Question

The geostationary orbit of the earth is at a distance of about 36000 km from the earth's surface. Find the weight of a 120-kg equipment placed in a geostationary satellite. The radius of the earth is 6400 km.

Concept of Physics - 1 , HC VERMA , Chapter "The Force"

Answer 1

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.

Answer 2

HEY!!

______________________________

✔The geostationary orbit of the Earth is at a distance of about 36000 km.


✔Value of the acceleration due to the gravity above the surface of earth:- g'=Gm/(R+h)^2

▶At h = 36000 km, we have:-

g'=Gm/(36000+6400)^2

▶At surface, we have:-

g=Gm(6400)^2

▶▶g'/g= 6400×6400 / 42400×42400   

256 / 106×106= 0.0228

✔g'= 0.0227× 9.8= 0.223            

((Taking g=9.8 m/s2^2 at the surface of the earth))

✔For a 120 kg equipment placed in a geostationary satellite, its weight will be:-

▶mg' = 120 × 0.233

▶▶26.76 ≈ 27 N