Gravitational potential energy of asteroid

[quietrain]

hi, i don't really understand the meaning to GPE.

if we define a point at infinity to have 0 GPE, then any point before infinity would have -ve GPE.

so let's say an asteroid crashing to Earth at the surface of the Earth has initial kinetic energy of 1/2mv^2 and GPE of -GMM/r

so E(total) = KE + GPE = 1/2mv^2 - GMM/r

so since it is minus GMM/r, let's say GMM/r = 5 and KE = 10, then does it mean that the total energy is 10 -5 =5 J ?

but when the asteroid get closer to earth, won't there be stronger attraction due to the inverse square law . so shouldn't the energy at the surface be greater than when the asteroid was at infinity having only 1/2mv^2? in other words, at the surface, it should have 1/2mv^2 + GPE ?

so let's say when the asteroid at infinity has KE + PE, but gpe is 0 at r = infinity, so E = KE only.

so when the asteroid gets closer to earth, it starts to have the GPE term , E= 1/2mv^2 - GMM/r

so when r decreases, the E value decreases, which means the asteroid has less energy than before. this doesn't make sense right? or am i missing something?

thanks

 

[Doc Al]

hi, i don't really understand the meaning to GPE.

if we define a point at infinity to have 0 GPE, then any point before infinity would have -ve GPE.

so let's say an asteroid crashing to Earth at the surface of the Earth has initial kinetic energy of 1/2mv^2 and GPE of -GMM/r

so E(total) = KE + GPE = 1/2mv^2 - GMM/r

so since it is minus GMM/r, let's say GMM/r = 5 and KE = 10, then does it mean that the total energy is 10 -5 =5 J ?

Right.

but when the asteroid get closer to earth, won't there be stronger attraction due to the inverse square law . so shouldn't the energy at the surface be greater than when the asteroid was at infinity having only 1/2mv^2? in other words, at the surface, it should have 1/2mv^2 + GPE ?

Total energy is conserved. As the asteroid gets closer, its KE increases as its GPE goes down. But 1/2mv^2 + GPE remains fixed.

so let's say when the asteroid at infinity has KE + PE, but gpe is 0 at r = infinity, so E = KE only.

so when the asteroid gets closer to earth, it starts to have the GPE term , E= 1/2mv^2 - GMM/r

so when r decreases, the E value decreases, which means the asteroid has less energy than before. this doesn't make sense right? or am i missing something?

You are missing the fact (mentioned above) that total energy is conserved.

 

[Dale]

As r gets smaller GMm/r gets bigger. So let's say that the object falls from where GMm/r = 5 J to where GMm/r = 12 J. Then, in order to keep the total energy equal to 5 J the KE must have increased from 10 J to 17 J (17 J - 12 J = 5 J)

 

[D H]

so let's say when the asteroid at infinity has KE + PE, but gpe is 0 at r = infinity, so E = KE only.

so when the asteroid gets closer to earth, it starts to have the GPE term , E= 1/2mv^2 - GMM/r

so when r decreases, the E value decreases, which means the asteroid has less energy than before. this doesn't make sense right? or am i missing something?

thanks

What you are missing is that total mechanical energy, kinetic+potential, is conserved. Suppose the asteroid has some velocity at essentially infinite separation. This is such an important concept that it has a name: v-infinity, or in math, v_{\infty}. The total mechanical energy at some point in space is the sum of the kinetic and potential energy at that point, and since total mechanical energy is conserved,

\frac 1 2 v(r)^2 - \frac{GM_e}{r} = \frac 1 2 {v_{\infty}}^2

or

v(r)^2 = 2\,\frac{GM_e}{r} + {v_{\infty}}^2

As the asteroid gets closer to Earth, its velocity must increase.

 

[sophiecentaur]

hi, i don't really understand the meaning to GPE.

if we define a point at infinity to have 0 GPE, then any point before infinity would have -ve GPE.


thanks

You could say that the sign of the Potential is chosen so that you get the right result.
The potential at a distant point will be higher than at a near point (to a central mass). It is taken as zero at infinity (moving from place to place at a great distance will involve very little energy transfer). For an attractive force, the potential is negative (a 'potential well'). So an object loses PE as it gets closer. But the total energy will be the same.
As an object gets closer, its PE will be transferred to KE - it gets faster.
For a repulsive force, the Potential is positive, so an approaching object will slow down, lose KE and increase its PE.

 

[quietrain]

oh so , since KE + PE = Constant

so when r decreases, PE increases, so 1/2mv^2 - GMM/r = constant, so when the PE term increases, the KE term must increases to keep it the same constant? so let's say KE = 5, Pe = 3, constant will then be 5-3 = 2

so at R =infinity, PE term is approx 0.

so KE has to decrease to 2 to make the constant 2?

is this correct? thanks!

 

[sophiecentaur]

The PE, of course, increases in the negative direction. But, yes, your sums seem correct. The speed at which KE is equal to minus the PE would be the escape velocity. In your example, the rocket has more than enough. One day we may have engines which can achieve that economically. As it is, we need to use the slingshot effect to get away from the Sun.

 

[Doc Al]

oh so , since KE + PE = Constant

so when r decreases, PE increases, so 1/2mv^2 - GMM/r = constant, so when the PE term increases, the KE term must increases to keep it the same constant?

When r decreases, so does the gravitational PE. (Remember that it's negative.)

so let's say KE = 5, Pe = 3, constant will then be 5-3 = 2

so at R =infinity, PE term is approx 0.

so KE has to decrease to 2 to make the constant 2?

is this correct?

Yes, except that PE = -3, so KE + PE = 5 + (-3) = 2. (It's important to realize that the PE term is negative.)

 

[quietrain]

ah ic..

thanks!

 

[quietrain]

erm i met another hurdle

this time its about gravitational potential

apparently the gravitational potential on the Earth's surface is constant

so the equation given is -Gmm/r(equator) - 1/2mr(equator)^2ω^2 = -Gmm/r(pole)

anyone know what equation is this? i know the 1st and 3rd term are the potential at the equator and poles respectively but what is the 2nd term? it looks like rotational centripetal force * r(equator). but why?

thanks a lot!

 

[sophiecentaur]

It must represent the rotational Kinetic Energy, I suppose. That would be, for example, the energy 'helping' rockets to take off when they are pointed Eastwards, rather than upwards or backwards. I have a feeling that the sign of the term should be positive rather than negative, as shown, because, if you are using this KE then your potential is effectively less negative.

 

[quietrain]

issn't rotational kinetic energy 1/2Iw^2? the moment of inertia I of the Earth is 2/5MR^2 right? so it wouldn't make sense if we sub in ?

so how did it become 1/2mr^2w^2?

 

[sophiecentaur]

Surely it's the KE of the object and not the rotational energy of the Earth that counts in this one?? It's the KE of the rocket what helps it 'get off' and not the KE of the launchpad etc.
It would help if we used M and m, I think.

 

[quietrain]

Surely it's the KE of the object and not the rotational energy of the Earth that counts in this one?? It's the KE of the rocket what helps it 'get off' and not the KE of the launchpad etc.
It would help if we used M and m, I think.

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

 

[Doc Al]

That second term (-½mω²r², where r is the distance from the axis of rotation) represents the potential energy due to the centrifugal force that appears when you view things from a rotating frame. Note that since the force is outward, the centrifugal PE becomes smaller (more negative) as r increases, reaching its minimum value at the equator where r = R_e. (At the poles, the centrifugal PE term is zero since r = 0.)

An equipotential surface would be squashed at the poles and stretched at the equator.

 

[sophiecentaur]

Yes. The small m and big M in the equation make perfect sense now. It's like I said. The fact that you're spinning around the Earth when you are at the Equator means that you need less energy to get away to infinity (which is what is meant by your Potential Energy). They all use it in real space shots.

 

[rcgldr]

If we define a point at infinity to have 0 GPE, then any point before infinity would have negative GPE.

The choice of using ∞ as a reference points is due to the gravitational inverse square relationship between distance and a point source.

GPE from a point source is relative to 1/d, so the logical choice is to define d = ∞ as the reference point where GPE = 0. Link to integral form of the equation:

http://hyperphysics.phy-astr.gsu.edu/Hbase/gpot.html#ui

GPE from an infinite line source is relative to ln(d), so the logical choice is to define d = 1 as the reference point where GPE = 0.

GPE from an infinite disk (plate or plane) is relative to d, so the logical choice is to define d = 0 as the reference point where GPE = 0. This is commonly done when considering the Earth as an infinite disk source, where GPE is defined as mgh, (h is distance from surface of earth).

It doesn't matter if GPE is negative or positive, just that it increases with distance between objects, as it would with any attractive force between objects.

 

[quietrain]

That second term (-½mω²r², where r is the distance from the axis of rotation) represents the potential energy due to the centrifugal force that appears when you view things from a rotating frame. Note that since the force is outward, the centrifugal PE becomes smaller (more negative) as r increases, reaching its minimum value at the equator where r = R_e. (At the poles, the centrifugal PE term is zero since r = 0.)

An equipotential surface would be squashed at the poles and stretched at the equator.

issn't centrifugal force a fictitious force? how did that have a potential energy now?
erm so why does taking the centripetal force term ½mω²r multiplying with r gives the potential forcE? its like force times distance which is work done? same units but why is this potential energy? is there a derivation or something? because i only know that rotational KE is 1/2mw^2 ?

 

[rcgldr]

isn't centrifugal force a fictitious force?

It's 'real' as observed in a rotational frame. The confusing issue here is that GPE uses ∞ as the reference, while the CPE (centrifugal potential energy in this rotating frame) uses 0 as the reference.

For GPE:

<br /> F = -\frac{GMm}{r^2}<br />

<br /> U = -\int_\infty^r \frac{-GMm}{r^2} dr<br />

<br /> U = \frac{-GMm}{r}<br />

For CPE:

<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 />

 

[sophiecentaur]

issn't centrifugal force a fictitious force? how did that have a potential energy now?
erm so why does taking the centripetal force term ½mω²r multiplying with r gives the potential forcE? its like force times distance which is work done? same units but why is this potential energy? is there a derivation or something? because i only know that rotational KE is 1/2mw^2 ?

If you're on Earth, at the equator, you are in a rotating frame and you will measure a difference in local g - implying a slightly higher (less negative) potential. That will make it easier to get away from Earth (in practical terms). It will be because of your extra Kinetic Energy (as viewed by a remote observer). The units are OK throughout, of course.

btw, I advise you: don't get embroiled in the centrifugal thing - tempers can get very short in that direction!
:-0

 

[rcgldr]

If you're on Earth, at the equator, you are in a rotating frame and you will measure a difference in local g - implying a slightly higher (less negative) potential.

But the equation in the attached image in the previous post is compensating for the increased GPE due to larger radius, by subtacting |CPE|, implying some exact relationship between the radius of the Earth versus the latitude and the associated centrifugal force in the rotating frame.

CPE (centrifugal potential energy in a rotating frame) decreases (becomes more negative) as distance increases. GPE increases (becomes less negative) as distance increases. Although including the CPE component decreases the total PE (GPE + CPE) at the equator, I doubt the equation in the attached image is correct, unless it's implies some abstract model of the earth, just flexible enough to flatten at the poles and expand at the equator for that equation to hold true. The equation fails in the case of a rigid Earth where the radius at the poles and equator are the same: R_p = R_e.

In a non-rotating frame of reference, the total mechanical energy of a two body system is normally defined as E = GPE + KE = -GMm/r + 1/2 M V² + 1/2 m v². The CPE from the rotating frame of reference = -1/2 m v² = -KE of the smaller object (since the rotating Earth is the frame of reference, V=0). I'm not sure what the usage of a rotating frame of refernce is trying to accomplish.

 

[quietrain]

wow.. i mean this is the first time i come across such a thing as centrifugal potential energy. all along , the only energies i learned are PE KE and spring PE. so how come all of a sudden we have a CPE?

does it mean that i can integrate any other forces to get their revelant energy? like integrating centripetal force to get Centripetal potential energy? also why is it not centripetal PE in this case?

 

[Doc Al]

I doubt the equation in the attached image is correct, unless it's implies some abstract model of the earth, just flexible enough to flatten at the poles and expand at the equator for that equation to hold true.

I'm sure that that's what they are talking about--a plastic Earth that is able to conform to the equipotential surface, something of an oblate spheroid.

 

[Doc Al]

wow.. i mean this is the first time i come across such a thing as centrifugal potential energy. all along , the only energies i learned are PE KE and spring PE. so how come all of a sudden we have a CPE?

How much experience do you have working in rotating frames of reference?

does it mean that i can integrate any other forces to get their revelant energy? like integrating centripetal force to get Centripetal potential energy? also why is it not centripetal PE in this case?

"Centripetal force" is just the name given to the net radial force required to produce the centripetal acceleration (viewed from an inertial frame). In this case, gravity provides the centripetal force--its energy is already included in the gravitational PE term.

 

[rcgldr]

i mean this is the first time i come across such a thing as centrifugal potential energy. So how come all of a sudden we have a CPE?

I don't know if this is an invention from the instructor, or a common practice when discsussing rotating frames of reference. Based on Doc Al's responses, I get the idea that this is common practice for rotating frames of reference.

Does it mean that i can integrate any other forces to get their revelant energy?

One definition of potential energy change is the negative of the work done by a force when an object is moved between two points. Making one of those points a reference point allows an absolute potential energy to be defined. Wiki link:

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

Why is it not centripetal PE in this case?

In the case of a rotating frame of reference, the direction of what appears to be a gravitational like force is outwards, not inwards, so it's called a centrifugal force. Wiki link:

http://en.wikipedia.org/wiki/Centrifugal_force_(rotating_reference_frame)

 

[mikeph]

When you are rotating you will experience strange potentials, this is why your arms fly out- at larger distances from the axis of rotation the potential is lower, so your arm will gain kinetic energy when it moves outward. It's not a ground breaking concept, but I guess you have to study fundamentals of rotation to understand it.

Consider the rotation matrix (cos theta, sin theta; -sin theta, cos theta), where the angular rotation theta = w*t, so it is changing over time. This is essentially a coordinate transformation from (x,y) onto (x cos wt - x sin wt, y sin wt + y sin cos t), or a rotating coordinate system (with angular velocity w), a different time-dependant orthogonal basis for R^2. Differentiate this with respect to time twice to find what the acceleration vector looks like in this rotating frame, and you'll see where the centrifugal "inertial" force comes from.

There is no magic, just a consequence of the transformation.Question to other members- since the Coriolis is a curl term, does that mean we can't express the Coriolis force as the gradient of a scalar potential?

 

[quietrain]

erm i have learned rotational mechanics but using potential in rotational mechanics is kind of new to me.

i only know of rotational KE = 1/2IW^2 but never really touched on potentials in rotational frames? ??

in fact, what is a rotational frame ?

also, the arms flinging out fast is due to v = rw? so when r increases, the tangential velocity increases? but how come potential is dragged into the explanation?

 

[D H]

But the equation in the attached image in the previous post is compensating for the increased GPE due to larger radius, by subtacting |CPE|, implying some exact relationship between the radius of the Earth versus the latitude and the associated centrifugal force in the rotating frame.

That is correct. The name for that surface is the geoid. Another name is mean sea level.

The equation fails in the case of a rigid Earth where the radius at the poles and equator are the same: R_p = R_e.

But the Earth isn't a perfect rigid sphere, Jeff. The Earth is a deformable plastic blob. Earth's surface is close to being in isostatic equilibrium. In other words, the Earth's surface is fairly close to the geoid.

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.

 

[sophiecentaur]

"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.

 

[Dale]

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.

 

[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?