Newton’s Law of Gravitation Questions in English

(A) Gravitational force is a universal force. According to Newton's law of universal gravitation,every particle of matter 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 their centers. Therefore,gravitational forces operate among all objects in the universe.

(A) Given:
Mass of each galaxy,$m = 8 \times 10^{11} \text{ solar masses} = 8 \times 10^{11} \times 2.0 \times 10^{30} \text{ kg} = 16 \times 10^{41} \text{ kg}$.
Distance between galaxies,$d = 2 \times 10^6 \text{ light-years} = 2 \times 10^6 \times 10^{16} \text{ m} = 2 \times 10^{22} \text{ m}$.
Gravitational force,$F = \frac{G m^2}{d^2}$.
Acceleration of the galaxy,$a = \frac{F}{m} = \frac{G m}{d^2}$.
Substituting the values:
$a = \frac{10^{-10} \times (16 \times 10^{41})}{(2 \times 10^{22})^2} = \frac{16 \times 10^{31}}{4 \times 10^{44}} = 4 \times 10^{-13} \text{ m/s}^2$.

(A) The long-range force experienced by a neutral particle with a finite mass is the gravitational force.
In this universe,every body attracts every other body with a force that is directly proportional to the product of their masses ($m_1$ and $m_2$) and inversely proportional to the square of the distance $(r)$ between them.
The formula is given by $F = \frac{G m_1 m_2}{r^2}$.
Since gravitational force acts over infinite distances and depends only on mass,it is the only fundamental force that acts on neutral particles with finite mass over long ranges.

(C) The four fundamental forces in nature,in decreasing order of their relative strengths,are as follows:
$1$. Strong nuclear force $(S)$: The strongest force,acting between nucleons.
$2$. Electromagnetic force $(E)$: Acts between charged particles.
$3$. Weak nuclear force $(W)$: Responsible for radioactive decay processes.
$4$. Gravitational force $(G)$: The weakest force,acting between all masses.
Comparing their relative magnitudes,we have $S > E > W > G$.
Therefore,the correct statement is $S > E > W > G$.

(C) Let the rod lie along the $x$-axis with its near end at $x = r$ and its far end at $x = r + L$ from the point mass '$m$' located at the origin.
Consider a small element of the rod of length '$dx$' at a distance '$x$' from the point mass '$m$'.
The mass of this element is $dm = \lambda dx = \frac{M}{L} dx$.
The gravitational force '$dF$' exerted by this element on the point mass '$m$' is given by:
$dF = \frac{G m dm}{x^2} = \frac{G m (M/L) dx}{x^2} = \frac{G M m}{L} \frac{dx}{x^2}$.
To find the total force '$F$',we integrate '$dF$' from $x = r$ to $x = r + L$:
$F = \int_{r}^{r+L} \frac{G M m}{L} \frac{dx}{x^2} = \frac{G M m}{L} \left[ -\frac{1}{x} \right]_{r}^{r+L}$.
$F = \frac{G M m}{L} \left( -\frac{1}{r+L} - (-\frac{1}{r}) \right) = \frac{G M m}{L} \left( \frac{1}{r} - \frac{1}{r+L} \right)$.
$F = \frac{G M m}{L} \left( \frac{r+L-r}{r(r+L)} \right) = \frac{G M m}{L} \left( \frac{L}{r(r+L)} \right) = \frac{G M m}{r(r+L)}$.

(C) The gravitational force acting between the two masses $m_{1}$ and $m_{2}$ separated by a distance $r$ is given by Newton's law of gravitation:
$F = \frac{G m_{1} m_{2}}{r^{2}}$
According to Newton's second law of motion,the force on mass $m_{1}$ is $F = m_{1} a_{1}$,where $a_{1}$ is the acceleration of mass $m_{1}$.
Equating the two expressions:
$m_{1} a_{1} = \frac{G m_{1} m_{2}}{r^{2}}$
$a_{1} = \frac{G m_{2}}{r^{2}}$
Since $G$ and $r$ are constant for the system at any given instant,we find that $a_{1} \propto m_{2}$.
Similarly,for mass $m_{2}$,$a_{2} = \frac{G m_{1}}{r^{2}}$,which implies $a_{2} \propto m_{1}$.

(A) The gravitational force $F$ between two masses $m_1$ and $m_2$ separated by a distance $r$ is given by $F = \frac{G m_1 m_2}{r^2}$.
Here,$m_1 = m$ and $m_2 = (M - m)$.
Substituting these values,we get $F = \frac{G m(M - m)}{r^2} = \frac{G}{r^2} (Mm - m^2)$.
For the force to be maximum,the derivative of $F$ with respect to $m$ must be zero,i.e.,$\frac{dF}{dm} = 0$.
$\frac{d}{dm} [\frac{G}{r^2} (Mm - m^2)] = 0$.
$\frac{G}{r^2} (M - 2m) = 0$.
Since $\frac{G}{r^2} \neq 0$,we have $M - 2m = 0$,which implies $m = \frac{M}{2}$.
Thus,the mass of the second piece is $(M - m) = M - \frac{M}{2} = \frac{M}{2}$.
The ratio of the masses is $\frac{m}{M-m} = \frac{M/2}{M/2} = \frac{1}{1}$.