# Gravitational Potential

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#1
I'm really confused with one of the answers given in the mark scheme for question 4bii) in this paper:

http://filestore.aqa.org.uk/subjects...2-QP-JUN12.PDF

Where it says - 'beyond X, gravitational potential decreases as Moon isapproached'

My confusion lies with the fact that if gravitational potential is the work done per unit mass in moving a small test mass from infinity to a point in a gravitational field, then why doesn't any more work have to be done to move the probe from point X to the surface of the moon?

Thanks!
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5 years ago
#2
(Original post by PhyM23)
I'm really confused with one of the answers given in the mark scheme for question 4bii) in this paper:

http://filestore.aqa.org.uk/subjects...2-QP-JUN12.PDF

Where it says - 'beyond X, gravitational potential decreases as Moon isapproached'

My confusion lies with the fact that if gravitational potential is the work done per unit mass in moving a small test mass from infinity to a point in a gravitational field, then why doesn't any more work have to be done to move the probe from point X to the surface of the moon?

Thanks!
Gravitational force provided by moon > that of the Earth.
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5 years ago
#3
(Original post by PhyM23)
I'm really confused with one of the answers given in the mark scheme for question 4bii) in this paper:

http://filestore.aqa.org.uk/subjects...2-QP-JUN12.PDF

Where it says - 'beyond X, gravitational potential decreases as Moon isapproached'

My confusion lies with the fact that if gravitational potential is the work done per unit mass in moving a small test mass from infinity to a point in a gravitational field, then why doesn't any more work have to be done to move the probe from point X to the surface of the moon?

Thanks!
The change in gravitational potential from point X to the surface of the moon is negative and since energy is conserved, the change in kinetic energy between point X and the surface of the moon is positive (i.e. gravity does work on the probe). Another way of thinking about it is by sketching a graph of how V changes with r and thinking about the equations for V and g (field strength) - you'll notice that the field direction points towards the moon beyond X.
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#4
(Original post by Plagioclase)
The change in gravitational potential from point X to the surface of the moon is negative and since energy is conserved, the change in kinetic energy between point X and the surface of the moon is positive (i.e. gravity does work on the probe). Another way of thinking about it is by sketching a graph of how V changes with r and thinking about the equations for V and g (field strength) - you'll notice that the field direction points towards the moon beyond X.

I'm really sorry but I am still very confused. I don't understand that, if no more work has to be done from point X onward, then why does the magnitude of the gravitational potential increase as the surface of the Moon is approached?
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5 years ago
#5
(Original post by PhyM23)

I'm really sorry but I am still very confused. I don't understand that, if no more work has to be done from point X onward, then why does the magnitude of the gravitational potential increase as the surface of the Moon is approached?
Just to make sure we're on the same wavelength, do you agree with this potential diagram showing how V varies with increasing distance from the earth's surface?

Gravitational potentials of fields superimpose and add. If we're talking about the earth-moon system, we only need to consider the fields of the earth and the moon. On the diagram, you can see how the gravitational potentials for the earth (red) and moon (black) add to make the actual potential (blue).

Also remember that we're talking about motion away from the earth (i.e. r increasing) - we're talking about vector quantities here, so the direction (i.e. sign) of the quantities are important.
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#6
(Original post by Plagioclase)
Just to make sure we're on the same wavelength, do you agree with this potential diagram showing how V varies with increasing distance from the earth's surface?

Gravitational potentials of fields superimpose and add. If we're talking about the earth-moon system, we only need to consider the fields of the earth and the moon. On the diagram, you can see how the gravitational potentials for the earth (red) and moon (black) add to make the actual potential (blue).

Also remember that we're talking about motion away from the earth (i.e. r increasing) - we're talking about vector quantities here, so the direction (i.e. sign) of the quantities are important.
I think I understand the diagram, but I think my main confusion lies with the definition of gravitational potential. I really don't understand why it's WD per unit mass to move a test mass from infinity to a point rather than the other way round, if no more energy has to be applied after X, but energy would have to be applied to get to X from the surface?
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5 years ago
#7
(Original post by PhyM23)
I think I understand the diagram, but I think my main confusion lies with the definition of gravitational potential. I really don't understand why it's WD per unit mass to move a test mass from infinity to a point rather than the other way round, if no more energy has to be applied after X, but energy would have to be applied to get to X from the surface?
If we say something has a gravitational potential of -50MJ then that means we need to give it 50MJ in order to escape the field (since the change in potential from the original position to infinity is positive, e.g. by supplying it with 50MJ of kinetic energy). Things always want to have the lowest energy possible which is why something will always go down a potential gradient. If we consider the point X vs the surface of the moon (M), X has a higher potential than M which means the change in potential in negative, which means that the object will go down the potential gradient (and as its PE becomes more negative, its KE becomes more positive, which is why it moves).
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#8
(Original post by Plagioclase)
If we say something has a gravitational potential of -50MJ then that means we need to give it 50MJ in order to escape the field (since the change in potential from the original position to infinity is positive, e.g. by supplying it with 50MJ of kinetic energy). Things always want to have the lowest energy possible which is why something will always go down a potential gradient. If we consider the point X vs the surface of the moon (M), X has a higher potential than M which means the change in potential in negative, which means that the object will go down the potential gradient (and as its PE becomes more negative, its KE becomes more positive, which is why it moves).
This is becoming clearer. So is the value for V below infinity negative because the work done per unit mass is referring to the GPE lost per unit mass? Or is this totally wrong?
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5 years ago
#9
(Original post by PhyM23)
This is becoming clearer. So is the value for V below infinity negative because the work done per unit mass is referring to the GPE lost per unit mass? Or is this totally wrong?
Potential is always relative to something and when we're talking about gravity, we take x=infinity to be the zero potential point which is why GPE is always negative. And yes, you can think of it like that. The important thing to understand is that when you've got a field (e.g. a gravitational field), the force is always in the direction of decreasing potential since everything in the universe wants to decrease its potential energy to increase its stability. So yes, as an objects moves towards an object away from infinity, its GPE becomes more negative (and in a conservative system, its KE hence becomes more positive).
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#10
(Original post by Plagioclase)
Potential is always relative to something and when we're talking about gravity, we take x=infinity to be the zero potential point which is why GPE is always negative. And yes, you can think of it like that. The important thing to understand is that when you've got a field (e.g. a gravitational field), the force is always in the direction of decreasing potential since everything in the universe wants to decrease its potential energy to increase its stability. So yes, as an objects moves towards an object away from infinity, its GPE becomes more negative (and in a conservative system, its KE hence becomes more positive).
Ah so the in the definition of V 'the WD per unit mass to move a small test mass from infinity to a point in a gravitational field', the 'WD per unit mass' is the 'WD by gravity per unit mass to convert GPE to KE', so after point X, only the gravitational force by the Moon is needed to bring the probe to its surface?
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5 years ago
#11
(Original post by PhyM23)
Ah so the in the definition of V 'the WD per unit mass to move a small test mass from infinity to a point in a gravitational field', the 'WD per unit mass' is the 'WD by gravity per unit mass to convert GPE to KE', so after point X, only the gravitational force by the Moon is needed to bring the probe to its surface?
Yes
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5 years ago
#12
(Original post by PhyM23)
Ah so the in the definition of V 'the WD per unit mass to move a small test mass from infinity to a point in a gravitational field', the 'WD per unit mass' is the 'WD by gravity per unit mass to convert GPE to KE', so after point X, only the gravitational force by the Moon is needed to bring the probe to its surface?
When you've got a negative change in V, gravity is doing work on the object. When you've got a positive change in V, work is being done on the object by something else (against gravity). You've got a negative change in V between X and M which is why gravity is doing work on the object (in contrast, from the earth to X, you've got a positive change in V which is why you're having to do work on the object against gravity, i.e. giving it KE).
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#13
(Original post by Plagioclase)
When you've got a negative change in V, gravity is doing work on the object. When you've got a positive change in V, work is being done on the object by something else (against gravity). You've got a negative change in V between X and M which is why gravity is doing work on the object (in contrast, from the earth to X, you've got a positive change in V which is why you're having to do work on the object against gravity, i.e. giving it KE).
It's all perfectly clear now! Thank you so much for taking time to help me, I really appreciate it!
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