The relation between gravitational potential and gravitational field is

$V_{r_2} – V_{r_1} = – \int^{r_2}_{r_1}\vec E\cdot d\vec r$

Where terms $V$ stands for potential and $E$ for gravitational field.

Now, I know by calculation that $E$ inside a thin spherical shell is $0$ where as V inside a shell is non zero and is fixed at

$$ V= -{GM\over a} $$

where $a =$ radius of shell

and

$M =$ mass of shell

To find $V$ is non zero and fixed at $-{GM\over a}$ inside a shell we have taken reference point as infinity and $V(\infty)=0$. So modifying the first equation we can write

$$V(r) – V(\infty) = – \int_\infty^r\vec E\cdot d\vec r$$

Since $V(\infty)$ is taken to be zero

$$V(r) = – \int_\infty^r\vec E\cdot d\vec r \tag{1}$$

I am unable to use (1) to verify relationship between $V$ and $E$ inside a shell. I have proved $E$ inside a shell is $0$. Then I should get $V(r) = 0$ from (1)? but my $V(r)$ must be fixed at $-{GM\over a}$. What am I missing here?

By sorting this concept out I also want to answer the following question-

Q) Let $V$ and $E$ denote the gravitational potential and fie;d at a point. It is possible to have-

A) $V=0$ and $E=0$

B) $V=0$ and $E\ne 0$

C) $V\ne0$ and $E =0$

D) $V\ne0$ and $E\ne0$

(The question does not specify if reference point is infinity. All it says is that given above)

The answer given is ALL OF THE OPTIONS ARE TRUE.

I have taken reference point as infinity and i came to following observations-

D) is obviously true and is a valid statement by generally thinking of V and E at a point and also by (1).

A) is also true if you assume a massless space and at all points E and V =0 ;; AND ALSO BY (1) { If i put either 0 in the equation shouldn't I get other =0??}

For C) If you consider case of shell it becomes true but I am NOT ABLE TO OBTAIN IT THROUGH (1)

For B) I am unable to think of such a situation nor am I able to obtain result by (1)

(PLEASE NOTE I HAVE USED REFERENCE POINT AS INFINITY.)

So I would like to clear the concept of relationship between the two through (1) and then use it to solve the problem at hand. How can ALL OPTIONS BE TRUE. If I take reference as some other point, will it give be different results leading to ALL 4 BEING TRUE?

Or are only 3 true and B) false because when i tried to obtain a solution I encountered sources where it was reported B) can never happen.

## Best Answer

To get the potential inside the spherical shell, note that the gravitational field is discontinuous at the shell, since it drops suddenly to zero. So, you need to do the integral in (1) in two parts: $$V(r < a) = -\int_\infty^r \vec{E}\cdot d\vec{r} = -\int_\infty^a \vec{E}_{outside}\cdot d\vec{r} - \int_a^r \vec{E}_{inside}\cdot d\vec{r}$$ since $\vec{E}_{inside} = \vec{0},$ this simplifies to $$V(r < a) = -\int_\infty^a \vec{E}_{outside}\cdot d\vec{r} = -\frac{GM}{a}$$ Since the result does not depend on $r$, it is constant inside the shell.

As for the multiple choice question, B is actually impossible because gravity is always attractive. To see this, start with an empty universe, were $V = 0$ everywhere. Now add any mass, and you get a gravitational potential: $$V(r) = -\frac{GM}{r}.$$ The terms $G$, $r$, and $M$ are all positive (we have not found negative mass anywhere), so this expression is always negative. If the gravitational field is not zero, then there must be some mass nearby, which means the gravitational potential is negative everywhere and cannot be zero.