration is $$GM/r^2$$, then when I set these equal, I find that the distance from the Earth to the Moon and the square of the Moon's orbital speed are inversely related. It would seem that increasing my speed will decrease my altitude.

Mayhap, but it is not significant to your question.

Ah. I'm not quite willing to concede that point. I will continue to reflect upon it. I think that the transfer of momentum can only occur because of the loss of mechanical energy. Consider the famous experiment of sitting on a spinning swivel chair with barbells in my hands. When I pull in my arms, my rotation rate changes, but only because I have done work.

Also, I should explain where my original skepticism/difficulty comes from. If the Earth's changing rotation rate transfers momentum to its satellite, then it's hard for me to understand how objects without satellites can have their rotation rates change. And yet the star system offers many examples of moons with no satellites of their own whose "days" and "years" are in lockstep, so such changes must have occurred.

 

[DaveC426913]

But if the Moon's centripetal acceleration is $$v^2/r$$ and the gravitational acceleration is $$GM/r^2$$, then when I set these equal, I find that the distance from the Earth to the Moon and the square of the Moon's orbital speed are inversely related.

If you add energy to the orbit, what's going to happen? Is it going to move inward?

It would seem that increasing my speed will decrease my altitude.

Well that would be wrong. I'm not sure what I can do to convince you of this.

How does the space shuttle achieve a higher orbit? It fires its rockets forward, increasing its orbital velocity, which raises it to a higher orbit.

Higher orbits require more energy. If they didn't, Moon shots would be easy.

Ah. I'm not quite willing to concede that point. I will continue to reflect upon it. I think that the transfer of momentum can only occur because of the loss of mechanical energy. Consider the famous experiment of sitting on a spinning swivel chair with barbells in my hands. When I pull in my arms, my rotation rate changes, but only because I have done work.

Not sure what I've said that goes against this.

Also, I should explain where my original skepticism/difficulty comes from. If the Earth's changing rotation rate transfers momentum to its satellite, then it's hard for me to understand how objects without satellites can have their rotation rates change. And yet the star system offers many examples of moons with no satellites of their own whose "days" and "years" are in lockstep, so such changes must have occurred.

It works nicely with the parent star too. It's not specific to satellites - it's about tides. Suns cause tides too. Tides act to slow (or speed up) a body's rotation until it is in lock-step with the parent body. In Pluto/Charon's case, because they are so close in mass, it has worked both ways.

 

[Janus]

Thanks for your patience. I am enjoying this exchange!



But if the Moon's centripetal acceleration is $$v^2/r$$ and the gravitational acceleration is $$GM/r^2$$, then when I set these equal, I find that the distance from the Earth to the Moon and the square of the Moon's orbital speed are inversely related. It would seem that increasing my speed will decrease my altitude.

If you start with an object in orbit, and give it a forward thrust, it will be moving faster than its proper orbital speed for that altitude. It will start to climb upward, into a higher orbit, as it does so, it begins to lose kinetic energy in exchange for the potential energy it gains by climbing. It slows as it climbs, ending up in a higher but slower orbit.

The Earth is applying a constant forward thrust to the Moon, in response, the moon climbs and sheds speed. But, since the Earth is constantly applying a forward thrust, the Moon is always moving just a bit faster than it should for its distance, So it just keeps climbing.

And yes, the Earth does lose a fair amount of energy to heat. So the Earth-Moon system does lose net energy. But there is also a transfer of energy from Earth to Moon. IOW, the amount of energy the Moon gains is less than the total energy the Earth loses.

 

[Cantab Morgan]

Janus, thanks for explaining so crisply what I was unable to articulate. What I really was going for was that if I were to change the altitude of my circular orbit -- however I accomplish that -- my orbital speed at the higher orbit will be smaller.

Of course Dave is correct. Adding energy does not make a satellite move inward. But ultimately it does make it move more slowly. The devil is in the details of course, because I would need at least two rocket burns. One to gain the altitude and the other to restore circularity to my orbit. I think where I was getting so confused is that the first of these burns does increase my speed, which is where Dave was so gently trying to steer my thinking.

The Earth is applying a constant forward thrust to the Moon, in response, the moon climbs and sheds speed.

Earth applying thrust to the Moon is exactly what is so difficult for me to visualize, given that gravity is a central force. But I guess it must somehow be true, or else, how could the moon's orbit be climbing?

 

[DaveC426913]

Earth applying thrust to the Moon is exactly what is so difficult for me to visualize, given that gravity is a central force. But I guess it must somehow be true, or else, how could the moon's orbit be climbing?

Pretend Earth is a merry-go-round.

Every mountain that protrudes from the Earth is a kid on the merry-go-round with his hand sticking outward.

Pretend the Moon is you, standing next the the merry-go-round.

Every time a kid with an outstuck hand comes around (a mountain), he grabs your shirt for a second. Slowly, you are propelled forward, tangential to the motion of the merry-go-round (i.e. Moon achieves greater velocity in its orbit).

 

[monty37]

the Earth has had protrusions from the time it had existed(oceans etc) so,why should it
suddenly slow now,and when referring to earth,we call its angular'velocity'
and not speed ,right,since it has a fixed orbit or direction,elliptical if iam right.so
if i calculate Earth's velocity sitting on Earth itself,it will change from
the velocity of Earth calculated wrt space or another planet.how can u call
it tidal 'friction'?and are the revolutions constant,will the moon's tidal friction affect Earth's revolution??

 

[mgb_phys]

Tidal friction is what slows Earth down, it's a little bit mis-named you can have tidal locking due to any irregularities - it's not necessarily due to water tides (mercury is tidally locked in a resonance to the sun)

The rotation rate of the Earth has been slowing down since the moon was formed. For dinosaurs the year was 4-500days, it was the same length of time (in seconds) but the faster rotating Earth meant they got more days.

 

[Count Iblis]

This is a nice exercise:

Suppose that in the future we use all of the Moon's materials here on Earth so that nothing is left of the Moon and all the matter from the Moon ends up here on Earth. What will then be the length of a day?

 

[Janus]

Earth applying thrust to the Moon is exactly what is so difficult for me to visualize, given that gravity is a central force. But I guess it must somehow be true, or else, how could the moon's orbit be climbing?

The moon raises tidal bulges on the Earth. Friction between the rotating Earth and the tidal bulges tends to drag these bulges out of alignment with the Moon. The bulge nearest the Moon then leads the Moon a little bit. The gravitational attraction between this bulge acts to pull forward on the Moon in its orbit (The bulge on the other side of the Earth will tend to pull backwards on the Moon, but since it is further away by the diameter of the Earth, it has a smaller effect).

The drag between tidal bulges and rotating Earth also slow the rotation of the Earth. Thus the Earth transfers rotational energy to the Moon.

If the Moon orbited the Earth in less time than the Earth rotates (Like the situation between Phobos & Mars), The opposite would occur, the tidal bulges would lag behind the Moon, exerting a force against its orbital direction, and the Moon would spiral in towards the Earth while the Earth gained rotational speed(the drag between tidal bulges and Earth would pull it forward with respect to its rotation).

If the Moon orbited retrograde( in the opposite direction of Earth's rotation), then the bulges would pull backward on the Moon, drawing it into a lower and lower orbit, and the drag between tidal bulges and Earth would slow its rotation.

 

[Cantab Morgan]

Hi, monty37. There's quite a lot going on in your post, but let me try to parse it out.

the Earth has had protrusions from the time it had existed(oceans etc) so,why should it
suddenly slow now

I don't think anyone is suggesting that the Earth has suddenly started to get slower, but that it has been doing so all along.

and when referring to earth,we call its angular'velocity'
and not speed ,right,since it has a fixed orbit or direction,elliptical if iam right.so
if i calculate Earth's velocity sitting on Earth itself,it will change from
the velocity of Earth calculated wrt space or another planet.

I'm afraid I'm too obtuse to understand what you have written there. Can you be a little clearer, please?

how can u call
it tidal 'friction'?

Friction is a non-conservative force that produces heat. Gravity is a force whose strength falls off with distance. (Actually with the square of the distance.) The Earth is so large that the side of the Earth closest to the Moon experiences a greater gravitational pull than the side farthest from it. This causes the tides. The continual deformation of the shape of the Earth by the tides produces heat, so the phrase tidal "friction" seems appropriate.

and are the revolutions constant,will the moon's tidal friction affect Earth's revolution??

That is a very intelligent question. We've been talking about the slowing of the Earth's rotation, but it's interesting to speculate about whether there's also an effect on its revolution.

 

[DaveC426913]

That is a very intelligent question. We've been talking about the slowing of the Earth's rotation, but it's interesting to speculate about whether there's also an effect on its revolution.

There's no reason why it would. The revolution about the sun is treated not merely as the Earth but as the Earth-Moon system i.e. it can be treated as one unit so any forces within will not have an affect outside the unit.

 

[monty37]

can you please explain why it does not affect revolution again?
since the Earth moves in a fixed(elliptical)orbit we can refer to it as velocity ,right.
so we can calculate angular velocity using ω =2Πn or F=(m.r.r.ωω)\r ,right

 

[DaveC426913]

can you please explain why it does not affect revolution again?
since the Earth moves in a fixed(elliptical)orbit we can refer to it as velocity ,right.
so we can calculate angular velocity using ω =2Πn or F=(m.r.r.ωω)\r ,right

Oh, OK, well if you plot the Earth's position on its path arond the sun and ignore the Moon you will see the Earth bobbing in a circular motion all the way around its orbit. Every 28 days or so the Earth will alternately go a little faster along its orbit and then a little slower. (Of course, there will be no effecton the length of the year.)

But that's a kind of myopic way of looking at it. It's analagous to plotting the motion of only one of the skaters in a two-skater "death-spiral" and pretending the other skater is invisible.

 

[monty37]

you say,it is circular, is not the motion elliptical,rather?ok,but does Earth have a value as its velocity,has it been found ,i have
never read any particular value known to be the velocity of Earth till date,so as i asked
can it be found using the angular velocity formula??

 

[mgb_phys]

you say,it is circular, is not the motion elliptical,rather?

Technicaly it's an elipse - but it's only 1.5% eccentric, so it's a circle in common terms.

,but does Earth have a value as its velocity,has it been found ,i have
never read any particular value known to be the velocity of Earth till date,so as i asked
can it be found using the angular velocity formula??

The velocity is simply the length of the orbit divided by the time
semimajor axis 149,597,887.5 km
semiminor axis 147 098 074 km
time = 1 year = 365.256366 days
You can then get the speed from the circumference of the elipse (= 29.783 km/s)

Or you can just rememebr the Monty Python song
Z2JU4gX6rg8[/youtube]

 

[D H]

I don't think anyone is suggesting that the Earth has suddenly started to get slower, but that it has been doing so all along.

Actually, the slowing of the Earth's rotation rate has increased in the recent past (recent being relative to the 4.6 billion year age of the Earth). A simple linear extrapolation of the current Earth-Moon recession rate yields the Moon being within the Roche limit a billion years or so ago. This flies in the face of evidence that the Moon has been the Earth's moon for 4.5 billion years or so. The obvious conclusion is that the simple extrapolation is wrong. The current lunar recession rate is anomalously high, and hence so is the current rate at which length of day is increasing. The Earth presently has two anomalous features that explain this anomalously high rate: Africa is connected to Eurasia and North America is connected to South America. This prevents a world-girdling tidal flow at the equator, and doubly so.

 

[monty37]

well but we are not talking of linear velocity to calculate distance\time, angular
velocity(omega) ,earth's velocity has to be calculated in terms of omega and frequency
since it is 'angular ' and not linear.by considering time and semi-major or minor axis
we arrive at linear velocity

 

[mgb_phys]

Wasn't clear what velocity you were taking about.
The angular velocity is measured all the time by a few observatories around the world.
All you do is look at a star overhead, look at the clock, wait nearly 24hours, when the star is overhead again check the clock.
Interestingly the answer is top secret - it only gets released 3-6months ater.

 

[lazypast]

im just curious, D H said something about the rate of radius increase to be the same as that of 1bn year ago.

what sort of evidence is there for this. for example we can see iron filings line up and so can see the direction of Earth's magnetic field.

what can we measure which gives a relationship between era and the rate of recession?

 

[russ_watters]

It was a calculation, lazypast - if you extrapolate backwards, you get a physically impossible result to the calculation (moon being closer to Earth than is physically possible), therefore the calculation must be wrong.