What is the gravitational potential energy of earth-moon system relative to the potential energy at infinite separation? Given: \(M_{Earth}=\left(5.98\times10^{24}\right)kg\) , \(M_{Moon}=\left(7.36\times10^{22}\right)kg\). Mean separation between earth and moon \(=\left(3.82\times10^{6}\right)m\).
Ans: \(U=-\left(7.68\times10^{30}\right)J\).
Two bodies of mass \(10kg\) and \(1000kg\) are separated at a distance of \(1m\) apart. At which point on the line joining the two will the gravitational field intensity be zero? Also calculate: the gravitational potential at this point, \(G=\left(6.67\times10^{-11}\right)Nm^{2}kg^{-2}\).
Mass of the earth is 81 times the mass of Moon. The distance between earth and moon is \(\left(3.8\times10^{8}\right)m\). Calculate: the position of the point on the line joining the centres of earth and moon where the gravitational field is zero.
Ans: \(x=\left(3.42\times10^{5}\right)km\).
Two masses \(400kg\) and \(300kg\) are at a distance of \(0.25m\) apart. Calculate: the magnitude of the intensity of gravitational field at a point at a distance \(0.20m\) from \(400kg\) mass and \(0.15\) from the \(300kg\) mass.
Ans: \(E=\left(1.11\times10^{-6}\right)N\).
Calculate: the work done to bring four particles each of mass \(400gram\) from large distance to the vertices of a square of side \(20cm\). Given: \(G=\left(6.67\times10^{11}\right)Nm^{2}kg^{-2}\).
Ans: \(W=-\left(9.756\times10^{-11}\right)J\).
Three point masses, each of mass m and placed at the vertices of an equilateral triangle of side l. What is the gravitational field and gravitational potential at the centroid of triangle?
Ans: \(E=\frac{3Gm}{l^{2}}\).
A mass M is split into two parts m and (M-m), which are then separated by a certain distance. What ratio \(\left(\frac{m}{M}\right)\) maximises the gravitational force between the parts?
Ans: \(\frac{m}{M}=\frac{1}{2}\).
What is the minimum energy required to launch a satellite of mass \(m\) kg from the earth’s surface of radius \(R\) in circular orbit at an altitude of [latex0]2R[/latex]?
Ans: \(E_{total}=\frac{5}{6}mgR\).
At a point above the surface of the earth , the gravitational potential energy is \(-\left(5.12\times10^{7}\right)\frac{J}{kg}\) and acceleration due to gravity is, \(6.4\frac{m}{s^{2}}\). Assume the mean radius of earth to be \(6400km\). Calculate: the height of the point above the surface of earth.
Ans: \(h=1600km\).
A body is released at a distance \(r\left(r>R\right)\) from the center of the earth. What will be the velocity when it strikes the surface of the earth?
The radius of the earth is \(R\) and acceleration due to gravity is \(g\) is at the surface of the earth. Calculate: the work required in raising a body of mass \(m\) to a height \(h\) from the surface of the earth.
Ans: \(\frac{mgh}{1+\frac{h}{R}}\).
The gravitational field is given by \(\vec{F_{g}}=-4\hat{k}\). What is the work done in transporting a mass of \(2kg\) from \(\left(0,0,0\right)\) to \(\left(4,3,5\right)\) metre by extrenal force?
Ans: \(20J\).
(i) Find: the potential energy of the system of four particle masses of \(m\) kg each, placed at the vertices of a square of side \(l\). Also; (ii) obtain the potential at centre of square?
Ans:(i) \(-\frac{5.414}{l}\) and (ii) \(-4\sqrt{2}\frac{Gm}{l}\).
