Gravitation & Celestial Mechanics

[kingwinner]

I am confused with some astrophysics stuff (esepically Q1), could anyone help me? I would really appreciate if you do. :warped:

 

1) Calculate the gravitational potential energy of the moon relative to the earth given that the distance between their centers is 3.94x10^8 m.

 

Solution:

Eg = -G(5.98x10^24)(7.35x10^22) / (3.94x10^8)

Eg = -7.40x10^28J

 

This is an example that I have seen. But I really really don't understand why Eg will have a negative value. (the moon is ABOVE the earth, right? So shouldn't the value of Eg be POSITIVE?) Where is the reference point of the Eg? (at the center of the earth?) For near earth surface case, Eg=mgh, and Eg of objects above the earth will be positive if you set the reference as ground level.

 

2) Calculate the change in gravitational potental energy for a 1.0-kg mass lifted 1.0x10^2 km above the surface of Earth.

I got an answer of 9.6x10^5 J, but the answer in my textbook is 1.0x10^6J. I am having less confidence on this topic. :) Did I calculate anything wrong or is there a problem with the provided answer?

[sebbysteiny]

Okay, I'll explain Q1)

 

Your thinking about it slightly wrong. If your going to think about having a reference point (ie a zero energy), where would the most sensible option be? It certainly will not be the Earth's surface, because you could just dig. Further, that won't make any sense for a planet like Jupiter.

 

Instead, scientists have defined potential energy as the WORK DONE TO GET AN OBJECT FROM INFINITY TO THAT POINT. This means the "reference point" is at infinity. Since gravity is an attractive force, you would need -ve energy (ie gravity does work on the object) to get the object to that point. Therefore, the answer is -ve. Ie, that is the work needed to push the moon out of the Earth's gravitational influence.

 

This is very similar to electromagnetic potential energy, defined as the energy required to get the object from infinity to that point. Clearly for two identical charges, they will have a +ve potential energy (ie +ve work needed to bring them together) while two opposite charges will have a -ve potential energy.

 

2) The way you calculate it is you find the potential energy at the surface of the earth (PE = - 1 * Me G / Re), find the potential energy 100km above the earth's serfice (PE = - 1 * Me G / R(e + 200km) ).

 

Then subtract one from the other being careful to get the signs right. The mass gains energy, so change in PE should be +ve.

 

Do you get the radius of the Earth or do you have to approximate it out by expanding the terms as a series of the small term 200 / Re?

[kingwinner]

Hi,

 

1) But the question says "Calculate the gravitational potential energy of the moon relative to the earth...", RELATIVE to the earth. Um...does that mean the zero Eg point is set on the earth's surface or the center of the earth?

 

And why do we always want to make Eg to be negative, in general? (i.e. Eg=-GMm/r) Why don't keep it positive?

 

2) Mass of earth=5.98x10^24 kg

Radius of earth=6.38x10^6 m

 

The question says it's 1.0x10^2 km above the surface of Earth, so r=(6.38x10^6 + 1.0x10^5)

 

I am still getting 9.6x10^5 J as the answer.

[sebbysteiny]

Okay, I'm going to admit, it's been a while since I've done A level style questions.

 

However, Q1) I guess it is badly phrased. The phrase "relative to the Earth" should probably mean the potential energy of the moon - the potential energy on the Earth's surface. Calculating the potential energy at the Earth's centre is a nightmare and will only ever be asked in an S-level Paper or beyond. However, if this was really what was needed, then, you're right, you would get a +ve answer (as work would need to be done to bring an object to the moon from anywhere on the Earth).

 

However, looking at the solution, it means just calculate the moon's potential energy in the Earth's gravitational field. Therefore, reference point is at infinity. Always resist moving the reference point from infinity unless the question unambigously forces you to do so, and if it does, it's a bad question. It is not scientifically useful to think of the reference point as being anywhere else. If it says, what is the DIFFERENCE between the energyies at etc, then in might meen what you thought it meant.

 

Q2) Correct me if I'm wrong, but arn't you just 0.4x10^5 out? This means you essentially have the right answer. The answer of 1.0 x 10 ^ 6 given could be due to the answer book rounding up the final answer, or using a more accurate (or less accurate) value for G, Re or Me.

 

I wouldn't worry about it too much.

[ronthepon]

For question 1, you are using the equation:

 

[math] {E_{g}} = -{\frac{GMm}{R}} [/math]

where G is what you know

M and m are masses or earth and moon

R is the distance between them

 

The negative sign of the RHS is what you seem not happy with.

 

TO explain why it's negative potential, understand that high potential is a state that objects 'like' to avoid. (The proof is a little long, I'm not bothering)

 

So all objects attempt to move to lower potential states. (Because the potentials create forces, that are of that kind)

 

Since when the objects are at infinity there is zero potential.

And objects will attempt to come closer (or to lower potential states) due to gravity.

 

Sorry if I'm confusing you, but I guess I'm a bad explainer.

[UncleAl]

Binding energies are always negative. The bound system must be less energetic than its unbound parts or it would spontaneously dissociate. When you bring together a bound system you must have a clever way to dump the binding energy - an expelled third body or radiative removal.

[cwes99_03]

Okay, I'm going to admit, it's been a while since I've done A level style questions.

 

However, Q1) I guess it is badly phrased. The phrase "relative to the Earth" should probably mean the potential energy of the moon - the potential energy on the Earth's surface. Calculating the potential energy at the Earth's centre is a nightmare and will only ever be asked in an S-level Paper or beyond. However, if this was really what was needed, then, you're right, you would get a +ve answer (as work would need to be done to bring an object to the moon from anywhere on the Earth).

 

However, looking at the solution, it means just calculate the moon's potential energy in the Earth's gravitational field. Therefore, reference point is at infinity. Always resist moving the reference point from infinity unless the question unambigously forces you to do so, and if it does, it's a bad question. It is not scientifically useful to think of the reference point as being anywhere else. If it says, what is the DIFFERENCE between the energyies at etc, then in might meen what you thought it meant.

 

Q2) Correct me if I'm wrong, but arn't you just 0.4x10^5 out? This means you essentially have the right answer. The answer of 1.0 x 10 ^ 6 given could be due to the answer book rounding up the final answer, or using a more accurate (or less accurate) value for G, Re or Me.

 

I wouldn't worry about it too much.

 

Seb, the questions poor young Kingwinner asks on this site are always poorly worded, through no fault of his own we've come to understand.

As far as I can tell, you've purported the best answer among them all.

 

Congrats and Cheers mates.

[arkain101]

There has been a few topics on gravity and black holes.

 

What products would be calculated if one was to calculate the rotational velocity of a sun that shrunk.

 

If a sun of a mass (M) was spinning at a cycle (hz) with a diamter (d) and began to collapse on itself it would speed up in rotation, thus increase in relativistic mass. Would and can this combination reach values needed to freeze time?

 

As it increases in velocity nearing the speed of light, and increasing in mass because it of its increased rotation, then also compacting space-time around it because of the increased V. Does this combination make an acception for achieving the results one would get for a mass traveling at C?

[sebbysteiny]

How do stars become black wholes?

 

Good question. This was one of the parts of physics I found most interesting.

 

The answer depends on statistical mechanics, if I remember right.

 

I'm only going to give a partial answer.

 

In stars, there are always two opposing forces, only one of which is gravity. That's how stars stay one size. If you think about it, gravity pushes the star together. If there was no counter force, the star would just keep on collapsing and collapsing, yet the stars we see don't do that.

 

The second force is a Fermi 'force'. Because fermions cannot be in the same state at the same time, protons and electrons cannot be pushed together unless one occupies a higher energy state than the other. Therefore, there is a limit in which matter in stars can push their contents together. In effect, eventually gravity runs out of "power" to push the fermions into higher energy states.

 

However, when the force of gravity is so high, the fermions have so much energy that they become relativistic (v close to c), and, if you follow the equations, the Fermi 'force' stops acting as a force and so the star collapses ..... until it is so packed together that the protons and electrons form neutrons. This is a neutron star. Because neutrons are heavier fermions, they will not be at a relativistic.

 

However, for even bigger stars, even the neutron becomes relativistic in speed and the star just collapses and collapses and keeps on collapsing. This is what is called a black hole.

 

The sun is too small even to become a neutron star.

[kingwinner]

Thanks for explaining!

 

1) So if we use the formula Eg=-GMm/r, the refernce point have to be at infinity because that's how to formual is derived, right? That is, when we use the formula Eg=-GMm/r, the gravitational potential energy becomes absolute, not relative......

[sebbysteiny]

That's right. I don't know how much intergration you know, but

 

F = G M1 M2 / R^2.

 

(Potential) energy = Integral (between R= infinity and r) of F dR (ie W = Fd for constant force)

= Integral (between R= infinity and r) of - G M1 M2 / R^2 dR

Note -ve sign, because force is in -ve direction (ie towards R = 0)

= [G M1 M2 / R^1] evaluated at infinity and r

At this stage, you subtract the energy at R = r from the energy at infinity.

= -G M1 M2 / r

 

Thus the gravitational potential energy has been derived using infinity as the reference point.