# A satellite orbits the earth at a distance of 1.50 × 10^{7} m above the planetʹs surface and takes 8.65 hours for each revolution about the

Question

A satellite orbits the earth at a distance of 1.50 × 10^{7} m above the planetʹs surface and takes 8.65 hours for each revolution about the earth. The earthʹs radius is 6.38 × 10^{6} m. The acceleration of this satellite is closest to

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1 year 2021-08-20T17:32:47+00:00 1 Answers 44 views 0

a = 0.43 m/s²

Explanation:

First, we need to find the velocity of the satellite:

Velocity = V = Distance Covered/Time Taken

here,

Distance = 1 revolution = 2π(1.5 x 10⁷ m) = 9.43 x 10⁷ m

Time = (8.65 hours)(3600 s/1 hour) = 31140 s

Therefore,

V = (9.43 x 10⁷ m)/(31140 s)

V = 3028.26 m/s

Now, the acceleration of the satellite will be equal to the centripetal acceleration, with the center of circular motion as the center of earth:

a = V²/R

where,

R = 1.5 x 10⁷ m + 0.638 x 10⁷ m

R = 2.138 x 10⁷ m

Therefore,

a = (3028.26 m/s)²/(2.138 x 10⁷ m)

a = 0.43 m/s²