Question
A satellite is revolving round the earth in circular orbit at some height above surface of earth. It takes $5.26 \times 10^{3}$ seconds to complete a revolution while its centripetal acceleration is $9.92 \mathrm{~m} / \mathrm{s}^{2}$. Height of satellite above surface of earth is (Radius of earth $6.37 \times 10^{6} \mathrm{~m}$ )(a) $70 \mathrm{~km}$(b) $120 \mathrm{~km}$(c) $170 \mathrm{~km}$(d) $220 \mathrm{~km}$
Step 1
Step 1: The centripetal acceleration of a satellite revolving around the earth in a circular orbit is given by $a_c = \frac{v^2}{r}$, where $v$ is the velocity of the satellite and $r$ is the radius of the circular orbit. Show more…
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A satellite is revolving round the earth in circular orbit at some height above surface of earth. It takes $5.26 \times 10^{3}$ seconds to complete a revolution while its centripetal acceleration is $9.92 \mathrm{~m} / \mathrm{s}^{2}$. Height of satellite above surface of earth is (Radius of earth $6.37 \times 10^{6} \mathrm{~m}$ ) (a) $70 \mathrm{~km}$ (b) $120 \mathrm{~km}$ (c) $170 \mathrm{~km}$ (d) $220 \mathrm{~km}$
An Earth Satellite An Earth satellite moves in a circular orbit $640 \mathrm{~km}$ above Earth's surface with a period of $98.0 \mathrm{~min}$. What are (a) the speed and (b) the magnitude of the centripetal acceleration of the satellite?
(a) Find the radius of the circular orbit of a satellite moving with an angular speed equal to the angular speed of earth's rotation. (b) If the satellite is directly above the north pole at some instant, find the time it takes to come over the equatorial plane. Mass of the earth $=6 \times 10^{24} \mathrm{~kg}$
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