Question

(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.

Answer 1

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Answer 2

Answer:

Explanation:

Angular speed of earth and the satellite will be same

2πTe=2πTs

or 124×3600=12–√π(R+h)3gh2−−−−−−−√

or 12×3600=3.14(R+h)3gR2−−−−−−−−√

or (R+h)3gR2=((12×3600)62(3.14)2

or(6400+h3)×1099.8×(6400)2×106=(12×3600)2(3.14)2

or (6400+h)×1096272×09=432×104

or (6400+h)3=6272x432×104

or 6400+h=(6272×432×104)13−6400

b. Time taken from north to equator

=12T

=12×6.28((42300+6400)310×(6400)2×106)

=3.14 (479)3×106−−−−−−−−−−√(64)2×1011

=3.14497×497×49764×64×105−−−−−−−−−−−−−−√

=6hours