Gravitational Field and Potential at certain point

In summary, the problem involves finding the gravitational field at point A, which is inside a larger sphere and outside a smaller one. The removed mass is ##\frac{1}{8}M## and the relevant equations are ##g=\frac{GM}{R^3}r##, ##g=\frac{GM}{r^2}##, and ##V=-\frac{GM}{r}##. The correct formula for the potential at point A is ##V=-\frac{GM(3R^2-r^2)}{2R^3}-\left(-\frac{G(\frac{1}{8}M)}{R}\right)##.
  • #1
songoku
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Homework Statement
A planet of mass ##M## has a part of it removed as shown in the diagram. Find the gravitational field at point A (which is 1/2 R from O) and point B (which is 2R from surface)
Relevant Equations
##g=\frac{GM}{r^2}##

##g=\frac{GM}{R^3}r##

##V=-\frac{GM}{r}##
1644635773984.png

The removed mass is ##\frac{1}{8}M##

My idea is to find ##g## from large sphere then minus it with ##g## from small sphere (because of the removed mass):

##g## at A =
$$\frac{GM}{R^3}\left(\frac{1}{2}R\right)-\frac{G\left(\frac{1}{8}M\right)}{R^2}$$

Is this correct? Thanks
 

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  • #2
songoku said:
Homework Statement:: A planet of mass ##M## has a part of it removed as shown in the diagram. Find the gravitational field at point A (which is 1/2 R from O) and point B (which is 2R from surface)
Relevant Equations:: ##g=\frac{GM}{r^2}##

##g=\frac{GM}{R^3}r##

##V=-\frac{GM}{r}##

View attachment 296979
The removed mass is ##\frac{1}{8}M##

My idea is to find ##g## from large sphere then minus it with ##g## from small sphere (because of the removed mass):

##g## at A =
$$\frac{GM}{R^3}\left(\frac{1}{2}R\right)-\frac{G\left(\frac{1}{8}M\right)}{R^2}$$

Is this correct? Thanks
Looks right.
 
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  • #3
Just remember the gravitational field is a vector.
 
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  • #4
When calculating potential gravity at point D due to large sphere, is it correct to use R as the distance since the potential gravity inside the sphere is constant and the same as at the surface?

Thanks
 
  • #5
What about point B?
 
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  • #6
songoku said:
When calculating potential gravity at point D due to large sphere, is it correct to use R as the distance since the potential gravity inside the sphere is constant and the same as at the surface?

Thanks
Where is point D?
 
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  • #7
songoku said:
When calculating potential gravity at point D due to large sphere, is it correct to use R as the distance since the potential gravity inside the sphere is constant and the same as at the surface?

Thanks
Not sure what you are asking here. Is this a new problem or did you mean point ##B##? Please draw it. Thanks.
 
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  • #8
kuruman said:
What about point B?
kuruman said:
Where is point D?
bob012345 said:
Not sure what you are asking here. Is this a new problem or did you mean point ##B##? Please draw it. Thanks.
I am really sorry, I mean point A (got mixed up with other question I am doing)

Thanks
 
  • #9
songoku said:
I am really sorry, I mean point A (got mixed up with other question I am doing)

Thanks
since the potential gravity inside the sphere is constant and the same as at the surface?

Still not sure what you mean here. You did point ##A## correctly in the first post.
 
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  • #10
bob012345 said:
since the potential gravity inside the sphere is constant and the same as at the surface?

Still not sure what you mean here. You did point ##A## correctly in the first post.
This is how I calculate gravitational potential at point A:

V at A =
$$-\frac{GM}{R}-\left(-\frac{G\left(\frac{1}{8}M\right)}{R}\right)$$

Is this correct? Thanks
 
  • #11
songoku said:
This is how I calculate gravitational potential at point A:

V at A =
$$-\frac{GM}{R}-\left(-\frac{G\left(\frac{1}{8}M\right)}{R}\right)$$

Is this correct? Thanks
Sorry, that does not look correct.
 
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  • #12
songoku said:
When calculating potential gravity at point D due to large sphere, is it correct to use R as the distance since the potential gravity inside the sphere is constant and the same as at the surface?

Thanks
No. This is only possible inside a spherical shell.
 
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  • #13
bob012345 said:
Sorry, that does not look correct.
Orodruin said:
No. This is only possible inside a spherical shell.
1644646925037.png

Is that the formual I need to use? So maybe like this: V at A =

$$V=-\frac{GM(3R^2-(\frac{1}{4}R^2)}{2R^2}-\left(-\frac{G(\frac{1}{8}M}{R}\right)$$

Thanks
 
  • #14
This is not dimensionally correct. How did you get it? Maybe your method is correct but not the execution. We can't tell unless you show us what you did.
 
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  • #15
kuruman said:
This is not dimensionally correct. How did you get it? Maybe your method is correct but not the execution. We can't tell unless you show us what you did.
I googled the formula for potential inside solid sphere but I got the wrong formula. This is the correct one:
$$V=-\frac{GM(3R^2-r^2)}{2R^3}$$

So V at A should be:
$$V=-\frac{GM(3R^2-\frac{1}{4}R^2)}{2R^3}-\left(-\frac{G(\frac{1}{8}M)}{R}\right)$$

Is that correct? Thanks
 
  • #16
How do you know that this formula that you googled is correct? In post #4
songoku said:
When calculating potential gravity at point D due to large sphere, is it correct to use R as the distance since the potential gravity inside the sphere is constant and the same as at the surface?
you seem to think that the potential inside the sphere is constant and the same as at the surface. The formula you found says that the potential is not constant inside and that it is equal to the potential at the surface only at the center. Which is correct and why?

I think that the intent here is for you to do the derivation yourself, not google it. Then you will have learned something other than "I can always google the answer." The googled answer should be a check whether you did it right.

The problem is asking you to find the gravitational field at a point inside the larger sphere and outside the smaller one (point A) and also outside both spheres at point B which you haven't done so far. Knowing the expressions for the fields, the idea then is to find the potential at points inside and outside by doing the appropriate line integrals.
 
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  • #17
kuruman said:
How do you know that this formula that you googled is correct?
I don't, I just compare several links to check

kuruman said:
you seem to think that the potential inside the sphere is constant and the same as at the surface. The formula you found says that the potential is not constant inside and that it is equal to the potential at the surface only at the center. Which is correct and why?
Orodruin told me in post#12 that potential inside the sphere is constant and the same as at the surface is only for spherical shell so I tried to find formula for solid sphere. The correct one is ##V=-\frac{GM(3R^2-r^2)}{2R^3}## because the question is about solid sphere, not shell.

kuruman said:
I think that the intent here is for you to do the derivation yourself, not google it. Then you will have learned something other than "I can always google the answer." The googled answer should be a check whether you did it right.
I am not sure whether the formula derivation is part of the lesson / curriculum because the students have never been asked to derive any formulas since the beginning of the class. I know the derivation but only because I looked it up, not because I did it myself.

kuruman said:
The problem is asking you to find the gravitational field at a point inside the larger sphere and outside the smaller one (point A) and also outside both spheres at point B which you haven't done so far. Knowing the expressions for the fields, the idea then is to find the potential at points inside and outside by doing the appropriate line integrals.
##g## at B:
$$g_B=\frac{GM}{9R^2}-\frac{G(\frac{1}{8}M)}{\frac{49}{4}R^2}=\frac{89}{882}\frac{GM}{R^2}$$

V at A:
$$V_A=-\frac{GM(3R^2-\frac{1}{4}R^2)}{2R^3}-\left(-\frac{G(\frac{1}{8}M)}{R}\right)$$

V at B:
$$V_B=-\frac{GM}{3R}-\left(-\frac{G(\frac{1}{8}M)}{\frac{7}{2}R}\right)$$

Is the idea correct? Thanks
 
  • #18
songoku said:
I don't, I just compare several links to checkOrodruin told me in post#12 that potential inside the sphere is constant and the same as at the surface is only for spherical shell so I tried to find formula for solid sphere. The correct one is ##V=-\frac{GM(3R^2-r^2)}{2R^3}## because the question is about solid sphere, not shell.I am not sure whether the formula derivation is part of the lesson / curriculum because the students have never been asked to derive any formulas since the beginning of the class. I know the derivation but only because I looked it up, not because I did it myself.##g## at B:
$$g_B=\frac{GM}{9R^2}-\frac{G(\frac{1}{8}M)}{\frac{49}{4}R^2}=\frac{89}{882}\frac{GM}{R^2}$$

V at A:
$$V_A=-\frac{GM(3R^2-\frac{1}{4}R^2)}{2R^3}-\left(-\frac{G(\frac{1}{8}M)}{R}\right)$$

V at B:
$$V_B=-\frac{GM}{3R}-\left(-\frac{G(\frac{1}{8}M)}{\frac{7}{2}R}\right)$$

Is the idea correct? Thanks
Without doing the math myself it seems all right but you can check it by deriving the field from the potential and see if they match.
 
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  • #19
bob012345 said:
Without doing the math myself it seems all right but you can check it by deriving the field from the potential and see if they match.
##g## at A =
$$\frac{GM}{R^3}\left(\frac{1}{2}R\right)-\frac{G\left(\frac{1}{8}M\right)}{R^2}=\frac{3}{8}\frac{GM}{R^2}$$

V at A:
$$V_A=-\frac{GM(3R^2-\frac{1}{4}R^2)}{2R^3}-\left(-\frac{G(\frac{1}{8}M)}{R}\right)=-\frac{5}{4}\frac{GM}{R}$$

Then:
$$-\frac{dV_A}{dR}=\frac{5}{4}\frac{GM}{R^2}$$

which is not the same as ##g_A##, but I can't find where I went wrong. Where is my mistake? Thanks
 
  • #20
songoku said:
Where is my mistake?
Differentiating wrt R is not going to give the field at A. You want how the potential changes if you move a little from A, but R is not a measure of displacement from A. If you change R the whole setup changes.
 
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  • #21
haruspex said:
Differentiating wrt R is not going to give the field at A. You want how the potential changes if you move a little from A, but R is not a measure of displacement from A. If you change R the whole setup changes.
I understand. I am just crazy to take derivative wrt R because R is constant.

Thank you very much for all the help and explanation haruspex, bob012345, kuruman, Orodruin
 
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FAQ: Gravitational Field and Potential at certain point

What is a gravitational field?

A gravitational field is a region in space where a mass experiences a force due to the presence of another mass. This force is known as the gravitational force and it is responsible for the attraction between objects with mass.

How is the strength of a gravitational field measured?

The strength of a gravitational field is measured by the force per unit mass, also known as the gravitational acceleration. This is typically denoted by the symbol g and is measured in units of meters per second squared (m/s²).

What is the difference between gravitational potential and gravitational potential energy?

Gravitational potential is a measure of the potential energy per unit mass at a certain point in a gravitational field. It is a scalar quantity and is measured in units of joules per kilogram (J/kg). Gravitational potential energy, on the other hand, is the energy an object possesses due to its position in a gravitational field. It is a vector quantity and is measured in units of joules (J).

How does the distance between two masses affect the strength of the gravitational field?

The strength of the gravitational field is inversely proportional to the square of the distance between two masses. This means that as the distance between two masses increases, the strength of the gravitational field decreases. This relationship is described by Newton's law of universal gravitation.

Can the gravitational potential at a certain point be negative?

Yes, the gravitational potential at a certain point can be negative. This means that the object at that point has a lower potential energy compared to an object at infinity. However, the gravitational potential energy is always negative, as it is a measure of the energy required to move an object from infinity to that point in the gravitational field.

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