Satellites A and B are in circular orbits at a height of 900 km and 300 km, respectively, above the Earth's surface. The velocity of satellite B will be ________ times the velocity of satellite A (Rounded off to three decimal places).
Assume radius of Earth to be 6378 km, acceleration due to gravity to be 9.81 \(m\,s^{-2}\), the product of universal gravitational constant and mass of Earth is \(3.98601\times10^{14}\ m^3\,s^{-2}\).
Hint: For a satellite to remain in circular orbit around the Earth, the Earth's gravitational force must be balanced by the centrifugal force of the orbiting satellite.
Show Hint
Balance gravitational force against the centripetal requirement for a circular orbit to get v equal to the square root of GM over r, then compare the two orbital radii.
Step 1: State the orbital velocity formula and the orbital radii.
$v = \sqrt{GM/r}$, with $GM = 3.98601\times10^{14}\ m^3\,s^{-2}$, $r_A = 7278\ km = 7.278\times10^6\ m$ (satellite A, 900 km altitude) and $r_B = 6678\ km = 6.678\times10^6\ m$ (satellite B, 300 km altitude).
Step 2: Compute the absolute velocity of satellite A.
\[ v_A = \sqrt{\frac{3.98601\times10^{14}}{7.278\times10^6}} = \sqrt{5.4761\times10^{7}} \approx 7400.1\ m\,s^{-1} \]
Step 3: Compute the absolute velocity of satellite B.
\[ v_B = \sqrt{\frac{3.98601\times10^{14}}{6.678\times10^6}} = \sqrt{5.9684\times10^{7}} \approx 7725.5\ m\,s^{-1} \]
Step 4: Divide to obtain the required ratio.
\[ \frac{v_B}{v_A} = \frac{7725.5}{7400.1} \approx 1.04398 \]
Step 5: Round off and confirm.
Rounded to three decimal places, $\dfrac{v_B}{v_A} \approx 1.044$, matching the ratio obtained directly from $\sqrt{r_A/r_B}$ and confirming that the lower, faster satellite B moves about $1.044$ times as fast as the higher satellite A.
