The orbital speed of a satellite depends only on the mass of the central body (in this case, Earth) and the orbital radius. The mass of the satellite itself is irrele vant. Use the formula for orbital speed to find the ratio of speeds
The velocity of a satellite in orbit is given by:
\( v = \sqrt{\frac{GM}{r}} \)
Rearranging the formula, we see that velocity is inversely proportional to the square root of the radius:
\( v \propto \frac{1}{\sqrt{r}} \)
For two different radii, \( r_1 \) and \( r_2 \):
\( \frac{v_1}{v_2} = \sqrt{\frac{r_2}{r_1}} \)
If \( r_2 = 3r_1 \):
\( \frac{v_1}{v_2} = \sqrt{\frac{r_2}{r_1}} = \sqrt{3} \)
The velocity at the smaller radius (\( v_1 \)) is \( \sqrt{3} \) times the velocity at the larger radius (\( v_2 \)).
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 the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by: