A mass of 50 kg is placed at the centre of a uniform spherical shell of mass 100 kg and radius 50 m.

Q: Why isn't the potential due to the spherical shell zero if the gravitational field inside is zero?

Gravitational field is the derivative of potential (negative gradient). If the field is zero, it means the potential is constant, not necessarily zero. The potential is the work done to bring a unit mass from infinity. You still have to do work to reach the surface of the shell, and since there's no force inside, no additional work is done, keeping the potential constant at the surface value.

Q: Can I just treat the spherical shell as a point mass at the center?

No, you can only treat a spherical shell as a point mass at its center for points that are OUTSIDE the shell. For points inside the shell, the potential is constant and depends on the radius of the shell, not the distance from the center.

Q: Why do we add the potentials directly without considering direction?

Gravitational potential is a scalar quantity, meaning it only has magnitude and no direction. Unlike gravitational force or field (which are vectors), you can simply add potentials algebraically.

Q: What does the negative sign in the answer mean?

The negative sign indicates that the gravitational force is attractive. It means that work is done BY the gravitational field as a mass is brought from infinity to that point. A mass at that point is in a 'potential well'.

Q: What would happen if point P was outside the shell, say at 75 meters?

If point P was outside the shell, then both the point mass and the shell would behave like point masses located at the center. The total potential would simply be $-G(m + M_s)/r$, where $r = 75$ m.

Question

A mass of $50 	ext{ kg}$ is placed at the centre of a uniform spherical shell of mass $100 	ext{ kg}$ and radius $50 	ext{ m}$. If the gravitational potential at a point, $25 	ext{ m}$ from the centre is $V 	ext{ kg/m}$. The value of $V$ is

Options:

(A) $-60G$

(B) $+2G$

(C) $-20G$

(D) $-4G$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine a massive, hollow sphere—a uniform spherical shell—floating in space. It has a mass of $100 	ext{ kg}$ and a radius of $50 	ext{ m}$. Right at the very center of this hollow sphere, we place a dense point mass of $50 	ext{ kg}$. Our mission is to find the total gravitational potential at a specific point $P$, which is located $25 	ext{ m}$ away from the center.

Because point $P$ is at a distance of $25 	ext{ m}$ and the shell's radius is $50 	ext{ m}$, it is clear that point $P$ lies **inside** the spherical shell. This geometric detail is the key to unlocking the problem.

### The Principle of Superposition
Gravitational potential is a scalar quantity. This means we don't have to worry about complex vector additions or angles. The net gravitational potential at any point is simply the algebraic sum of the potentials created by all individual masses in the system.

$$V_{net} = V_{point} + V_{shell}$$

Let's break this down and calculate the contribution from each mass independently.

### Potential Due to the Point Mass
The first contributor is the $50 	ext{ kg}$ point mass located at the center. The formula for the gravitational potential due to a point mass $m$ at a distance $r$ is straightforward:

$$V_{point} = -\frac{Gm}{r}$$

Here, $m = 50 	ext{ kg}$ and the distance to point $P$ is $r = 25 	ext{ m}$.

### Potential Due to the Spherical Shell
Now comes the beautiful part of the physics: the spherical shell. According to the Shell Theorem, the gravitational field inside a uniform spherical shell is exactly zero. Because there is no gravitational force inside, no work is done moving a mass around within the shell.

Consequently, the gravitational potential everywhere inside the shell is **constant** and is exactly equal to the potential at its surface. The potential at the surface of a shell of mass $M_s$ and radius $R$ is:

$$V_{shell} = -\frac{GM_s}{R}$$

For our shell, $M_s = 100 	ext{ kg}$ and $R = 50 	ext{ m}$. Notice that we use the radius $R$, not the distance $r$, because the potential is constant everywhere inside!

### Final Calculation
Now, we bring it all together. We substitute our individual potential expressions into the superposition equation:

$$V_{net} = -\frac{Gm}{r} - \frac{GM_s}{R}$$

Plugging in the given numerical values:

$$V_{net} = -G \left( \frac{50}{25} + \frac{100}{50} ight)$$

Simplifying the fractions inside the parenthesis:

$$V_{net} = -G (2 + 2)$$

$$V_{net} = -4G$$

The net gravitational potential at point $P$ is **$-4G$**. The negative sign reminds us that gravitational potential is always negative, representing a bound state where work must be done against gravity to move a mass to infinity.

Animated Solution for Physics - Gravitation: A mass of $50 kg$ is placed at the centre of a uniform spherical shell of mass $100 kg$ and radius $50 m$ . If the gravitational potential at a point, $25 m$ from the centre is $V kg/m$ . The value of $V$ is

Select Answer:

Visualized Solution

$V_{n e t} = V_{p o in t} + V_{s h e l l}$

$Principle of Superposition$

$V_{p o in t} = - \frac{G m}{r}$

$V_{s h e l l} = - \frac{G M _{s}}{R}$

$V_{n e t} = - \frac{G m}{r} - \frac{G M _{s}}{R}$

$Substitution$

$Final Calculation$

The Sigma Insight: Gravitational Potential and Potential Energy

Analyzing the Setup Imagine a massive, hollow sphere—a uniform spherical shell—floating in space

The Principle of Superposition Gravitational potential is a scalar quantity

Potential Due to the Point Mass The first contributor is the $50 kg$ point mass located at the center

Potential Due to the Spherical Shell Now comes the beautiful part of the physics: the spherical shell

Final Calculation Now, we bring it all together

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Vnet​=Vpoint​+Vshell​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Principle of Superposition

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Vpoint​=−rGm​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Vshell​=−RGMs​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Vnet​=−rGm​−RGMs​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation