Suppose the gravitational force varies inversely as the \(n^{th}\) power of distance. Then, find the expression for the time period of a planet in a circular orbit of radius \(r\) around sun.
Ans: \(T\propto r^{\frac{\left(n+1\right)}{2}}\).
Two bodies of masses \(m_{1}\) and \(m_{2}\) are initially at rest placed infinite distance apart. They are then allowed to move towards each other under the mutual gravitational attraction. Show that their relative velocity of approach at separation \(r\) between them is \(v=\sqrt{\frac{2G}{r}}\left(\sqrt{m_{1}}+\sqrt{m_{2}}\right)\).
If a satellite is revolving around a planet of mass \(M\) in an elliptical orbit of semi-major axis \(a\), Show that: orbital speed of the satellite when it is at a distance \(r\) from the focus will be given by; \(v_{o}^{2}=GM\left[\frac{2}{r}-\frac{1}{a}\right]\).
Calculate the speed with which the earth have to rotate on its axis so that a person on equator would weigh \(\frac{4}{5}^{th}\) as much as present. Consider the equatorial radius \(6400km\).
Two satellites \(S_{1}\) and \(S_{2}\) revolve around a planet in coplaner circular orbit in the same sense. Their periods of revolutions are \(1hour\) and \(8hours\) respectively. The radius of the orbit of \(S_{1}\) is \(10^{4}\) when \(S_{2}\) is closest to \(S_{1}\), Find: (I) the speed of \(S_{2}\) with respect to \(S_{1}\). (II) The angular speed of \(S_{2}\) actually observed by an astronaut in \(S_{1}\).
Ans:(I) \(|v_{2}-v_{1}|=\left(3.14\times10^{4}\right)\frac{km}{h}\), and (II) \(\omega=\left(2.91\times10^{-4}\right)\frac{rad}{s}\).
A particle is projected vertically upwards from the surface of the earth having radius \(R\) with a kinetic energy equal to half of the kinetic energy required it to escape. To what height does it rise above the surface of the earth?
Ans: \(h=r\).
An artificial satellite is moving in a circular orbit around the earth with a speed equal to the half of the magnitude of the escape velocity from the surface of the earth. (I) Determine: height of the satellite above the surface of the earth? (II) If, the satellite is stopped suddenly in its orbit and allowed to fall freely on the earth. Take: \(g=9.8\frac{m}{s^{2}}\) and Radius of the earth \(R_{e}=6400km\).
Ans:(I) \(h=R\) and (II) \(v=7.92\frac{km}{s}\).
Two bodies of masses \(m_{1}\) and \(m_{2}\) are placed at a distance \(r\) apart. Show that at the position where the gravitational feild due to them is Zero the potential is given by; \(V=-\frac{G}{r}\left(m_{1}+m_{2}+2\sqrt{m_{1}.m_{2}}\right)\).
Three particles each of mass \(m\) are situated at the vertices of an equilateral triangle of side \(a\). The forces acting on the particles are only their mutual gravitational forces. It is desired that each particle moves in circle while maintaining the original separation \(a\). find the initial velocity that should be given to each particle and also the time period of the circular motion.
Ans: \(v=\sqrt{\frac{Gm}{a}}\) and \(T=2\pi\sqrt{\frac{a^{2}}{3Gm}}\).
A simple pendulum has a time period \(T_{1}\) on the earth’s surface and when taken to height of \(R\) above the earth’s surface, where \(R\) is the radius of the earth. What is the value of \(\frac{T_{2}}{T_{1}}\)?
Ans: \(\frac{T_{1}}{T_{2}}=2\).
Imagine a light planet revolving around a very massive star in a circular orbit of radius \(R\) with a period of revolution \(T\). If the gravitational force of attraction between the planet and the star varies as \(R^{-3}\),then how \(T\) is related to \(R\).
Ans: \(T\propto R^{2}\).
Imagine a light planet is revolving around a star on circular orbit of radius \(R\) with a period of revolution \(T\). If, the gravitational force of attraction between the planet and the star is proportional to \(R^{-\frac{5}{2}}\), then find the relation between \(T\) and \(r\).
Ans: \(T^{2}\propto R^{\frac{7}{2}}\)‘
What is the change in potential energy when a body of mass \(m\) is raised to a height \(h=nR\) (R=radius of the earth) from the surface of the earth?
Ans: \(\Delta U=\frac{1}{2}mgR\).
A body is projected vertically upwards from the surface of the earth with an initial velocity \(v\). Show that it will go upto height \((h)\) given by; \(h=\frac{v^{2}R}{\left(2gr-v^{2}\right)}\) and hence deduce the expression for escape velocity.
Calculate: the distance from the earth’s surface at which acceleration due to gravity is same below and above the surface of earth.
Ans: \(0.62R\).
If the moon describes a circular orbit of radius \(r\) around the earth with a uniform angular velocity \(\omega\) so that \(\omega^{2}r^{3}=gR^{2}\), where \(R\) is the radius of the earth and \(g\) is the acceleration due to gravity. If, \(r=60R\) and the period of revolution of moon around the earth is \(27.3days\). Then; find \(r\).
