Question 284: Gravitation

Question

Two particles of equal mass ' $m$ ' move in a circle of radius ' $r$ ' under the action of their mutual gravitational attraction. The speed of each particle will be ( $\mathrm{G}=$ Universal gravitational constant)

Accepted Answer

Let the two particles, each of mass $m$, move in a circle of radius $r$ due to their mutual gravitational attraction.
Step 1: Gravitational Force
The gravitational force between the particles separated by distance $2r$ is:
$$ F = \frac{G m^2}{(2r)^2} = \frac{G m^2}{4 r^2} $$
Step 2: Required Centripetal Force
Each particle moves in a circle of radius $r$ about the center of mass. The centripetal force for each is:
$$ F_{	ext{centripetal}} = \frac{m v^2}{r} $$
Step 3: Equate Gravitational and Centripetal Force
The only force providing this centripetal acceleration is the gravitational force:
$$ \frac{m v^2}{r} = \frac{G m^2}{4 r^2} $$
Step 4: Solve for $v$
Divide both sides by $m$:
$$ \frac{v^2}{r} = \frac{G m}{4 r^2} $$
Multiply both sides by $r$:
$$ v^2 = \frac{G m}{4 r} $$
Take the square root:
$$ v = \sqrt{\frac{G m}{4 r}} $$
So, the correct answer is:
$ \boxed{\sqrt{\frac{\mathrm{Gm}}{4 \mathrm{r}}}} $
Option A is correct.

Answer

$\sqrt{\frac{\mathrm{Gm}}{4 \mathrm{r}}}$

Answer

$\sqrt{\frac{\mathrm{Gm}}{\mathrm{r}}}$

Answer

$\sqrt{\frac{\mathrm{Gm}}{2\mathrm{r}}}$

Answer

$\sqrt{\frac{4 \mathrm{Gm}}{\mathrm{r}}}$

Gravitation

Two particles of equal mass ' m ' move in a circle of radius ' r ' under the action of their mutual gravitational attraction. The speed of each particle will be ( \mathrm{G}= Universal gravitational constant)

Choose the correct answer: