Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action…

Q: Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction, the speed of each particle is

(a) $\large \sqrt{\frac{G M}{R}}$

(b) $\large \sqrt{2\sqrt{2}\frac{G M}{R}}$

(c) $\large \sqrt{(1+2\sqrt{2})\frac{G M}{R}}$

(d) $\large \frac{1}{2}\sqrt{(1+2\sqrt{2})\frac{G M}{R}}$

Ans: (d)

Sol: Net force acting on any one of the particle is

$\large F = \frac{GM^2}{(2R)^2} + \frac{GM^2}{(\sqrt{2}R)^2} cos45^o + \frac{GM^2}{(\sqrt{2}R)^2} cos45^o$

$\large F = \frac{GM^2}{R^2}(\frac{1}{4}+\frac{1}{\sqrt{2}}) $

This force will provide required centripetal force .

$\large \frac{M v^2}{R} = \frac{GM^2}{R^2}(\frac{1}{4}+\frac{1}{\sqrt{2}}) $

What is the minimum energy required to launch a satellite of mass m from the surface of a planet…

Q: What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R ?

(a)5GmM/6R

(b)2GmM/3R

(c)GmN/2R

(d)GmN/3R

Ans: (a)

Sol: Energy required , E = Energy of satellite – Energy of mass m on the surface of planet

$\large E = -\frac{G M m}{2 r} – (-\frac{G M m}{R})$

here , r = R + 2R = 3R

$\large E = -\frac{G M m}{6R} – (-\frac{G M m}{R})$

$\large E = \frac{5 G M m}{6R} $

A satellite is moving with a constant speed v in a circular orbit about the earth. An object of mass m…

Q: A satellite is moving with a constant speed v in a circular orbit about the earth. An object of mass m is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time objects is

(a)(1/2)mv2

(b) mv2

(c)(3/2) mv2

(d) 2 mv2

Ans: (b)

Sol: In circular orbit of satellite ,

Potential Energy = -2 × Kinetic Energy

$\large P.E = -2 \times \frac{1}{2}mv^2 = – m v^2$

To escape from gravitational Pull , its total mechanical Energy should be zero .

Therefore , its K.E should be + mv2

A double star system consists of two stars A and B which have time periods TA and TB…

Q: A double star system consists of two stars A and B which have time periods TA and TB. Radius RA and RB and mass MA and MB. Choose the correct option.

(a)If TA > TB then RA > RB

(b)If TA > TB then MA > MB

(c)(TA/TB ) = (RA/RB )

(d)TA = TB

Ans: (d)

Sol: In case of binary star system angular velocity and hence time period of both the stars are equal .

A geostationary satellite orbits around the earth in a circular orbit of radius 36,000 km…

Q: A geostationary satellite orbits around the earth in a circular orbit of radius 36,000 km. Then, the time period of a spy satellite orbiting a few hundred km above the earth’s surface (Re = 6400 km) will approximately be

(a)(1/2) h

(b)1 h

(c)2 h

(d)4 h

Ans: (c)

Sol: The time period of a satellite very close to the surface of Earth is 84.6 min.

As the distance of satellite from the surface increases its time period increases . So , time period of spy satellite orbiting a few 100 km above the surface of earth should be slightly greater than 84.6 min . Therefore most appropriate option is (c)