A satellite is orbiting just above the surface of the earth with period T. If d is the dependent of the earth and G is the universal constant of gravitation, the quantity 3π/Gd respectively:
T2
T3
\(\sqrt{T}\)
T
The correct option is (A): T2
\(T=\frac{2\pi}{\sqrt{GM}}r^{\frac{3}{2}}\)
\(\Rightarrow T^2=\frac{4\pi^2R^3}{G(\frac{4}{3}\pi R^3d)}(r=R)\)
\(T^2=\frac{3\pi}{Gd}\)
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:
The height from Earth's surface at which acceleration due to gravity becomes \(\frac{g}{4}\) is \(\_\_\)? (Where \(g\) is the acceleration due to gravity on the surface of the Earth and \(R\) is the radius of the Earth.)
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : The potential (V) at any axial point, at 2 m distance(r) from the centre of the dipole of dipole moment vector
\(\vec{P}\) of magnitude, 4 × 10-6 C m, is ± 9 × 103 V.
(Take \(\frac{1}{4\pi\epsilon_0}=9\times10^9\) SI units)
Reason R : \(V=±\frac{2P}{4\pi \epsilon_0r^2}\), where r is the distance of any axial point, situated at 2 m from the centre of the dipole.
In the light of the above statements, choose the correct answer from the options given below :
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].