Gravitation MCQ Questions & Answers in Basic Physics | Physics
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121.
A planet revolves about the sun in elliptical orbit. The areal velocity $$\left( {\frac{{dA}}{{dt}}} \right)$$ of the planet is $$4.0 \times {10^{16}}\,{m^2}/s.$$ The least distance between planet and the sun is $$2 \times {10^{12}}m.$$ Then the maximum speed of the planet in $$km/s$$ is
122.
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 of its ejection, the kinetic energy of the object is-
A
$$\frac{1}{2}m{V^2}$$
B
$$m{V^2}$$
C
$$\frac{3}{2}m{V^2}$$
D
$$2m{V^2}$$
Answer :
$$m{V^2}$$
$$V$$ is the orbital velocity. If $${V_C}$$ is the escape velocity then $${V_e} = \sqrt 2 V.$$
The kinetic energy at the time of ejection $$KE = \frac{1}{2}mV_e^2 = \frac{1}{2}m{\left( {\sqrt 2 V} \right)^2} = m{V^2}$$
123.
A satellite of mass $$m$$ revolves around the earth of radius $$R$$ at a height $$x$$ from its surface. If $$g$$ is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is-
A
$$\frac{{g{R^2}}}{{R + x}}$$
B
$$\frac{{gR}}{{R - x}}$$
C
$$gx$$
D
$${\left( {\frac{{g{R^2}}}{{R + x}}} \right)^{\frac{1}{2}}}$$
Gravitational force provides the necessary centripetal force
$$\eqalign{
& \therefore \frac{{m{v^2}}}{{\left( {R + x} \right)}} = \frac{{GmM}}{{{{\left( {R + x} \right)}^2}}}{\text{ also }}g = \frac{{GM}}{{{R^2}}} \cr
& \therefore {v^2} = \frac{{g{R^2}}}{{R + x}} \Rightarrow v = {\left( {\frac{{g{R^2}}}{{R + x}}} \right)^{\frac{1}{2}}} \cr} $$
124.
The gravitational field in a region is given by $$\overrightarrow g = \frac{{5N}}{{kg\hat i}} + \frac{{12N}}{{kg\hat j}}.$$ The change in the gravitational potential energy of a particle of mass $$1\,kg$$ when it is taken from the origin to a point $$\left( {7\,m, - 3\,m} \right)$$ is :
126.
The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is $$R,$$ the radius of the planet would be
A
$$\frac{1}{2}R$$
B
$$2\,R$$
C
$$4\,R$$
D
$$\frac{1}{4}R$$
Answer :
$$\frac{1}{2}R$$
$$g = \frac{{GM}}{{{R^2}}}\,{\text{also}}\,M = d \times \frac{4}{3}\pi {R^3}$$
$$\therefore g = \frac{4}{3}\;d\pi R$$ at the surface of planet
$$\eqalign{
& {g_p} = \frac{4}{3}\left( {2d} \right)\pi R',{g_e} = \frac{4}{3}\left( d \right)\pi R \cr
& {g_e} = {g_p} \Rightarrow dR = 2dR' \Rightarrow R' = \frac{R}{2} \cr} $$
127.
The satellite of mass $$m$$ is orbiting around the earth in a circular orbit with a velocity $$v.$$ What will be its total energy?
Let satellite of mass $$m$$ be revolving closely around the earth of mass $$M$$ and radius $$R.$$
Total energy of satellite $$ = PE + KE = - \frac{{GMm}}{R} + \frac{1}{2}m{v^2}$$
$$\eqalign{
& = - \frac{{GMm}}{R} + \frac{m}{2}\frac{{GM}}{R}\,\,\left[ {{\text{as}}\,v = \sqrt {\frac{{GM}}{R}} } \right] \cr
& = - \frac{{GMm}}{{2R}} \cr} $$
∴ Total energy of satellite $$ = - \frac{1}{2}m{v^2}$$
128.
The height of the point vertically above the earth's surface, at which acceleration due to gravity becomes $$1\% $$ of its value at the earth's surface is (Radius of the earth = $$R$$ )
129.
$$R$$ is the radius of the earth and $$\omega $$ is its angular velocity and $${g_p}$$ is the value of $$g$$ at the poles. The effective value of $$g$$ at the latitude $$\lambda = {60^ \circ }$$ will be equal to
130.
For a satellite escape velocity is $$11\,km/s.$$ If the satellite is launched at an angle of $${60^ \circ }$$ with the vertical, then escape velocity will be
A
$$11\,km/s$$
B
$$11\sqrt 3 \,km/s$$
C
$$\frac{{11}}{{\sqrt 3 }}km/s$$
D
$$33\,km/s$$
Answer :
$$11\,km/s$$
Escape velocity on earth (or any other planet) is defined as the minimum velocity with which the body has to be projected vertically upwards from the surface of the earth (or any other planet). So, that it just crosses the gravitational field of earth and never returns on its own. The escape velocity of the earth is given by
$${v_e} = \sqrt {\frac{{2GM}}{R}} = \sqrt {2gR} = \sqrt {\frac{{8\pi \rho G{R^2}}}{3}} $$
From above equation it is clear that value of escape velocity of a body does not depend upon the mass $$\left( m \right)$$ of the body and its angle of projection from the surface of the earth or the planet. So, escape velocity remains same.