Gravitation MCQs for NEET — Physics Questions with Answers

$ {T_2 \over T_1 } = \left( {r_1 \over r_2} \right ) ^ {3/2} \Rightarrow T_2 = 24 \left( {6400 \over 3600} \right)^{3/2} = 2hours $

The gravitational law (Newton's law of gravitation) and Coulomb's law (which describes the force between two charges) are both inverse square laws. This means that the force decreases proportionally to the square of the distance between the two objects or charges.

The gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. This relationship is described by Newton's Universal Law of Gravitation. Hence, mass is the property responsible for gravitational force.

$ {Electronmagnetic force \over Gravational force} = {10 ^ {-2} \over 10 ^ {-38 } } = 10 ^ {36 } $

Sir Isaac Newton is known for unifying the terrestrial and celestial domains under a common law of gravitation. Newton's law of universal gravitation states that every mass attracts every other mass in the universe with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

$ { T_2 ^ 2 \over T_1^2 } = { r_2^3 \over r_1^3 } $

We know, 

$dV=-\overrightarrow{I}.\overrightarrow{dr} \Rightarrow dV=-\frac{k}{x^2}dx \Rightarrow \int dV=-k \int \frac{dx}{x^2} \Rightarrow V=\frac{k}{x}+C$

When $x = \infty , V=0 \Rightarrow C=0$

$\Rightarrow V=\frac{k}{x} \therefore At x = a, V= \frac{k}{a}$

$$\begin{gathered} T=\frac{2\mathrm{\pi R}}{v}=\frac{2\mathrm{\pi R}}{\sqrt{{\displaystyle \frac{GM}{R}}}}=\frac{2{\mathrm{\pi R}}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{\sqrt{GM}} \\ T=\frac{2\pi R^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{\sqrt{G{\displaystyle \frac{4}{3}}{\mathrm{\pi R}}^3\mathrm{\rho }}}=\frac{2{\mathrm{\pi R}}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{2\sqrt{{\displaystyle \frac{\mathrm{\pi }}{3}}\rho G}{R}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}=\sqrt{\frac{3\mathrm{\pi }}{\rho G}} \end{gathered}$$

$$\begin{gathered} \left(2\right) \\ {K}_i+U_i=K_f+U_f \\ \frac{1}{2}m{\left(k{v}_e\right)}^2-\frac{GMm}{R}=0-\frac{GMm}{r} \\ \frac{1}{2}m{\left(k\sqrt{\frac{2GM}{R}}\right)}^2-\frac{GMm}{R}=0-\frac{GMm}{r} \\ \frac{k^2}{R}-\frac{1}{R}=-\frac{1}{r} \\ r=\frac{R}{1-k^2} \end{gathered}$$

$$\begin{gathered} Potential difference at 10m=5J/kg \\ Potential difference at 8m=\frac{5}{10}\times 8=4J/kg \\ Work done=m∆V=2\times 4=8J \end{gathered}$$