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 (\( v_2 \)).
Net gravitational force at the center of a square is found to be \( F_1 \) when four particles having masses \( M, 2M, 3M \) and \( 4M \) are placed at the four corners of the square as shown in figure, and it is \( F_2 \) when the positions of \( 3M \) and \( 4M \) are interchanged. The ratio \( \dfrac{F_1}{F_2} = \dfrac{\alpha}{\sqrt{5}} \). The value of \( \alpha \) is 

A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer)
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.