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

Answer 1

Let us consider a small portion of length dx situated at a distance x from the centre. (please see the attached image)

mass of this portion:

dm = (M/L)dx

Hence gravitational field due to this portion =

dE₁ = $\frac{G dm}{D^2 + x^2}$

the vertical component of this field = dE₁ sinФ

similarly, another component present in the opposite direction of the rod will exert the same gravitational field, dE₁ = dE₂

the vertical component of this field = dE₂ sinФ

Hence net vertical field due to both the portion,

dE =dE₁ sinФ + dE₂ sinФ = 2dE₁ sinФ

= 2 x $\frac{G dm}{D^2 + x^2}  * \frac{D}{\sqrt{D^2 +x^2} }$

putting the value of dm = Mdx/L we get,

dE = 2M/L x $\frac{G dx}{D^2 + x^2}  * \frac{D}{\sqrt{D^2 +x^2} }$

integrating on both the sides where x varies from 0 to L/2,

E = 2GMD/L x $\int\limits^\frac{L}{2} _0 {\frac{1}{\sqrt[3/2]{D^2 + x^2} } } \, dx$

Simplifying this we will get ,

E = $\frac{2GM}{D\sqrt{L^2 + 4D^2} }$

please ignore all the A^ present in all the equations.

Answer 2

Explanation:

• Let us find the gravitational field for a small element of the uniform rod of length L and mass M at a point on its perpendicular bisector at a distance d from the centre.  
• We know that $\mathrm{dm}=\left(\frac{M}{L}\right) * d x$, where m is the mass of the element. The gravitational field due to this element can be written as $d E=\frac{G(d m)+1}{\left(d^{2}+x^{2}\right)}$ .
• The components of the gravitational field in numerator and denominator cancel each other due to the symmetrical mass element along the length of the rod.  
• Therefore, the resultant gravitational field can be written as  $2 d E \sin \theta=\frac{2+G M+d d x}{L\left(d^{2}+x^{2}\right)(\sqrt{d^{2}+x^{2}})}$  .  On integrating the above equation, the gravitational field due to a uniform rod is $E=\frac{2 G m}{d \sqrt{L^{2}+4 d^{2}}}$ .