Gravitational Field at outside point is given by
Answer
$=\frac{GM}{r^2}$
Gravitational Field at internal point is given by
$=\frac{GM}{b^3}r$
So all the statement are true
Electric field inside the spherical shell is zero but Potential is not zero
Answer
Potential is same inside the shell
So a and c are correct
Let us consider a small element of length dx at a distance x from the origin
Answer
Mass of the element=(M/L)dx
Field due to this mass element at point P and it will be directed towards x axis
$dE=\frac{GM}{L}\frac{dx}{(a-x)^2}$
Total field strength
$E=\int_{0}^{L}{\frac{GM}{L}\frac{dx}{(a-x)^2}}$
$=\frac{GM}{a(a-L)}$
Let us consider a small element of length dx at a distance x from the origin
Answer
Mass of the element=(M/L)dx
Potential due to this mass element at point P
$dV=\frac{GM}{L}\frac{dx}{(a-x)}$
Total Potential
$V=\int_{0}^{L}{\frac{GM}{L}\frac{dx}{(a-x)}}$
$=\frac{GM}{L}ln{\frac{a-L}{a}}$
Potential energy of the system
Answer
U=Potential energy AB + Potential energy BC + Potential energy AC
$U=\frac{-3Gm^2}{a}$
Potential at point A = Potential due to B + Potential due to C
Answer
$ =\frac{-Gm}{a}-\frac{Gm}{a}$
$ =\frac{-2Gm}{a}$
Potential energy of the system after the change
Answer
$ U=\frac{-3Gm^2}{3a}=\frac{-Gm^2}{a}$
So work done
$=\frac{-Gm^2}{a}-\frac{-3Gm^2}{a}$
$=\frac{2Gm^2}{a} $
For $R_1 < d < R_2$
Answer
Shell A will exert force on mass m
$d < R_1$
mass is inside all the shell .So no force
$R_1 < R_2 < d < R_3$
Shell A and B will exert force on mass m
$R_1 < R_2 < R_3 < d$
Shell A , B and C will exert force on mass m
Kinetic energy=$\frac{1}{2}mv^2$
Answer
We know that
$v=\sqrt{\frac{GM}{R}}$
Where M is the mass of the earth
So KE varies with 1/R
Angular momentum=$mvR=m\sqrt{GMR}$
So it varies with $\sqrt R$
Linear momentum=mv
So it varies as $\frac{1}{\sqrt R}$
Frequency of revolutions
$=\frac{1}{T}=\frac{1}{2\pi}\sqrt{\frac{GM}{R^3}}$
So frequency varies as $\frac{1}{R^{3/2}}$
So c is correct
(c)
Answer
Answer is (a)
Answer
$F \alpha bm \times (1-b)m= bm^2(1-b)$
For maximum force $\frac {dF}{dx}=0$
$\frac {dF}{dx}=m^2-2bm^2=0$
or b=1/2
(a) is correct
Answer
Correct Answer: C
Answer
$\frac {T_2}{T_1}=(\frac {r_2}{r_1})^{3/2}$
$T_2=24(\frac {6400}{36000})^{3/2} =2 \ hour$
(a) and (c) are correct
Answer
Let velocities of these masses at r distance from each other be $v_1$ and $v_2$ respectively.
Answer
Law of conservation of Momentum
$m_1v_1-m_2v_2=0$ or $m_1v_1=m_2v_2$
By conservation of energy
change in P.E.=change in K.E.
$\frac {Gm_1m_2}{r}=\frac {1}{2}m_1v_1^2+\frac {1}{2}m_1v_2^2 $
On solving equation ,we get
$v_1=\sqrt {\frac {2Gm_2^2}{r(m_1+m+2)}}$
and $v_2=\sqrt {\frac {2Gm_1^2}{r(m_1+m+2)}}$
Now relative velocity of approach at a separation distance r = $|v_1| + |v_2| = \sqrt {2G \frac {m_1 + m_2}{r}}$
(a), (c), (d)
Answer
At the minimum separation with the sun,the cosmic body s velocity is perpendicular to its position vector relative to the sun
Answer
Let v be the velocity of the cosmic at minimum separation
And r be minimum separation distance
Now Angular momentum about Sun at starting
$=mv_0d$
Angular momentum at minimum separation
$= mvr$
Since no torque acts through out the motion.,angular momentum is conserved
$mv_0d= mvr$
or
$v=\frac{v_0d}{r}$
Now applying law of conservation of energy
$\frac{1}{2}mv_0^2=\frac{1}{2}mv^2-\frac{GM_sm}{d}$
Substituting the value of v in the above equation
$v_0^2r^2+2GM_sr-v_0^2d^2=0$
Or
$r=\frac{GM_s}{v_0^2}\left(\sqrt{1+\frac{d^2v_0^2}{G^2M_s^2}}-1\right)$
So velocity at minimum separation
$v=\frac{v_0^3d}{GM\left(\sqrt{1+\frac{d^2v_0^2}{G^2M_s^2}}-1\right)}$
At minimum and maximum distance positions of the planet,the velocity vector and position vector makes right angles with each other
Answer
Let $v_1$ and $v_2$ be the velocities at minimum and maximum position
Since the force is central,Angular momentum is conserved
Applying law of conservation of angular momentum at minimum and maximum distance
$Mv_1a =Mv_2b$
Or $v_1a=v_2b$ -----(1)
Applying law of conservation of energy at minimum and maximum distance
$\frac{-GMm}{a}+\frac{1}{2}Mv_1^2=\frac{-GMm}{b}+\frac{1}{2}Mv_2^2$ ----(2)
Solving equation 1 and 2,we get
$v_1=\sqrt{\frac{2Gmb}{a(a+b)}}$
$v_2=\sqrt{\frac{2Gma}{b(a+b)}}$
Total energy of the system
$=\frac{-GMm}{a}+\frac{1}{2}Mv_1^2$
Substituting the value $v_1$
We get
Total energy
$=\frac{-GMm}{(a+b)}$
Angular momentum of the planet about Sun
$L=Mav_1$
Substituting the value of v1
$L=M\sqrt{\frac{2Gmab}{a+b}}$
Applying law of conservation of angular momentum at Point P and minimum distance point
$Mr_0vsin{\theta}=Mav_1$
Substituting the value of v1,
$v=\frac{a}{r_0sin{\theta}}\sqrt{\frac{2Gmb}{a(a+b)}}$
$v=\sqrt{\frac{2Gmab}{r_0^2{sin}^2{\theta}(a+b)}}$
Ecentricty is given by
$e=\frac{b-a}{b+a}$
Semi major axis=$\frac{a+b}{2}$
Suppose the earth is divided in spherical shell.
Answer
For the spherical shell whose radii is less than r,Force on the particle m would be
$dF=\frac{-GmdM}{r^2}$
The force will be towards the center so the minus sign
For the spherical shell whose radii are greater then r,the force would be zero as the point will be internal to them
So total Force
$F=\frac{-Gm}{r^2}\int d M$
Where $\int d M$ is the mass of the sphere up to the radii r
Now mass per unit volume of Earth
$=\frac{M}{\frac{4}{3}\pi R^3}$
So
$\int{dM=\frac{M}{\frac{4}{3}\pi R^3}\frac{4}{3}\pi r^3}$
So force would be
$F=-\frac{GMmr}{R^3}$
Now the above equation can be written as
$ma=\frac{-GMm}{R^3}r$
Or
$a=-kr$
Where $k=\frac{GM}{R^3}$
Since acceleration is directed toward center and proportional to r, The particle will follow a Simple harmonic motion
Time period
$T=2\pi\sqrt{\frac{m}{k}}$
From Keplers law we know that
Answer
$T^2\alpha R^3$
So
$\frac{T_1}{T_2}=\left(\frac{R_1}{R_2}\right)^{3/2}$
Or
$R_1=\left(\frac{T_1}{T_2}\right)^{2/3}R_2$
$=\left(\frac{2}{16}\right)^{2/3}3.2*10^4$
$=.8*10^4$
Orbital Velocity of M
$v_1=\frac{2\pi R_1}{T_1}=\frac{2\pi X.8*10^4}{2}=.8\pi X10^4 \ Km/hr$
Orbital Velocity of N
$v_2=\frac{2\pi R_{21}}{T_2}=\frac{2\pi X3.2*10^4}{16}=.4\pi X10^4 \ Km/hr$
At t=8 Hour,the satellite are closest to each other.
So Angular velocity
$\omega=\frac{v_1+v_2}{R_2-R_1}=.5\pi \ rad/sec$