Question:

The velocity ($V_s$) of a satellite moving in a circular orbit at a height of 1000 km above earth surface is ___________ km s$^{-1}$ (rounded off to 2 decimal places).
{($G = 6.67 \times 10^{-11}$ m$^3$ kg$^{-1}$ s$^{-2}$, $M_e = 5.972 \times 10^{24}$ kg and $r_e = 6378$ km)}

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To calculate satellite velocity in a circular orbit, use \( V = \sqrt{GM/r} \), ensuring that the radius \( r \) includes the Earth's radius plus the satellite's height.
Updated On: Jan 31, 2026
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Correct Answer: 7.25

Solution and Explanation

To find the orbital velocity of a satellite, we use the formula: \[ V_s = \sqrt{\frac{GM_e}{r}} \] where: $G = 6.67 \times 10^{-11}$ m$^3$kg$^{-1}$s$^{-2}$ $M_e = 5.972 \times 10^{24}$ kg $r = r_e + h = (6378 + 1000) \times 10^3 = 7378000$ m Now plug in the values: \[ V_s = \sqrt{ \frac{6.67 \times 10^{-11} \times 5.972 \times 10^{24}}{7378000} } \approx \sqrt{5.403 \times 10^7} \approx 7349 { m/s} = 7.35 { km/s} \] So, the final velocity of the satellite is: \[ \boxed{7.35 { km/s}} \]
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