A body of mass m is placed on earth surface which is taken from earth surface to a height of h = 3R then change in gravitational potential energy is :
\(\frac{mgR}{4}\)
\(\frac{2}{3}mgR\)
\(\frac{3}{4}mgR\)
\(\frac{mgR}{2}\)
The correct option is (C) : \(\frac{3}{4}mgR\)
Change in G.P.E. = final energy – initial energy
= \(-\frac{GMm}{4R}+\frac{GMm}{R}\) = \(\frac{GMm}{R}[\frac{1}{\frac{1}{4}}]\)
\(\frac{3}{4}\frac{GMm}{R}=\frac{3}{4}\frac{GM}{R^2}mR=\frac{3}{4}gmR\)
Match the LIST-I with LIST-II
\[ \begin{array}{|l|l|} \hline \text{LIST-I} & \text{LIST-II} \\ \hline \text{A. Gravitational constant} & \text{I. } [LT^{-2}] \\ \hline \text{B. Gravitational potential energy} & \text{II. } [L^2T^{-2}] \\ \hline \text{C. Gravitational potential} & \text{III. } [ML^2T^{-2}] \\ \hline \text{D. Acceleration due to gravity} & \text{IV. } [M^{-1}L^3T^{-2}] \\ \hline \end{array} \]
Choose the correct answer from the options given below:
A small point of mass \(m\) is placed at a distance \(2R\) from the center \(O\) of a big uniform solid sphere of mass \(M\) and radius \(R\). The gravitational force on \(m\) due to \(M\) is \(F_1\). A spherical part of radius \(R/3\) is removed from the big sphere as shown in the figure, and the gravitational force on \(m\) due to the remaining part of \(M\) is found to be \(F_2\). The value of the ratio \( F_1 : F_2 \) is: 
In mechanics, the universal force of attraction acting between all matter is known as Gravity, also called gravitation, . It is the weakest known force in nature.
According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,
On combining equations (1) and (2) we get,
F ∝ M1M2/r2
F = G × [M1M2]/r2 . . . . (7)
Or, f(r) = GM1M2/r2
The dimension formula of G is [M-1L3T-2].