41.
Radius of moon is $$\frac{1}{4}$$ times that of earth and mass is $$\frac{1}{81}$$ times that of earth. The point at which gravitational field due to earth becomes equal and opposite to that of moon, is (Distance between centres of earth and moon is $$60R,$$ where $$R$$ is radius of earth)
42.
Two satellites of the earth, $${S_1}$$ and $${S_2}$$ are moving in the same orbit. The mass of $${S_1}$$ is four times the mass of $${S_2}.$$ Which one of the following statements is true?
A.
The time period of $${S_1}$$ is four times that of $${S_2}$$
B.
The potential energies of the earth and satellite in the two cases are equal
C.
$${S_1}$$ and $${S_2}$$ are moving with the same speed
D.
The kinetic energies of the two satellites are equal
Answer :
$${S_1}$$ and $${S_2}$$ are moving with the same speed
When two satellites of the earth are moving in same orbit, then time period of both are equal.
From Kepler’s third law $${T^2} \propto {r^3}$$
Time period is independent of mass, hence their time periods will be equal.
The potential energy and kinetic energy are mass dependent, hence the potential energy and kinetic energy of satellites are not equal.
But, if they are orbiting in a same orbit, then they have equal orbital speed.
43.
A space vehicle approaching a planet has a speed $$v,$$ when it is very far from the planet. At that moment tangent of its trajectory would miss the centre of the planet by distance $$R.$$ If the planet has mass $$M$$ and radius $$r,$$ what is the smallest value of $$R$$ in order that the resulting orbit of the space vehicle will just miss the surface of the planet?
A.
$$\frac{r}{v}{\left[ {{v^2} + \frac{{2GM}}{r}} \right]^{\frac{1}{2}}}$$
B.
$$vr\left[ {1 + \frac{{2GM}}{r}} \right]$$
C.
$$\frac{r}{v}\left[ {{v^2} + \frac{{2GM}}{r}} \right]$$
From the principle of conserving angular momentum, we have
$$mvR = mv'r\,.....\left( {\text{i}} \right)$$
[$${v'}$$ = speed when spaceship is just touching the plane]
From conserving of energy, we have
$$\frac{1}{2}m{v^2} = \frac{1}{2}m{{v'}^2} - \frac{{GMm}}{r}\,......\left( {{\text{ii}}} \right)$$
Solving Eqs. (i) and (ii), we get
$$R = \frac{r}{v}{\left[ {{v^2} + \frac{{2GM}}{r}} \right]^{\frac{1}{2}}}$$
44.
An artificial satellite is first taken to a height equal to half the radius of earth. Assume that it is at rest on the earth’s surface initially and that it is at rest at this height. Let $${{E_1}}$$ be the energyrequired. It is then given the appropriate orbital speed such that it goes in a circular orbit at that height. Let $${{E_1}}$$ be the energy required. The ratio $$\frac{{{E_1}}}{{{E_2}}}$$ is
Energy required $$=$$ (Potential energy of the Earth $$-$$ mass system when mass is at distance $$3R$$ ) $$-$$ (Potential energy of the Earth $$-$$ mass system when mass is at distance $$2R$$ )
$$\eqalign{
& = \frac{{ - GMm}}{{3R}} - \left( {\frac{{ - GMm}}{{2R}}} \right) \cr
& = \frac{{ - GMm}}{{3R}} + \frac{{GMm}}{{2R}} \cr
& = \frac{{ - 2GMm + 3GMm}}{{6R}} \cr
& = \frac{{GMm}}{{6R}} \cr
& \cr} $$
46.
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 $$2R?$$
47.
Three particles $$P,Q$$ and $$R$$ are placed as per given figure. Masses of $$P,Q$$ and $$R$$ are $$\sqrt 3 m,\sqrt 3 m$$ and $$m$$ respectively. The gravitational force on a fourth particle $$S$$ of mass $$m$$ is equal to
A.
$$\frac{{\sqrt 3 G{M^2}}}{{2{d^2}}}$$ in $$ST$$ direction only
B.
$$\frac{{\sqrt 3 G{M^2}}}{{2{d^2}}}$$ in $$SQ$$ direction and $$\frac{{\sqrt 3 G{M^2}}}{{2{d^2}}}$$ in $$SU$$ direction
C.
$$\frac{{\sqrt 3 G{M^2}}}{{2{d^2}}}$$ in $$SQ$$ direction only
D.
$$\frac{{\sqrt 3 G{M^2}}}{{2{d^2}}}$$ in $$SQ$$ direction and $$\frac{{\sqrt 3 G{M^2}}}{{2{d^2}}}$$ in $$ST$$ direction
Answer :
$$\frac{{\sqrt 3 G{M^2}}}{{2{d^2}}}$$ in $$SQ$$ direction only
48.
The ratio of escape velocity at earth $$\left( {{v_e}} \right)$$ to the escape velocity at a planet $$\left( {{v_p}} \right)$$ whose radius and mean density are twice as that of earth is
Since, the escape velocity of earth can be given as
$$\eqalign{
& {v_e} = \sqrt {2gR} = R\sqrt {\frac{8}{3}\pi G\rho } \,\,\left[ {\rho = {\text{density of earth}}} \right] \cr
& \Rightarrow {v_e} = R\sqrt {\frac{8}{3}\pi G\rho } \,......\left( {\text{i}} \right) \cr} $$
As it is given that the radius and mean density of planet are twice as that of earth. So, escape velocity at planet will be
$${v_p} = 2R\sqrt {\frac{8}{3}\pi G2\rho } \,......\left( {{\text{ii}}} \right)$$
Divide, Eq. (i) by Eq. (ii), we get
$$\eqalign{
& \frac{{{v_e}}}{{{v_p}}} = \frac{{R\sqrt {\frac{8}{3}\pi G\rho } }}{{2R\sqrt {\frac{8}{3}\pi G2\rho } }} \cr
& \Rightarrow \frac{{{v_e}}}{{{v_p}}} = \frac{1}{{2\sqrt 2 }} \cr} $$
49.
A ball is dropped from a satellite revolving around the earth at a height of $$120\,km.$$ The ball will
A.
continue to move with same speed along a straight line tangentially to the satellite at that time
B.
continue to move with the same speed along the original orbit of satellite
C.
fall down to the earth gradually
D.
go far away in space
Answer :
continue to move with the same speed along the original orbit of satellite
The ball, when dropped from the orbiting satellite will not reach the surface of the earth. When ball is dropped from the satellite, the ball also starts moving with the same speed due to inertia. As the orbit of a satellite does not depend upon its mass, the ball continues to move along with the satellite in the same orbit.
50.
A seconds pendulum is mounted in a rocket. Its period of oscillation decreases when the rocket
When rocket accelerates upward with acceleration $$a,$$ then effective acceleration of rocket is $$\left( {g + a} \right).$$
As, $$T = 2\pi \sqrt {\frac{l}{g}} = 2\pi \sqrt {\left( {\frac{l}{{g + a}}} \right)} $$
Hence, period of oscillation of seconds pendulum decreases when the rocket moves up with uniform acceleration.