Gravitational Field and Potential at certain point

[songoku]

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

$g=\frac{GM}{r^2}$

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

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

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

 

[haruspex]

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.

 

[bob012345]

Just remember the gravitational field is a vector.

 

[songoku]

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

 

[kuruman]

What about point B?

 

[kuruman]

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?

 

[bob012345]

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.

 

[songoku]

What about point B?

Where is point D?

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

 

[bob012345]

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.

 

[songoku]

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

 

[bob012345]

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.

 

[Orodruin]

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.

 

[songoku]

Sorry, that does not look correct.

No. This is only possible inside a spherical shell.

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

 

[kuruman]

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.

 

[songoku]

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

 

[kuruman]

How do you know that this formula that you googled is correct? In post #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?

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.

 

[songoku]

How do you know that this formula that you googled is correct?

I don't, I just compare several links to check

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.

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.

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

 

[bob012345]

I don't, I just compare several links to check


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.


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.

 

[songoku]

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

 

[haruspex]

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.

 

[songoku]

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