Solve the following :
Question: The moon takes about $27.3$ days to revolve round the earth in a nearly circular orbit of radius $3.84 \times 105 \mathrm{~km}$. Calculate the mass of the earth from these data. Solution: Time period of the moon around the earth is given by, $T=2 \pi \sqrt{\frac{r^{\mathrm{g}}}{G M}}$ $\mathrm{m}=$ mass of earth $r=$ Distance between center of the moon and earth $27.3 \times 86400=2 \times 3.14 \sqrt{\frac{\left(3.84 \times 10^{5} \times 10^{\mathrm{s}}\right)^{\mathrm{s}}}{6.67 \times...
Read More →Solve the following :
Question: The time taken by Mars to revolve round the sun is $1.88$ years. Find the ratio of average distance between Mars and the sun to that between the earth and the sun. Solution: Kepler's law of planetary motion says, $T^{2} \propto R^{3}$ $\frac{T_{\mathrm{m}}^{\mathrm{z}}}{T_{e}^{\mathrm{Z}}}=\frac{R_{\mathrm{m}}^{\mathrm{I}}}{R_{e}^{\mathrm{I}}}$ $\left(\frac{1.88}{1}\right)^{2}=\left(\frac{R_{m}}{R_{e}}\right)^{3}$ $\frac{R_{m}}{R_{e}}=\left(\frac{1.88}{1}\right)^{\frac{\mathrm{}}{\math...
Read More →Solve the following :
Question: A pendulum having a bob of mass $m$ is hanging in a ship sailing along the equator from east to west. When the ship is stationary with respect to water the tension in the string is To. (a) Find the speed of the ship due to rotation of the earth about its axis. (b) Find the difference between To and the earth's attraction on the bob. (c) If the ship sails at speed $v$, what is the tension in the string? Angular speed of earth's rotation is $\omega$ and radius of the earth is $R$. Soluti...
Read More →Solve the following :
Question: At what rate should the earth rotate so that the apparent $g$ at the equator becomes zero? What will be the length of the day in this situation? Solution: Apparent acceleration due to gravity at equator becomes 0 . $G^{\prime}=g-\omega^{2} R=0$ $g=\omega^{2} R$ $\omega=\sqrt{\frac{g}{R}}$ $=\sqrt{\frac{9.8}{6400 \times 10^{\mathrm{s}}}}$ Time period $=\mathrm{T}=\frac{2 \pi}{\omega}=\frac{2 \times 3.14}{1.2 \times 10^{\mathrm{s}}}=1.41 \mathrm{hrs}$...
Read More →Solve the following :
Question: A body stretches a spring by a particular length at the earth's surface at equator. At what height above the south pole will it stretch the same spring by the same length? Assume the earth to be spherical. Solution: At equator, let stretch in spring be $x$ Spring force $={ }^{m} g_{e}$ $k x=m\left(g-R \omega^{2}\right)-(1)$ Now, at height $h$ above surface, stretch in spring is $x$ Sprina force $=m g_{h}$ $k x=m g\left(1-\frac{2 h}{R}\right)-(2)$ From (1) and (2) $m\left(g-R \omega^{2}...
Read More →Solve the following :
Question: A body stretches a spring by a particular length at the earth's surface at equator. At what height above the south pole will it stretch the same spring by the same length? Assume the earth to be spherical. Solution: At equator, let stretch in spring be $x$ Spring force $={ }^{m} g_{e}$ $k x=m\left(g-R \omega^{2}\right)-(1)$ Now, at height h above surface, stretch in spring is $x$ Now, at height $h$ above surface, stretch in spring is $x$ Spring force $=m g_{h}$ $k x=m g\left(1-\frac{2 ...
Read More →Solve the following :
Question: A body is weighed by a spring balance to be $1000 \mathrm{~kg}$ at the north pole. How much will it weigh at the equator? Account for the earth's rotation only. Solution: Let $g_{p}$ be the gravity at the poles , $g_{e}$ be the gravity at the equator $g_{e}=g_{p}-\omega^{2} R$ $=9.81-\left(7.3 \times 10^{-5}\right)^{2} \times\left(7.3 \times 10^{-5}\right)^{2} \times 6400 \times 10^{3}$ $=9.766 \mathrm{~N} / \mathrm{m}^{2}$ Now, $\mathrm{mg}_{\mathrm{e}}=1 \mathrm{~kg} \times 9.766 \ma...
Read More →Solve the following :
Question: Find the acceleration due to gravity in a mine of depth $640 \mathrm{~m}$ if the value at the surface is $9^{\prime} 800 \mathrm{~m} / \mathrm{s}^{2}$. The radius of the earth is $6400 \mathrm{~km}$. Solution: $\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{\stackrel{d}{R}}{)}\right)$ g' be the acceleration due to gravity at surface $=9.8 \times\left(1-\frac{640}{6400 \times 10^{\mathrm{s}}}\right)$ $=9.79 \mathrm{~m} / \mathrm{s}^{2}$...
Read More →Solve the following :
Question: What is the acceleration due to gravity on the top of Mount Everest? Mount Everest is the highest mountain peak of the world at the height of $8848 \mathrm{~m}$. The value at sea level is $9^{\prime} 80 \mathrm{~m} / \mathrm{s}^{2}$. Solution: $g^{\prime}=g\left(1-\frac{2 \hbar}{R}\right)$ g' be the acceleration due to gravity at Mount Everest $h=8848$ $g^{\prime}=9.8 \times\left(1-\frac{2 \times 8848}{6400 \times 10^{\mathrm{s}}}\right)$ $=9.77 \mathrm{~m} / \mathrm{s}^{2}$...
Read More →Solve the following :
Question: Find the height over the earth's surface at which the weight of a body becomes half of its value at the surface. Solution: Let height be h $m \frac{G M}{(R+h)^{2}}=\frac{1}{2} m \frac{G M}{R^{2}}$ $2^{R^{2}}=(R+h)^{2}$ $\sqrt{2} R=R+h$ $\mathrm{~h}=(\sqrt{2}-1) \mathrm{R}$...
Read More →Solve the following :
Question: The gravitational field in a region is given by $E=\left(2 \hat{\imath}+3 j^{\wedge}\right) \mathrm{N} / \mathrm{kg}$. Show that no work is done by the gravitational field when a particle is moved on the line $3 y+2 x=5$ [Hint : If a line $y=m x+c$ makes angle $\theta$ with the $X$-axis, $m=\tan \theta$ ] Solution: Gravitational field in the region is given by $E=2 \hat{\imath}+3^{\hat{\jmath}}$ Slope of gravitational field, $\mathrm{m}_{1}=\tan \theta_{1}=3 / 2$ The line $3 y+2 x=5$ c...
Read More →Solve the following :
Question: The gravitational potential in a region is given by $\mathrm{V}=(20 \mathrm{~N} / \mathrm{kg})(\mathrm{x}+\mathrm{y})$. (a) Show that the equation is dimensionally correct. (b) Find the gravitational field at the point $(x, y)$. Leave your answer in terms of the unit vectors $i, j, k$. (c) Calculate the magnitude of the gravitational force on a particle of mass 500 Solution: $V=20 \frac{N}{k g}(x+y)$ Dimension of L.H.S $=[\mathrm{V}]$ $=\left[\frac{G M}{R}\right]$ $=\frac{\left[M^{-1} ...
Read More →Solve the following :
Question: The gravitational field in a region is given by $E=(5 \mathrm{~N} / \mathrm{kg}) \mathrm{i}+(12 \mathrm{~N} / \mathrm{kg}) \hat{\jmath})$. (a) Find the magnitude of the gravitational force acting on a particle of mass $2 \mathrm{~kg}$ placed at the origin. (b) Find the potential at the points $(12 \mathrm{~m}, 0)$ and $(0,5 \mathrm{~m})$ if the potential at the origin is taken to be zero. (c) Find the change in gravitational potential energy if a particle of mass $2 \mathrm{~kg}$ is ta...
Read More →Solve the following :
Question: A particle of mass $100 \mathrm{~g}$ is kept on the surface of a uniform sphere of mass $10 \mathrm{~kg}$ and radius $10 \mathrm{~cm}$. Find the work to be done against the gravitational force between them to take the particle away from the sphere. Solution: Work done against gravitational force $=$ change in potential energy $W=U_{f}-U_{i}$ $W=0-\left(-\frac{G M m}{r}\right)$ $W=\frac{G M m}{r}=\frac{6.67 \times 10^{-11} \times 0.1 \times 10}{0.1}$ $W=6.67 \times 10^{-10} \mathrm{~J}$...
Read More →Solve the following :
Question: A particle of mass $100 \mathrm{~g}$ is kept on the surface of a uniform sphere of mass $10 \mathrm{~kg}$ and radius $10 \mathrm{~cm}$. Find the work to be done against the gravitational force between them to take the particle away from the sphere. Solution: Work done against gravitational force $=$ change in potential energy $W=U_{f}-U_{i}$ $W=0-\left(-\frac{G M m}{r}\right)$ $W=\frac{G M m}{r}=\frac{6.67 \times 10^{-11} \times 0.1 \times 10}{0.1}$ $W=6.67 \times 10^{-10} \mathrm{~J}$...
Read More →Solve the following :
Question: Three particles of mass $\mathrm{m}$ each are placed at the three corners of an equilateral triangle of side a. Find the work which should be done on this system to increase the sides of the triangle to $2 a$. Solution: Work done $=\Delta U=$ final P.E - initial P.E $=U_{f}-U_{i}$ $=3\left(\frac{-G m m}{2 a}\right)-3\left(\frac{-G m m}{a}\right)$ $=\left(\frac{3 G m^{2}}{2 a}\right)$...
Read More →Solve the following :
Question: Two small bodies of masses $2.00 \mathrm{~kg}$ and $4.00 \mathrm{~kg}$ are kept at rest at a separation of $2.0 \mathrm{~m}$. Where should a particle of mass $0.10 \mathrm{~kg}$ be placed to experience no net gravitational force from these bodies? The particle is placed at this point. What is the gravitational potential energy of the system of three particles with usual reference level? Solution: Let mass $0.5 \mathrm{~kg}$ is placed at distance $x$ from $2 \mathrm{~kg}$ between both m...
Read More →Solve the following :
Question: A thin spherical shell having uniform density is cut in two parts by a plane and kept separated as shown in figure (11-E3). The point $A$ is the centre of the plane section of the first part and B is the centre of the plane section of the second part. Show that the gravitational field at A due to the first part is equal in magnitude to the gravitational field at B due to the second part. Solution: Resultant field inside any spherical shell is zero at all points. So, at point a and B, f...
Read More →Solve the following :
Question: A uniform metal sphere of radius a and mass $\mathrm{M}$ is surrounded by a thin uniform spherical shell of equal mass and radius $4 a$ (figure 11-E2). The center of the shell falls on the surface of the inner sphere. Find the gravitational field at the points $P_{1}$ and $P_{2}$ shown in the figure. Solution: (a) At point $P_{1}$ Net gravitational field=field due to shell+ field due to sphere $\vec{E}_{N}=0+\frac{G M}{(a+4 a-a)^{2}}$ $\vec{E}_{N}=\frac{G M}{16 a^{2}}$ (b) At point $P_...
Read More →Solve the following :
Question: A solid sphere of mass $m$ and radius $r$ is placed inside a hollow thin spherical shell of mass $M$ and radius $\mathrm{R}$ as shown in figure (11-E1). A particle of mass $\mathrm{m}^{\prime}$ is placed on the line joining the two centers at a distance $x$ from the point of contact of the sphere and the shell. Find the magnitude of the resultant gravitational force on this particle due to the sphere and the shell if (a) $rx2 r$, (b) $2 rx2 R$ and $(c) x2 R$. Solution: (a) Gravitationa...
Read More →Solve the following :
Question: A tunnel is dug along a chord of the earth at a perpendicular distance $R / 2$ from the earth's center. The wall of the tunnel may be assumed to be frictionless. Find the force exerted by the wall on a particle of mass $m$ when it is at a distance $x$ from the center of the tunnel. Solution: Distance of the particle from earth's center $r=\sqrt{\left(\frac{R}{2}\right)^{2}+x^{2}}$ Force on the particle=mass ${ }^{X}$ gravitational field intensity $F=m\left(\frac{G M r}{R^{3}}\right)$ $...
Read More →Solve the following :
Question: A tunnel is dug along a diameter of the earth. Find the force on a particle of mass m placed in the tunnel at a distance $x$ from the center. Solution: Method 1: The gravitational field inside a sphere of radius $R$ at a distance $x$ from center is So, force on particle of mass $m$ is $E=\frac{G M_{g} x}{R_{e}^{3}}$ So, force on particle of mass $m$ is $F=m E$ $F=\frac{G M_{e} m}{R^{3}} x$ Method 2: Acceleration due to gravity at depth $d$ from surface $g_{d}=g_{s}\left(1-\frac{d}{R_{e...
Read More →Solve the following :
Question: Two concentric spherical shells have masses $M_{1}, M_{2}$ and radii $R_{1}, R_{2}\left(R_{1}R_{2}\right)$. What is the force exerted by this system on a particle of mass $m 1$ if it is placed at a distance $\left(R_{1}+R_{2}\right) / 2$ from the centre? Solution: Here, $d=\frac{R_{1}+R_{2}}{2}$ The gravitational force of mass $\mathrm{m}$ due to shell of $M_{2}$ is zero as gravitational field inside shell is zero. Gravitational field due to shell of mass $M_{1}$ outside the shell at m...
Read More →Solve the following :
Question: Derive an expression for the gravitational field due to a uniform rod of length $L$ and mass $M$ at a point on its perpendicular bisector at a distance $d$ from the centre. Solution: Consider two small elements of length $\mathrm{dx}$ at a distance $x$ from center on both sides of center. Mass of each element, $\mathrm{dm}=\frac{M}{L} \mathrm{dx}$ Field due to each element at point $P$ $=\frac{G d m}{\sqrt{\left(d^{2}+x^{2}\right)^{2}}}$ $d E=\frac{G d m}{d^{2}+x^{2}}$ Resultant of the...
Read More →Solve the following :
Question: A semi-circular wire has a length $\mathrm{L}$ and mass $\mathrm{M}$. A particle of mass $\mathrm{m}$ is placed at the centre of the circle. Find the gravitational attraction on the particle due to the wire. Solution: Consider two small elements at an angle $\theta$ from horizontal length of the element $=\mathrm{Rd} \theta$ Mass of each element $\mathrm{dm}=\frac{M}{L}(\mathrm{Rd} \theta)$ Force by elemental mass on particle $\mathrm{dF}=\frac{\mathrm{Gm}(\mathrm{dm})}{R^{2}}$ $\mathr...
Read More →