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Q. Statement I : The speed of revolution of an artificial satellite revolving very near the earth is $8 kms^{-1}$
Statement II : Orbital velocity of a satellite, become in dependent of height of near satellite.

Gravitation

Solution:

$v_0 = R_e \sqrt{\frac{g}{R_e + h}}$
For satellite revolving very near to earth $Re + h = R_e$
As $(h < < R)$
$v_0 = \sqrt{R_eg} \simeq \sqrt{64 \times 10^5 \times 10}$
$ = 8 \times 10^3\,m/s$
$ = 8\,kms^{-1}$
Which is independent of height of a satellite.