Question 21:
easyThree equal masses of \(3\text{ kg}\) each are fixed at the vertices of an equilateral triangle ABC. The gravitational force acting on mass \(2\text{ kg}\) placed at the centroid of triangle is
Due to perfect symmetric distribution, the three gravitational pull forces on the mass at the centroid are equal in magnitude and separated by \(120^circ\), resulting in a net vector sum of zero.
Question 22:
moderateIf the mass of the Sun were ten times smaller and the universal gravitational constant were ten times larger in magnitude, which of the following is not correct?
(2018)
The value of acceleration due to gravity on Earth is $g = \frac{GM}{R^2}$. If $G$ increases by $10$ times, $g$ also increases by $10$ times since it depends on the mass of the Earth, not the Sun. Hence, the statement that '$g$' will not change is incorrect.
Question 23:
easyTwo 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 density $c$. The gravitational force will now be:
(2003)
The gravitational force between two point masses is completely independent of the intervening medium. Therefore, the force remains $F$ even when the space is filled with a liquid.
Question 24:
difficultTwo particles of equal mass $m$ go around a circle of radius $R$ under the action of their mutual gravitational attraction. The speed $v$ of each particle is:
(1995)
The gravitational force provides the necessary centripetal force. $\frac{mv^2}{R} = \frac{Gmm}{(2R)^2} = \frac{Gm^2}{4R^2}$. Solving for $v$, we get $v^2 = \frac{Gm}{4R}$, which means $v = \frac{1}{2}\sqrt{\frac{Gm}{R}}$.
Question 25:
moderateThe earth (mass $= 6 \times 10^{24}\text{ kg}$) revolves around the sun with an angular velocity of $2 \times 10^{-7}\text{ rad/s}$ in a circular orbit of radius $1.5 \times 10^8\text{ km}$. The force exerted by the sun on the earth, in newton, is:
(1995)
The force is the centripetal force $F = mR\omega^2$. Substituting the values: $F = (6 \times 10^{24}) \times (1.5 \times 10^{11}\text{ m}) \times (2 \times 10^{-7})^2 = 9 \times 10^{35} \times 4 \times 10^{-14} = 36 \times 10^{21}\text{ N}$.
Question 26:
moderateIf the gravitational force between two objects were proportional to $\frac{1}{R}$ (and not as $\frac{1}{R^2}$), where R is the distance between them, then a particle in a circular path (under such a force) would have its orbital speed v, proportional to:
(1994, 89)
Centripetal force is provided by the given gravitational force: $$\frac{mv^2}{R} = \frac{k}{R}$$.
Solving for $v$, we get $v^2 = \frac{k}{m}$.nSince $k$ and $m$ are constants, $v$ is independent of $R$, meaning $v \propto R^0$.
Question 27:
easyTwo astronauts are floating in gravitational free space after having lost contact with their spaceship. The two will:
(2017-Delhi)
Due to universal gravitational attraction between the two astronauts, they will experience a mutual attractive force and move towards each other in space.