Question

time period of a satellite revolving close to the surface of the Earth is (a= mean density of earth)
A) √3π/ag
B)√5π/ag
C)√2π/ag
d)√4π/ag





plz solve it stepwise​

Answer 1

Explanation:

as it is revolving close to the earth so we will consider the radius of circular path of the satellite to be almost equal to the radius of earth.

let the radius of earth be R.

let the mass of earth be M.

let the velocity of the revolving satellite be V.

as the movement of satellites is uniform circular motion so there will be only centripetal acceleration.

acceleration due to earth at a point near to earth, g = G×M/R²

centripetal acceleration = V² / R

g = V²/R

•°•, V = √(gR) ..................(1)

mean density of earth, a = M÷ (4/3πR³)

•°•, M/R = 4aπR² /3. ..................(2)

time period = 2πR/V

= 2π√(R/g)

= 2π√(R/ (G×a×4/3×π×R³/R²))

=

$2\pi { { { \sqrt{(\frac{r}{ \frac{G \times 4 \div 3 \times \pi {r}^{3} }{ {r}^{2} } }} )}}}$

now, by solving we will get the answer:-

$\sqrt{ \frac{3\pi}{aG} }$

here in the baith formulas g represents the G.

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