# Gravitational potential energy of asteroid

1. Nov 19, 2009

### 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 lets 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, lets 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 lets 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

2. Nov 19, 2009

### Staff: Mentor

Right.

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

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

3. Nov 19, 2009

### Staff: Mentor

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)

4. Nov 19, 2009

### D H

Staff Emeritus
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.

5. Nov 19, 2009

### sophiecentaur

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.

6. Nov 19, 2009

### 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 lets 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!

7. Nov 19, 2009

### 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.

8. Nov 19, 2009

### Staff: Mentor

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

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

9. Nov 20, 2009

### quietrain

ah ic..

thanks!

10. Nov 22, 2009

### 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!

11. Nov 22, 2009

### 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.

12. Nov 22, 2009

### 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?

13. Nov 23, 2009

### 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.

14. Nov 23, 2009

### quietrain

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

15. Nov 23, 2009

### Staff: Mentor

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 = Re. (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.

16. Nov 23, 2009

### 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.

17. Nov 23, 2009

### rcgldr

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.

Last edited: Nov 23, 2009
18. Nov 24, 2009

### quietrain

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 ?

19. Nov 24, 2009

### rcgldr

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:

$$F = -\frac{GMm}{r^2}$$

$$U = -\int_\infty^r \frac{-GMm}{r^2} dr$$

$$U = \frac{-GMm}{r}$$

For CPE:

$$F = \frac{m v^2}{r} = \frac{m \ \omega^2 r^2}{r} = m \ \omega^2 r$$

$$U = -\int_0^r \m \ \omega^2 r dr$$

$$U = -\frac{1}{2}\ m \ \omega^2 r^2$$

Last edited: Nov 24, 2009
20. Nov 24, 2009

### sophiecentaur

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