Question:
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.
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.
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.
Updated On: Jul 20, 2026
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Verified By Collegedunia
Correct Answer: 1.042
Solution and Explanation
Step 1: Set up the force balance for a circular orbit (per the hint).
For a satellite of mass \(m\) moving in a circular orbit of radius \(r\) about the Earth (mass \(M\)), gravitational attraction supplies the centripetal force: \[ \frac{GMm}{r^2} = \frac{mv^2}{r} \implies v = \sqrt{\frac{GM}{r}} \] where \(GM = 3.98601\times10^{14}\ m^3\,s^{-2}\) is given directly, and \(r\) is the distance from the Earth's centre to the satellite.
Step 2: Compute the orbital radii of A and B.
\[ r_A = R_{Earth} + h_A = 6378 + 900 = 7278\ km, \qquad r_B = R_{Earth} + h_B = 6378 + 300 = 6678\ km \]
Step 3: Form the velocity ratio symbolically.
Since \(v \propto \dfrac{1}{\sqrt{r}}\) for both satellites (the same \(GM\) applies to both), the ratio simplifies without needing the numeric value of \(GM\): \[ \frac{v_B}{v_A} = \sqrt{\frac{GM/r_B}{GM/r_A}} = \sqrt{\frac{r_A}{r_B}} \]
Step 4: Substitute the radii.
\[ \frac{v_B}{v_A} = \sqrt{\frac{7278}{6678}} = \sqrt{1.08985} \]
Step 5: Evaluate the square root and round off.
\[ \sqrt{1.08985} \approx 1.04396 \] Rounded to three decimal places, \(\dfrac{v_B}{v_A} \approx 1.044\), which lies in the accepted range of \(1.042\) to \(1.046\). This makes physical sense: satellite B orbits closer to Earth, so it must move faster to maintain its circular orbit.
\[ \boxed{\dfrac{v_B}{v_A} \approx 1.044} \]
For a satellite of mass \(m\) moving in a circular orbit of radius \(r\) about the Earth (mass \(M\)), gravitational attraction supplies the centripetal force: \[ \frac{GMm}{r^2} = \frac{mv^2}{r} \implies v = \sqrt{\frac{GM}{r}} \] where \(GM = 3.98601\times10^{14}\ m^3\,s^{-2}\) is given directly, and \(r\) is the distance from the Earth's centre to the satellite.
Step 2: Compute the orbital radii of A and B.
\[ r_A = R_{Earth} + h_A = 6378 + 900 = 7278\ km, \qquad r_B = R_{Earth} + h_B = 6378 + 300 = 6678\ km \]
Step 3: Form the velocity ratio symbolically.
Since \(v \propto \dfrac{1}{\sqrt{r}}\) for both satellites (the same \(GM\) applies to both), the ratio simplifies without needing the numeric value of \(GM\): \[ \frac{v_B}{v_A} = \sqrt{\frac{GM/r_B}{GM/r_A}} = \sqrt{\frac{r_A}{r_B}} \]
Step 4: Substitute the radii.
\[ \frac{v_B}{v_A} = \sqrt{\frac{7278}{6678}} = \sqrt{1.08985} \]
Step 5: Evaluate the square root and round off.
\[ \sqrt{1.08985} \approx 1.04396 \] Rounded to three decimal places, \(\dfrac{v_B}{v_A} \approx 1.044\), which lies in the accepted range of \(1.042\) to \(1.046\). This makes physical sense: satellite B orbits closer to Earth, so it must move faster to maintain its circular orbit.
\[ \boxed{\dfrac{v_B}{v_A} \approx 1.044} \]
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