Gravitational potential energy of asteroid

[mikeph]

what is a rotational frame ?

A frame in which a rotating body maintains a fixed point. It's a form of transformation.

eg. think of a coordinate frame where the Earth's position is always at (a,0), and the sun's position is always at (-b,0). Then this frame is rotating with the angular speed of the Earth around the sun. If you look at things in this frame, you find that satellites will accelerate away from the Earth, and also as they move away, they will follow curved paths.

 

[Frame Dragger]

Speaking of rotating frames... Sorry, I couldn't resist! Anyway, my username and Lense-Thirring and all of that.

 

[sophiecentaur]

No, DH has it right. If the Earth were a sphere instead of a geoid that would mean that it would be lower at the equator and higher at the poles than it is now. So, more ocean and atmosphere at the equator and less at the poles compared to now.

Because of the geometry and not because of the different g? That's fair enough. But is it true to say that the difference in sea depth would be as big as the difference in diameter? The implication seemed to be that there would be an enormous 'bulge' of sea and atmosphere. Would there be? (i.e about 40km difference implies some very deep water.)

But it would just be 'different' and it always would have been - so we would just see it as normal, I reckon. And what is a 'normal' Young modulus for a planet and what is a 'normal' amount of equator spread?

 

[D H]

"If the Earth was a rigid sphere, the oceans (and the atmosphere) would be piled up around the equator. The air at the poles would be very, very thin."
I think you need to reconsider that statement. The difference in radius is very small and the difference in g is, in fact, ' the wrong way round, with less g at the equator, not more than at the pole. The centrifugal (there, I said it and I don't care) effect is much less than the effect of g so the sea and atmosphere are merely thrown a bit 'uphill' towards the equator. There is plenty of restoring force to stop them all glooping to the equator.

The Earth's equatorial radius is 6,378.1 km while the polar radius is 6,356.8 km -- a difference of 21.3 km. That is not "very small". The effective height of the atmosphere is 19.3 km.

An article of interest: http://www.esri.com/news/arcuser/0703/geoid1of3.html
In particular, check out the "SPHEROID DEM" what-if scenario on page 2.

 

[Dale]

Because of the geometry and not because of the different g? That's fair enough. But is it true to say that the difference in sea depth would be as big as the difference in diameter? The implication seemed to be that there would be an enormous 'bulge' of sea and atmosphere. Would there be? (i.e about 40km difference implies some very deep water.)

I think the difference is around 21 km in diameter, so about 10 km in radius. So that is considerably higher than Mt. Everest. So, it would indeed be "enormous" from a human perspective, but less than 1% from Earth's perspective.

But it would just be 'different' and it always would have been - so we would just see it as normal, I reckon. And what is a 'normal' Young modulus for a planet and what is a 'normal' amount of equator spread?

Yes, we would see it as normal, but it would mean that weather would be considerably different and the habitable regions of the planet would be different from what we see as normal today.

 

[D H]

I think the difference is around 21 km in diameter, so about 10 km in radius. So that is considerably higher than Mt. Everest.

It's 21.3 km in radius, not diameter. That is several times that of Mt. Everest, and most of Everest's height results from Everest comprising comparatively light rock afloat on the lithosphere.

 

[Dadface]

hmm.. i don't really understand. my lecturer gave his answer as this , hope it helps.

The middle term is the ordinary translational KE of the rocket(v=rw)due to the Earth's spin.The size of the rocket can be considered as negligible when compared to the size of the Earth and to an excellent first approximation the rocket can be treated as a point object.

 

[sophiecentaur]

It's 21.3 km in radius, not diameter. That is several times that of Mt. Everest, and most of Everest's height results from Everest comprising comparatively light rock afloat on the lithosphere.

. . . and other posts. .
What I get my head around is the apparent coincidence that the Earth happens to have just the right stiffness modulus to make it the proportions that it is. Or would it still have the same shape if it were made of dough - with the appropriate density?
Was the suggestion of a possible 'rigid' Earth just not feasible?

 

[Dale]

It's 21.3 km in radius, not diameter.

Oh, my mistake.

 

[D H]

What I get my head around is the apparent coincidence that the Earth happens to have just the right stiffness modulus to make it the proportions that it is.

Where did you get that idea?

Or would it still have the same shape if it were made of dough - with the appropriate density?

The Earth is *old* and plastic. Suppose the Earth's shape were such that it was well out of isostatic equilibrium. This does happen occasionally. For example, kilometers-thick sheet of ice have covered vast parts of the Northern Hemisphere at times and then melted very quickly. While the Earth is still recovering from the last ice age, most of the isostatic rebound occurred during the first 5,000 years after the ice melted. 5,000 years is an incredibly short time compared to the age of the Earth.

Compared to the age of the Earth, that 5,000 year time constant is essentially instantaneous.

Was the suggestion of a possible 'rigid' Earth just not feasible?

Not feasible, given the temperatures in the Earth. It does however make for a neat what-if game. Did you read the article cited in post #34?

 

[quietrain]

A frame in which a rotating body maintains a fixed point. It's a form of transformation.

eg. think of a coordinate frame where the Earth's position is always at (a,0), and the sun's position is always at (-b,0). Then this frame is rotating with the angular speed of the Earth around the sun. If you look at things in this frame, you find that satellites will accelerate away from the Earth, and also as they move away, they will follow curved paths.

ok, if i visualize the Earth and sun to be at fixed points, then when i look at this frame, its as though i am looking at two holes in a vertical plane of glass rotating about the z-axis.

so why would the satellites be accelerating away from earth? am i suppose to visualize the satellites as not moving in that plane of glass? but as separate objects ?

 

[quietrain]

The middle term is the ordinary translational KE of the rocket(v=rw)due to the Earth's spin.The size of the rocket can be considered as negligible when compared to the size of the Earth and to an excellent first approximation the rocket can be treated as a point object.

how come rocket is brought into the picture? >< . the question is about estimating the flattening of the Earth , given by the equation 1-rp/re or sorts, and so my lecturer just set up this "equal potential on the surface of earth" and manipulate the equation to the form of rp/re, which is = to (w^2 R^3)/2GM. wow. i mean this seems sort of deep for me.

so, what exactly is potential?

i know potential energy is the work required to bring a point from infinity to a reference point

i know what gravitational force is , its the attraction of something onto something.

but what is potential? its like a mid way between the force and the energy?

 

[rcgldr]

The middle term is the ordinary translational KE

It works out to be that way, except for the negative sign, but that term is derived from the integral of CPE from 0 to r, not from the KE of an object at the surface of the earth. Repeating my previous post about this: derivation of term for CPE (centrifugal potential energy):

<br /> F = \frac{m v^2}{r} = \frac{m \ \omega^2 r^2}{r} = m \ \omega^2 r<br />

<br /> U = -\int_0^r \m \ \omega^2 r dr<br />

<br /> U = -\frac{1}{2}\ m \ \omega^2 r^2<br />

Earth's shape ... isostatic equilibrium.

I realize the Earth is spheroidal, but is its suface really equipotential as described in the attached images (ignoring surface issues like mountains)?

What exactly is potential?

Potential is essentially potential energy with the target object removed, so it's just based on the strength of the field of the source object. In the case of gravity close to the Earth surface where the Earth can be considered to be an infinitely large disk, gravitaional potential energy = m g h. Remove the target object and it's mass, and you end up with gravitational potential = g h. Since gravity near the Earth surface is a constant, gravitational potential is relative to height. Although not commonly used for gravity, potential is commonly used for electrical fields and electric potential uses the unit "volt".

http://en.wikipedia.org/wiki/Volt

http://en.wikipedia.org/wiki/Electric_potential

centrifugal force

For a real object (one with mass), centrifugal force only exists if there is some centripetal force causing the object to follow a circular path, regardless of the frame of reference. In this particular case, of an object at the equator of the earth, part of the gravitational force provides the centripetal force, the rest of the gravitational force provides the weight of the object (the force the object exerts onto the surface of the earth).

 

[Frame Dragger]

ok, if i visualize the Earth and sun to be at fixed points, then when i look at this frame, its as though i am looking at two holes in a vertical plane of glass rotating about the z-axis.

so why would the satellites be accelerating away from earth? am i suppose to visualize the satellites as not moving in that plane of glass? but as separate objects ?

Ultimately yes, that would help you in the future. Remember that all intertial frames have their own relative measures of all other frames. Approximations used to teach are just that... approximations. This ties right into your issue with potential energy: it's a way of measuring the potential of a system do work. Remember that it's something that exists in equations used to measure "things", and in the absence of the relevant interactions described therin, it has little to no meaning.

 

[mikeph]

ok, if i visualize the Earth and sun to be at fixed points, then when i look at this frame, its as though i am looking at two holes in a vertical plane of glass rotating about the z-axis.

so why would the satellites be accelerating away from earth? am i suppose to visualize the satellites as not moving in that plane of glass? but as separate objects ?

The satellites would be accelerating away from Earth because in your rotating frame you are forced to accept the existence of inertial forces. The satellites will probably move in that "plane of glass", or rotating frame, unless they are in very special orbits (there are five points where they will be stationary).I did a load of simulations on how satellites will move in this frame, and I still have a few videos of this but they're on my phone and I don't have a cable to connect it, I'll download them later because there are really some very weird paths that the satellites can take in this rotating frame. Think about a spirograph, for example- this is what you will see when you view an elliptical orbit in a rotating frame. It's pretty hard to work out why the object moves like it does, but it is essentially a combination of gravitational attraction, centrifugal repulsion, and the Coriolis force.

 

[quietrain]

oh erm, last question,

so if potential is just removing the target mass, then does it mean that potential at infinity is 0 just like GPE is defined? or is potential infinity at infinity since it is proportional to height only?

also my g term is GMe / r^3? since mgh = GMem/r^2

 

[D H]

I realize the Earth is spheroidal, but is its suface really equipotential as described in the attached images (ignoring surface issues like mountains)?

Yes and no.

The "no" response first: That equation is not correct. A quick check shows that it cannot be right. It implies that

\frac 1 {R_p} = \frac 1 {R_e} + \frac 1 2\,\frac{(R_e\Omega)^2}{\mu}

Plugging in the known values for the Earth's equatorial radius (6,378.137 km), the Earth's rotation rate (2*pi/23.9344696 hours), and the Earth's gravitational parameter (398,600.4418 km³/s²) yields a polar radius of 6367.1 km, and that is far from the mark. The correct value is 6,356.752 km.

The problem here is that the equation in that attached image implicitly assumes a spherical Earth by having potential as μ/r. The whole point of this exercise is to find out why the Earth is not spherical. A better model for the gravitational potential at some point on or above the surface of the Earth that accounts for the Earth's oblateness is

\Phi(r,\lambda) =<br /> \frac{\mu}{r}\left(<br /> 1-J_2 \left(\frac{R_e} r\right)^2 \frac{3\sin^2\lambda - 1} 2<br /> \right)

where

• r is the distance of the point in question from the center of the Earth,
• λ is the geocentric (not geodetic) latitude of the point in question, and
• J₂ is the first non-spherical term in the spherical harmonics expansion of gravitational potential. For the Earth J₂=0.00108263.

With this, the expression for the Earth's polar radius becomes

\frac 1 {R_p}\left(1-J_2\left(\frac{R_e}{R_p}\right)^2\right) =<br /> \frac 1 {R_e}\left(1+\frac{J_2}2\right) + \frac 1 2\,\frac{(R_e\Omega)^2}{\mu}

Solving this for the Earth's polar radius yields 6356.743 km. The accepted value is 6,356.752 km, so now the error is all of 9 meters. Compare this to 10 kilometer error that results from ignoring Earth's oblateness.Now the "yes" response: The Earth's surface is an equipotential surface of gravitational potential plus centrifugal potential.

Although not commonly used for gravity, potential is commonly used for electrical fields and electric potential uses the unit "volt".

Gravitational potential is used for gravity in several regimes. The best models for the Earth's gravitational field are spherical harmonics expansions of the Earth's gravitational potential. Some of the more recent models:

EGM96 (Earth gravity model, 1996), a 360x360 spherical harmonics model of the Earth's potential, see http://cddis.nasa.gov/926/egm96/egm96.html.

GGM02C (GRACE gravity model 02 constrained), a 200x200 spherical harmonics model of the Earth's potential, see http://www.csr.utexas.edu/grace/gravity/

EGM2008 (Earth gravity model, 2008), a 2190x2159 spherical harmonics model of the Earth's potential, see http://earth-info.nga.mil/GandG/wgs84/gravitymod/egm2008/index.html

 

[Dadface]

Look at it in a slightly different way.Imagine a mass m hanging from a string at the poles and then at the equator.In both cases there are two forces on the mass,the gravitational force(GMm/r^2) which acts downwards and the tension(T) in the string which acts upwards.
POLES Weight =T
EQUATOR Weight-T=mrw^2 At the equator the mass is accelerating towards the Earth's centre and the centripetal force=weight-T.It follows that T,the apparent weight is smaller at the equator, than at the poles.
I have just looked up the values of g at the poles and the equator.
POLES g=9.832m/s^2
EQUATOR g=9.780m/s^2
The book I used is old and with a search you may get some updated figures

 

[quietrain]

See post 19 for the discussion on defining zero points.

https://www.physicsforums.com/showpost.php?p=2457167&postcount=19

? that post doesn't say much about my question

 

[quietrain]

Look at it in a slightly different way.Imagine a mass m hanging from a string at the poles and then at the equator.In both cases there are two forces on the mass,the gravitational force(GMm/r^2) which acts downwards and the tension(T) in the string which acts upwards.
POLES Weight =T
EQUATOR Weight-T=mrw^2 At the equator the mass is accelerating towards the Earth's centre and the centripetal force=weight-T.It follows that T,the apparent weight is smaller at the equator, than at the poles.
I have just looked up the values of g at the poles and the equator.
POLES g=9.832m/s^2
EQUATOR g=9.780m/s^2
The book I used is old and with a search you may get some updated figures

OH. i see. but shouldn't the summation of forces be weight + mrw^2 = tension at the equator?

 

[Dadface]

OH. i see. but shouldn't the summation of forces be weight + mrw^2 = tension at the equator?

The weight acts vertically downwards and the tension acts vertically upwards.There is a resultant(centripetal) force which acts vertically downwards,in the same direction as the weight,which is equal to weight-T.It may be easier to see it if you sketch a free body force diagram and mark in the direction of the resultant force,

 

[quietrain]

oh. i forgot centripetal force is resultant. i was drawing the free body diagram including it as a force :X.

it kind of reminds me of the vertical swinging of a string attached to a mass.

thanks!

anyway, so if potential energy is defined to be 0 at infinity, is potential 0 at infinity too?

 

[Dadface]

oh. i forgot centripetal force is resultant. i was drawing the free body diagram including it as a force :X.

it kind of reminds me of the vertical swinging of a string attached to a mass.

thanks!

anyway, so if potential energy is defined to be 0 at infinity, is potential 0 at infinity too?

We can only calculate how potential energy changes with separation and the choice of an infinite separation and then the choice of calling the potential at infinity zero are both arbitary choices but nevertheless generally agreed to be the choices that are most useful and easiest to work with.Someone else may decide to choose a separation of two million and twelve point five metres and choose that the potential at such a separation is fifty three point zero seven J/kg.They might be able to calculate how the potential energy changes with separation but the calculations will be more messy and not easily followed by others.

 

[quietrain]

ic thanks

 

[mikeph]

This video was the final part of my dissertation last year, basically a minute long showing how you can have a satellite orbit around the sun in a very weird but stable path, half of the orbit it will move faster than the earth, then just before it reaches it, it slows down and the Earth almost catches up with it, like a funny cosmic cat and mouse.

The simulation is 2D but in a 3D analogue a similar thing happens. I think they've found an object that does this with Mars.edit- I didn't make it clear, the central circle is the sun, and the circle on the right is Jupiter, I think the mass ratio was about 10^3 in this simulation. It is of course in a rotating reference frame, basically using a multistep iterative method with a rotating coordinate transform added onto the end of it

 

[quietrain]

:X somehow that graffiti looked like marsian to me

thanks though