# Potential gradient and field strength

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#1
In my textbook and in many other sources it states that the gravitational field strength is the negative of the field strength. The field strength in a gravitational field is positive in all regions in that field. If you move a mass a positive displacement (from r to r+dr where the displacement vector is pointing outwards from the mass to infinity) then the gravitational potential also increases. Hence dv/dr is positive. This would then mean that the gravitational field strength is negative here since it is defined to be the negative of the potential gradient. But obviously, the field strength is not negative. I have read further into this and the explanation that I have come across mentions that the negative sign arises because the field is in the opposite direction of the displacement. I understand this, but purely in terms of the mathematics, why does this never work. Surely for the maths to work it should be dg=-dv/dr not g=-dv/dr. (d = capital delta, not a derivative in the cases stated above) If you do take the derivative of potential with respect to displacement you do get a positive g. (GM/r^2).

EDIT: Upon further reading, I have noticed that this does work in the case of a hypothetical uniform gravitational field. I think my main question now is where you measure the displacement vector from, because if I measure it against the field as positive it doesn't work. But the graphs of potential and field strength definitely do measure displacement (r) as positive from the surface of the mass to infinity, not the other way around.
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7 years ago
#2
(Original post by Protoxylic)
In my textbook and in many other sources it states that the gravitational field strength is the negative of the field strength. The field strength in a gravitational field is positive in all regions in that field. If you move a mass a positive displacement (from r to r+dr where the displacement vector is pointing outwards from the mass to infinity) then the gravitational potential also increases. Hence dv/dr is positive. This would then mean that the gravitational field strength is negative here since it is defined to be the negative of the potential gradient. But obviously, the field strength is not negative. I have read further into this and the explanation that I have come across mentions that the negative sign arises because the field is in the opposite direction of the displacement. I understand this, but purely in terms of the mathematics, why does this never work. Surely for the maths to work it should be dg=-dv/dr not g=-dv/dr. (d = capital delta, not a derivative in the cases stated above) If you do take the derivative of potential with respect to displacement you do get a positive g. (GM/r^2).

EDIT: Upon further reading, I have noticed that this does work in the case of a hypothetical uniform gravitational field. I think my main question now is where you measure the displacement vector from, because if I measure it against the field as positive it doesn't work. But the graphs of potential and field strength definitely do measure displacement (r) as positive from the surface of the mass to infinity, not the other way around.
Can you in one sentence ask the question that you're stuck on? I'm finding it hard to battle all this. Am I correct in thinking you're confused where the negative in gravitational field strength arises?
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#3
(Original post by Phichi)
Can you in one sentence ask the question that you're stuck on? I'm finding it hard to battle all this. Am I correct in thinking you're confused where the negative in gravitational field strength arises?
Sorry, I had different thoughts throughout typing this out. My main question is where does the negative in the potential gradient arise. The field strength is defined to be positive, at least this is what it says in the 3 textbooks I have read to try to understand this.
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7 years ago
#4
(Original post by Protoxylic)
Sorry, I had different thoughts throughout typing this out. My main question is where does the negative in the potential gradient arise. The field strength is defined to be positive, at least this is what it says in the 3 textbooks I have read to try to understand this.
Well, firstly we define gravitational field strength as: Where is in the direction away from the source, i.e centre of the earth. If you didn't have this minus sign, the acceleration would be in a direction away from the earth, which, well, is self explanatory.

So we have (1)

Using A point further away (larger r), has a less negative potential. Hence, if you were to move further away from the source, you would have a positive gain in potential, because you are doing work to move. Using (1), we'd have a negative gravitational field strength again, which due to the direction we take r as, would be acceleration towards earth for example, which is what we want.
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#5
(Original post by Phichi)
Well, firstly we define gravitational field strength as: Where is in the direction away from the source, i.e centre of the earth. If you didn't have this minus sign, the acceleration would be in a direction away from the earth, which, well, is self explanatory.

So we have (1)

Using A point further away (larger r), has a less negative potential. Hence, if you were to move further away from the source, you would have a positive gain in potential, because you are doing work to move. Using (1), we'd have a negative gravitational field strength again, which due to the direction we take r as, would be acceleration towards earth for example, which is what we want.
This is what is confusing me, every textbook I have read has stated the the gravitational field strength is a strictly positive quantity which is a vector in the direction of the field.
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7 years ago
#6
(Original post by Protoxylic)
This is what is confusing me, every textbook I have read has stated the the gravitational field strength is a strictly positive quantity which is a vector in the direction of the field.
Positive would define its magnitude, which is usually the only thing we care about.
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#7
(Original post by Phichi)
Positive would define its magnitude, which is usually the only thing we care about.
So is the field strength defined to be negative such that if you take dV/dr you would get the negative of the field strength.
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7 years ago
#8
(Original post by Protoxylic)
So is the field strength defined to be negative such that if you take dV/dr you would get the negative of the field strength.
No it's defined as negative as its direction is antiparallel to r. It just so happens that its the negative of the potential gradient, it's not defined negative for that reason.
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