The geo stationary orbit of the earth is at a distance of about 36000 km from the earth surface. Then the weight of 120 kg equipment placed in geo stationary satellite is(radius of earth is 6400 km,mass of the earth is
Answers
We have m=120 kg, r=6400+36000 km =42400 km =4.24x 107 m,
Time period T=24 h = 24x3600 s=86400 s
ω= 2π/T= 2π/86400 rad/s
Now weight = 120x(2π/86400)²x 4.24x 107 N
= 26.88 N ≈ 27 N
Hello Dear.
Given ⇒
Distance of the Geostationary Orbit from the Earth's Surface(h) = 36000 km.
Radius of the Earth(r) = 6400 km.
Total Distance between the Geostationary Orbit and the center of the Earth = h + r
= 36000 + 6400
= 42400 km.
We know, as the Height or the depth increases or decreases the value of the Acceleration due to gravity changes.
At Earth Surface,
Acceleration due to gravity = G m/r²
where, G = Gravitation Constant.
m = mass of the earth. and r = radius of the Earth.
∴ g = G m/(6400)²
G m = g × (6400)² ---------eq(i)
Now, At height (h+ r), the Acceleration due to gravity = G m/(h + r)²
g' = G m/(42400)²
∴ g' = g × (6400)² ÷ (42400)² [From eq(i)]
⇒ g' = g × 0.023
∵ g = 9.8 m/s².
∴ g' = 9.8 × 0.023
⇒ g' = 0.2254 m/s²
Hence, the Acceleration due to Gravity at the height (h + r) is 0.2254 m/s².
Mass of the Equipment placed in the Satellite = 120 kg.
∵ Weight = Mass × Acceleration due to gravity at that place.
∴ Weight = 120 × 0.2254
⇒ Weight = 27.048 N.
⇒ Weight ≈ 27 N.
Hence, the weight of the body placed in the Geostationary Satellite is 27 N.
Hope it helps.