A straight rod of length L extends from x=a to x=L+a. The gravitational force it exerts on a point m

Q: Why can't we assume the whole mass of the rod is concentrated at its center of mass?

Because the gravitational field is non-uniform over the length of the rod, and the mass distribution is also non-uniform. The center of mass shortcut only works for uniform spherical bodies or when the distance is extremely large compared to the size of the object.

Q: How did we get $dm = (A+Bx^2)dx$?

The linear mass density $\lambda$ is given as $A+Bx^2$. By definition, linear mass density is mass per unit length, so $\lambda = \frac{dm}{dx}$. Rearranging this gives $dm = \lambda dx$.

Q: Why are the limits of integration from $a$ to $a+L$?

The rod starts at a distance $a$ from the origin and has a total length $L$. Therefore, it ends at a distance $a+L$. We need to sum the forces from all elements along this specific length.

Q: What is the integral of $1/x^2$?

It is $-1/x$. Remember the power rule for integration: $\int x^n dx = \frac{x^{n+1}}{n+1}$. Here $n=-2$, so $\int x^{-2} dx = \frac{x^{-1}}{-1} = -\frac{1}{x}$.

Q: Does the force depend on the distance $a$?

Yes, absolutely. As seen in the final expression, the force depends on both the starting distance $a$ and the length $L$. If you move the rod further away (increase $a$), the force will decrease.

Question

A straight rod of length $L$ extends from $x=a$ to $x=L+a$. The gravitational force it exerts on a point mass $m$ at $x=0$, if the mass per unit length of the rod is $A+Bx^2$, is given by

Options:

(A) $Gm \left[ A\left(\frac{1}{a+L} - \frac{1}{a}\right) - BL \right]$

(B) $Gm \left[ A\left(\frac{1}{a+L} - \frac{1}{a}\right) + BL \right]$

(C) $Gm \left[ A\left(\frac{1}{a} - \frac{1}{a+L}\right) + BL \right]$

(D) $Gm \left[ A\left(\frac{1}{a} - \frac{1}{a+L}\right) - BL \right]$

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### The Challenge of Extended Bodies

When we first learn Newton's Law of Universal Gravitation, $F = \frac{G m_1 m_2}{r^2}$, it feels like a magic key to the cosmos. But there is a catch: this elegant formula strictly applies only to **point masses** or perfectly spherical bodies.

What happens when you are dealing with a straight rod? You cannot simply plug the total mass and the distance to its center into the formula. The gravitational pull from the closer end of the rod is significantly stronger than the pull from the farther end. To solve this, we must summon the power of calculus.

### Slicing the Rod into Elements

Imagine a point mass $m$ sitting at the origin, $x=0$. A rod of length $L$ lies along the x-axis, stretching from $x=a$ to $x=a+L$. The problem gives us a fascinating twist: the rod isn't uniform. Its mass per unit length (linear mass density, $\lambda$) varies with distance as $\lambda = A + Bx^2$.

To find the total force, we mentally slice the rod into infinitesimally small pieces. Let's pick one such piece of length $dx$ located at a distance $x$ from the origin.

The mass of this tiny element is:
$$dm = \lambda dx = (A + Bx^2)dx$$

Because this element is so small, we can treat it as a point mass. The tiny gravitational force $dF$ it exerts on our mass $m$ at the origin is:
$$dF = \frac{G m dm}{x^2}$$

Substituting our expression for $dm$, we get:
$$dF = \frac{G m (A + Bx^2) dx}{x^2}$$

Let's simplify this before integrating:
$$dF = Gm \left( \frac{A}{x^2} + B \right) dx$$

### The Grand Summation (Integration)

To find the total force $F$, we must add up all these tiny forces $dF$ from every element along the rod. This continuous summation is exactly what integration does. We integrate from the start of the rod ($x=a$) to the end of the rod ($x=a+L$).

$$F = \int_{a}^{a+L} Gm \left( \frac{A}{x^2} + B \right) dx$$

Since $Gm$ is a constant, we pull it out. Now we integrate the terms inside the parenthesis. The integral of $A/x^2$ is $-A/x$, and the integral of $B$ is $Bx$.

$$F = Gm \left[ -\frac{A}{x} + Bx \right]_{a}^{a+L}$$

### Applying the Limits

Now, we carefully substitute the upper and lower limits. This is where many students make a sign error, so proceed with caution!

$$F = Gm \left[ \left( -\frac{A}{a+L} + B(a+L) \right) - \left( -\frac{A}{a} + Ba \right) \right]$$

Let's group the terms with $A$ and the terms with $B$ to make it look cleaner:

$$F = Gm \left[ A\left(\frac{1}{a} - \frac{1}{a+L}\right) + B(a+L - a) \right]$$

The $a$ and $-a$ in the second term cancel out beautifully, leaving us with our final, elegant result:

**$$F = Gm \left[ A\left(\frac{1}{a} - \frac{1}{a+L}\right) + BL \right]$$**

This result perfectly captures how the variable density of the rod influences its gravitational pull on the point mass.

Animated Solution for Physics - Gravitation: A straight rod of length $L$ extends from $x = a$ to $x = L + a$ . The gravitational force it exerts on a point mass $m$ at $x = 0$ , if the mass per unit length of the rod is $A + B x^{2}$ , is given by

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Visualized Solution

$V i s u a l i z in g t h e S e t u p$

$E l e m e n t a l M a ss an d F or ce$

$S u b s t i t u t in g d m$

$S e tt in g u p t h e I n t e g r a l$

$P er f or min g I n t e g r a t i o n$

$A ppl y in g L imi t s$

$F ina l E x p r ess i o n$

The Sigma Insight: Gravitational Force

The Challenge of Extended Bodies

Slicing the Rod into Elements

The Grand Summation (Integration)

Applying the Limits

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

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

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

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

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

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

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