Name: Escape Velocity: Derivation
Uploaded: 2019-07-19
Duration: 7 min 7 s
Description: Do you know how much velocity is required to escape from Earth to space? In this video, we will discuss the escape velocity of bodies and their values.

Question

Derive an expression for the gravitational field due to a uniform rod of length

L

and mass

M

at a point on its perpendicular bisector at a distance

d

from the centre.

Accepted Answer

Consider the image representation of the given problem.

Here, the end points of the rod subtend an angle $d$ at the point of observation.

Consider small elements of mass $d m$ and length $d x$ placed at two opposite sides of the bisector, symmetrically, at a distance $x$ from the bisector.

Given,

The rod is uniform, the mass per unit length of the rod is $\frac{M}{L}$ .

Hence, the mass of the small element is $d m = \frac{M}{L} d x$ .

For any point mass $M$ , the gravitational field at a distance $r$ is given by $\vec{g} = - \frac{G M}{r^{2}} \hat{r}$ .

Hence, for this elemental mass, the magnitude of gravitational field is written as

$d g = \frac{G d m}{(d^{2} + x^{2})} = \frac{G}{(d^{2} + x^{2})} \frac{M}{L} d x$

The direction of the field is from the point of observation to the elemental mass.

This field is resolved into two mutually perpendicular components as shown in the image.

Let the elements subtend an angle $α$ at point $P$ .

We can observe that for the gravitational field, the parallel components to the rod cancel out, while the perpendicular components add up.

Hence, the net field at point $P$ is

$2 d g cosα = \frac{2 G}{(d^{2} + x^{2})} \frac{M}{L} cosα d x$

From the image, we have,

$cosα = \frac{adjacent side}{hypotenuse} = \frac{d}{\sqrt{x^{2} + d^{2}}}$

On substituting these values, we get,

$2 d g \cos α = \frac{2 G}{(d^{2} + x^{2})} \frac{M}{L} \frac{d}{{(x^{2} + d^{2})}^{\frac{1}{2}}} d x = \frac{2 G M d}{L {(d^{2} + x^{2})}^{\frac{3}{2}}} d x$

Hence, the net gravitational field will be the integration of the above equation between the limits $0$ and $\frac{L}{2}$ .

Let the net field be represented by $g_{net}$ .

Hence,

$g_{net} = \int_{0}^{\frac{L}{2}} \frac{2 G M d}{L {(d^{2} + x^{2})}^{\frac{3}{2}}} d x = \frac{2 G M d}{L d^{3}} \int_{0}^{\frac{L}{2}} \frac{1}{{(1 + {(\frac{x}{d})}^{2})}^{\frac{3}{2}}} d x$

Let $\frac{x}{d} = \tan α$ .

This gives,

$x = d \tan α \Rightarrow d x = d \sec^{2} α dα$

Accordingly, the limits of the integration change from $0$ to $θ$ .

Hence,

$g_{net} = \frac{2 G M d}{L d^{3}} \int_{0}^{θ} \frac{1}{{(1 + \tan^{2} α)}^{\frac{3}{2}}} d \sec^{2} α d α = \frac{2 G M}{L d} \int_{0}^{θ} \frac{1}{{(\sec^{2} α)}^{\frac{3}{2}}} d \sec^{2} α dα$

On solving this further, we get,

$g_{net} = \frac{2 G M}{L d} \int_{0}^{θ} \cos α dα = \frac{2 G M}{L d} {[\sin α]}_{0}^{θ} = \frac{2 G M}{L d} \sin θ = \frac{2 G M}{L d} \frac{\frac{L}{2}}{{[{(\frac{L}{2})}^{2} + d^{2}]}^{\frac{1}{2}}} = \frac{2 G M}{d {[L^{2} + 4 d^{2}]}^{\frac{1}{2}}}$

Derive an expression for the gravitational field due to a uniform rod of length $L$ and mass $M$ at a point on its perpendicular bisector at a distance $d$ from the centre.

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