Law of Gravitation
By The universal law of gravitation states that every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them i.e., \begin{align} F = \frac{G m_1 m_2}{ r^2} \end{align} where $G=6.67\times10^{-11}$ m3kg-1s-2 is the gravitational constant.
Problems from IIT JEE
Problems (JEE Mains 2020): Two identical spheres, each of radius $r$ and uniform density $\rho$, are touching each other. The gravitational force $F$ on a sphere by the other is
- $F\propto\frac{1}{r^4}$
- $F\propto\frac{1}{r^2}$
- $F\propto r^2$
- $F\propto r^4$
Solution: According to Newton's law of gravitation, the attractive force between two particles of masses $M$ and $m$, separated by a distance $r$ is given by \begin{align} F=\frac{GMm}{r^2}.\nonumber \end{align} This formula also gives the attractive force between two spherically symmetric objects (e.g., sphere, shell) of masses $M$ and $m$ with their centres separated by a distance $r$. This is known as Newton's shell theorem.
The masses of two spheres are $M=m=\rho\cdot\frac{4}{3}\pi r^3$ and separation between their centres is $2r$. Thus, gravitational force on one sphere by the other sphere is \begin{align} F &=\frac{G\left(\rho\cdot\frac{4}{3}\pi r ^3\right) \left(\rho\cdot\frac{4}{3}\pi r ^3\right)}{(2r)^2} \\ &=\frac{4}{9}G \pi^2 \rho^2 r^4.\nonumber \end{align}
Problem (JEE Mains 2014): From a sphere of mass $M$ and radius $R$, a smaller sphere of radius $R/2$ is carved out such that the cavity made in the original sphere is between its centre and the periphery (see figure). For the configuration in the figure where the distance between the centre of the original sphere and the removed sphere is $3R$, the gravitational force between the two spheres is
- $\frac{41GM^2}{3600 R^2}$
- $\frac{41GM^2}{450 R^2}$
- $\frac{59GM^2}{450 R^2}$
- $\frac{GM^2}{225 R^2}$
Solution: The mass of the sphere of radius $R/2$ is $M_2=M/8$ and that of the sphere having the cavity is \begin{align} M_1=M-M_2=7M/8. \end{align}
Let $F_1$ be the force on the sphere having the cavity, $F_2$ be the force on a sphere (of radius $R/2$) at the location of the cavity, and $F_3$ be the force on the sphere (of radius $R$) without a cavity. It can be seen that, \begin{align} F_3&=F_1+F_2,\nonumber\\ \frac{GM(M/8)}{(3R)^2} &=F_1+\frac{G(M/8)(M/8)}{(5R/2)^2}.\nonumber \end{align} Solve to get \begin{align} F_1={41GM^2}/{(3600R^2)}. \end{align}
Problem (IIT JEE 2014): A planet of radius $R=\frac{1}{10}\times$ (radius of Earth) has the same mass density as Earth. Scientists dig well of depth R/5 on it and lower a wire of the same length and of linear mass density 10-3 kg/m into it. If the wire is not touching anywhere, the force applied at the top of the wire by a person holding it in place is, (take the radius of Earth = 6×106 m and the acceleration due to gravity on Earth = 10 m/s2).
- 96 N
- 108 N
- 120 N
- 150 N
Answer: The answer is (B) i.e., 108 N.
Solution: Given, radius of the earth $R_e=10R=6\times{10}^{6}\;\mathrm{m}$ and acceleration due to gravity on earth $g_e={10}\;\mathrm{m/s^2}$. Consider a wire element of length $\mathrm{d}r$ placed at a distance $r$ from the planet's centre O (see figure).
Let $\rho$ be common mass density of the planet and the earth and $m$ be mass of the planet inside the sphere of radius $r$ i.e., $m=\frac{4}{3}\pi r^3 \rho$. The mass distribution has spherical symmetry. The gravitational force on the wire element by the planet is equal to the force by a point mass of magnitude $m$ placed at the centre O. Thus, force on the wire element is, \begin{align} \mathrm{d}F=\frac{Gm\lambda\mathrm{d}r}{r^2}=\frac{4}{3}G\pi\rho \lambda r\mathrm{d}r.\nonumber \end{align} Integrate $\mathrm{d}F$ from $r=4R/5$ to $r=R$ to get total force on the wire, \begin{align} F&=\frac{4}{3}G\pi\rho\lambda\int_{\frac{4}{5}R}^{R} r\mathrm{d}r \nonumber\\ &=\frac{4}{3}G\pi \rho\lambda \left(\frac{9R^2}{50}\right) \nonumber\\ &=\frac{\frac{4}{3}\pi\rho R_e^3 G}{R_e^2}\left(\frac{9\lambda R_e}{5000}\right)\nonumber\\ &=g_e \frac{9\lambda R_e}{5000}\nonumber\\ &=\frac{(10)(9)({10}^{-3})(6\times{10}^{6})}{5000}\nonumber\\ &={108}\;\mathrm{N}.\nonumber \end{align}
