a remote sensing satellite of the earth revolves in a circular orbit at a height of 250km above the earths surface. What is the1. orbital speed2. time period of the revolution of the satellite

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Solution

Given, h = 250 km = 250 X 10³ m
As we know radius of earth, R = 6.38 X 10⁶m
Mass of the earth = 6 X 10²⁴ kg
G = 6.67 X 10^-11 Nm²kg^-2
1.Orbital speed of the satellite is given by the formula :
^{$v_{0} = \sqrt{\frac{G M}{r}} w h e r e, r = R + h$}
here,
2. Time period of the revolution of the satellite is given by the formula :
T = $\frac{2 π r}{v_{0}}$ ..........(1)
where r = R + h, plug in the corresponding values in the eq. 1 to get your answer.

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