Question

What is a period of revolution of the earth satellite?
Ignore the height of satellite above the surface of the earth. Given,
(i) the value of gravitational acceleration, g = 10 ms ^-2.
(ii) radius of the earth, Re = 6400 km
(Take, $π$ = 314) [KCET 2014]

Moderate

Solution

Given, R_e = 6400 km:: 6.4 X 10⁶m, $π$ =3.14, 9 =10 ms^-2
We know that, the period of revolution of the earth satellite
$T = 2 π \sqrt{\frac{{(R_{e} + h)}^{3}}{g R_{e}^{2}}} [if h < R_{e}, then (R_{e} + h = R_{e})]$
$So, T = 2 π \sqrt{\frac{R_{e}^{3}}{g R_{e}^{2}}} = 2 π \sqrt{\frac{R_{e}}{g}} = 2 \times 3.14 \sqrt{\frac{6.4 \times 10^{6}}{10}}$
$= 5.024 \times 10^{3} = 5024 s$
= 83.73 min

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