Question

find the gravitional field intensity at a height of 3600km from the surface of the earth .The radius of the earth is 6400km add its mass is 6*10^24kg​

Answer 1

Answer:

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.

Explanation:

Answer 2

Height from centre of earth

=6400+3600=1000 km

or 1000×1000m

$10 {}^{6}m \\ \huge{we \: knw \: tht}$

$g = \frac{ \bold{g}m}{r {}^{2} }$

$= \frac{6.67 \times 10 {}^{ - 11} \times mass \: of \: earth}{(10 {}^{6} ){}^{2} }$

$= \frac{6.67 \times 10 {}^{ - 11} \times 6 \times 10 {}^{24} }{10 {}^{12} }$

$\frac{40.02 \times 10 {}^{13} }{10 {}^{12} } \\ = \it{40.02 \times 10 = 400.2ms {}^{ - 2} }$