A point mass 'm' is located at a distance 'r' | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the rod lie along the $x$-axis with its near end at $x = r$ and its far end at $x = r + L$ from the point mass '$m$' located at the origin.Con

Vedclass · Accepted Answer

$\frac{G M m}{r(r+L)}$

Vedclass · Answer

(C) Let the rod lie along the $x$-axis with its near end at $x = r$ and its far end at $x = r + L$ from the point mass '$m$' located at the origin.Consider a small element of the rod of length '$dx$' at a distance '$x$' from the point mass '$m$'.The mass of this element is $dm = \lambda dx = \frac{M}{L} dx$.The gravitational force '$dF$' exerted by this element on the point mass '$m$' is given by:$dF = \frac{G m dm}{x^2} = \frac{G m (M/L) dx}{x^2} = \frac{G M m}{L} \frac{dx}{x^2}$.To find the total force '$F$',we integrate '$dF$' from $x = r$ to $x = r + L$:$F = \int_{r}^{r+L} \frac{G M m}{L} \frac{dx}{x^2} = \frac{G M m}{L} \left[ -\frac{1}{x} \right]_{r}^{r+L}$.$F = \frac{G M m}{L} \left( -\frac{1}{r+L} - (-\frac{1}{r}) \right) = \frac{G M m}{L} \left( \frac{1}{r} - \frac{1}{r+L} \right)$.$F = \frac{G M m}{L} \left( \frac{r+L-r}{r(r+L)} \right) = \frac{G M m}{L} \left( \frac{L}{r(r+L)} \right) = \frac{G M m}{r(r+L)}$.

$A$ point mass '$m$' is located at a distance '$r$' from the end of a uniform thin rod of mass '$M$' and length '$L$' as shown in the figure. The magnitude of the gravitational force of attraction is

Similar Questions

$A$ black hole is a:

Two masses $m_1$ and $m_2$ $(m_1 < m_2)$ are released from rest from a finite distance. They start moving under their mutual gravitational attraction.

We can shield a charge from electric fields by putting it inside a hollow conductor. Can we shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means?

Gravitational forces operate among which of the following?

$A$ mass $M$ is divided into two parts $xM$ and $(1 - x)M$. For a given separation,the value of $x$ for which the gravitational attraction between the two pieces becomes maximum is

Vedclass Test Series

Exam Paper Generator

Online Exam Module