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, [itex]v_{\infty}[/itex]. 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,

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

or

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

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:

[tex]
F = -\frac{GMm}{r^2}
[/tex]

[tex]
U = -\int_\infty^r \frac{-GMm}{r^2} dr
[/tex]

[tex]
U = \frac{-GMm}{r}
[/tex]

For CPE:

[tex]
F = \frac{m v^2}{r} = \frac{m \ \omega^2 r^2}{r} = m \ \omega^2 r
[/tex]

[tex]
U = -\int_0^r \m \ \omega^2 r dr
[/tex]

[tex]
U = -\frac{1}{2}\ m \ \omega^2 r^2
[/tex]

 

[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.