Exercise 1

A satellite revolving around a planet in stationary orbit has time period 6 hours. The mass of planet is one-fourth the mass of earth. The radius orbit of planet is : (Given = Radius of geo-stationary orbit for earth is $4.2 \times {10}^4$ km)

$$\begin{gathered} \mathrm{A} \mathrm{small} \mathrm{point} \mathrm{of} \mathrm{mass} \mathrm{m} \mathrm{is} \mathrm{placed} \mathrm{at} \mathrm{a} \mathrm{distance} 2\mathrm{R} \mathrm{from} \mathrm{the} \mathrm{centre} \text{'}\mathrm{O}\text{'} \mathrm{of} \mathrm{a} \mathrm{big} \mathrm{uniform} \mathrm{solid} \mathrm{sphere} \mathrm{of} \mathrm{mass} \mathrm{M} \mathrm{and} \mathrm{radius} \mathrm{R}. \mathrm{The} \mathrm{gravitational} \mathrm{force} \mathrm{on} \text{'}\mathrm{m}\text{'} \mathrm{due} \mathrm{to} \mathrm{M} \mathrm{is} \mathrm{F}. \mathrm{A} \mathrm{spherical} \mathrm{part} \mathrm{of} \mathrm{radius} \mathrm{R}/3 \mathrm{is} \mathrm{removed} \mathrm{from} \mathrm{the} \mathrm{big} \mathrm{sphere} \mathrm{as} \mathrm{shown} \mathrm{in} \mathrm{the} \mathrm{figure} \mathrm{and} \mathrm{the} \mathrm{gravitational} \mathrm{force} \mathrm{on} \mathrm{m} \mathrm{due} \mathrm{to} \\ \mathrm{remaining} \mathrm{part} \mathrm{of} \mathrm{M} \mathrm{is} \mathrm{found} \mathrm{to} \mathrm{be} \mathrm{F}₂. \mathrm{The} \mathrm{value} \mathrm{of} \mathrm{ratio} {\mathrm{F}}_1 {\mathrm{F}}_2 \mathrm{is} \end{gathered}$$

A positive ion A and a negative ion B has charges $6.67 \times 10{ }^{-19} \mathrm{C}$ and $9.6 \times {10}^{-10} \mathrm{C}$, and masses $19.2 \times {10}^{-27} \mathrm{kg} \mathrm{and} 9 \mathrm{x} {10}^{-27} \mathrm{kg}$ respectively. At an instant, the ions are separated by a certain distance $\mathrm{r}$. At that instant the ratio of the magnitudes of electrostatic force to gravitational force is $\mathrm{P} \times {10}^{45}$, where the value of $\mathrm{P}$ is_____.

$\left(\text{Take }\frac{1}{4\pi {\epsilon }_0}= 9 \times {10}^9 {\text{ Nm}}^2{\text{C}}^{-1 }\text{and universal gravitational constant as }6.67\times {10}^{-11} {\text{ Nm}}^2{\text{ kg}}^{-2}\right)$

$$\begin{gathered} If a satellite orbiting the Earth is 9 times closer to t \\ he Earth than the Moon, what is the time period of \\ rotation of the satellite? Given rotational time period \\ of Moon = 27 days and gravitational attraction between \\ the satellite and the moon is neglected. \end{gathered}$$

$$\begin{gathered} A satellite of mass \frac{M}{2} is revolving around earth in a \\ a circular orbit at a height of \frac{R}{3} from earth \\ surface. The angular momentum of the satellite is \\ M\sqrt{\frac{GMR}{x}}. The value of x is \_\_\_\_\_\_\_\_\_\_. where M \\ and R are the mass and radius of earth, respectively. \\ (G is the gravitational cons\tan t) \end{gathered}$$

$\mathrm{A} \mathrm{satellite} \mathrm{is} \mathrm{launched} \mathrm{into} \mathrm{a} \mathrm{circular} \mathrm{orbit} \mathrm{of} \mathrm{radius} \text{'}\mathrm{R}\text{'} \mathrm{around} \mathrm{the} \mathrm{earth}. \mathrm{A} \mathrm{second} \mathrm{statellite} \mathrm{is} \mathrm{launched} \mathrm{into} \mathrm{an} \mathrm{orbit} \mathrm{of} \mathrm{radius} 1.03 \mathrm{R}. \mathrm{The} \mathrm{time} \mathrm{period} \mathrm{of} \mathrm{revolution} \mathrm{of} \mathrm{the} \mathrm{second} \mathrm{satellite} \mathrm{is} \mathrm{larger} \mathrm{than} \mathrm{the} \mathrm{first} \mathrm{one} \mathrm{approximately} \mathrm{by} :-$

$$\begin{gathered} Acceleration due to gravity on the surface of earth \\ is \text{'}g\text{'}. If the diameter of earth is reduced to one third \\ of its original value and mass remains unchanged, \\ then the acceleration due to gravity on the surface of t \\ he earth is \_\_\_\_\_\_g. \end{gathered}$$

$$\begin{gathered} \mathrm{Earth} \mathrm{has} \mathrm{mass} 8 \mathrm{times} \mathrm{and} \mathrm{radius} 2 \mathrm{times} \mathrm{that} \mathrm{of} \mathrm{a} \mathrm{planet}. \mathrm{If} \mathrm{the} \mathrm{escape} \mathrm{velocity} \mathrm{from} \mathrm{the} \mathrm{earth} \mathrm{is} 11.2 \mathrm{km}/\mathrm{s}, \mathrm{the} \mathrm{escape} \mathrm{velocity} \mathrm{in} \mathrm{km}/\mathrm{s} \mathrm{from} \mathrm{the} \mathrm{planet} \\ \mathrm{will} \mathrm{be} : \end{gathered}$$