Two particles of equal mass 'm' go around a circle of radius R under action of their mutual

English

Gujarati

Hindi

7.Gravitation

normal

Two particles of equal mass $'m'$ go around a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each particle with respect to their centre of mass is

A

$\sqrt {\frac{{Gm}}{R}} $

B

$\sqrt {\frac{{Gm}}{{4R}}} $

C

$\sqrt {\frac{{Gm}}{{3R}}} $

D

$\sqrt {\frac{{Gm}}{{2R}}} $

Solution

As gravitational force provides necessary centripetal force.

i.e., $\quad F=\frac{G m^{2}}{(2 R)^{2}}=\frac{m v^{2}}{R} \Rightarrow v=\sqrt{\frac{G m}{4 R}}$

Standard 11

Physics

Share

Similar Questions

The value of escape velocity on a certain planet is $2\, km/s$ . Then the value of orbital speed for a satellite orbiting close to its surface is

normal

A body of mass $m$ is situated at a distance equal to $2R$ ($R-$ radius of earth) from earth's surface. The minimum energy required to be given to the body so that it may escape out of earth's gravitational field will be

normal

At what altitude will the acceleration due to gravity be $25\% $ of that at the earth’s surface (given radius of earth is $R$) ?

normal

Two spheres of masses $m$ and $M$ are situated in air and the gravitational force between them is $F$ . the space around the masses is now filled with a liquid of specific gravity $3$ . The gravitational force between bodies will now be

normal

The angular speed of earth in $rad/s$, so that bodies on equator may appear weightless is : [Use $g = 10\, m/s^2$ and the radius of earth $= 6.4 \times 10^3\, km$]

normal