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.

Answer 1

The body weight in the Geostationary Satellite is 27 N.  

Explanation:

Step 1:

Given  

Length of the Earth's Geostationary Orbit(h) = 36000 km.

Earth's radius(r) = 6400 km.

Total distance from the Geostationary Orbit to Earth's core = h + r

= 36000 + 6400

= 42400 km.

Step 2:

We realize that because of variations in gravity, the height or depth increases or reduces the magnitude of the Acceleration.  

On the surface of the earth,

Acceleration due to gravity $\begin{equation}g=\frac{G m}{r^{2}}$

Where

G = Gravitation Constant.

m = mass of the earth.  

and  

r = radius of the Earth.

$\begin{equation}g=\frac{\mathrm{G} \mathrm{m}}{6400^{2}}$

6400² × g = G m    .......... equation 1

Now, At height (h+ r),

The Acceleration due to gravity  is

$\begin{equation}\begin{aligned}&g^{\prime}=\frac{\mathrm{G} \mathrm{m}}{(\mathrm{h}+\mathrm{r})^{2}}\\&g^{\prime}=\frac{\mathrm{G} \mathrm{m}}{(42400)^{2}}\end{aligned}$

6400² × g = G m .......... from equation 1

$\begin{equation}\begin{aligned}g^{\prime} &=\frac{6400^{2} \times g}{(42400)^{2}} \\g^{\prime} &=\frac{40960000 \times g}{1797760000}\end{aligned}\end$

g' = 0.02278 × g

g = 9.8 m/s² (earth surface).

g' = 0.02278 × 9.8

Step 3:

Acceleration at height (h + r) due to gravity is therefore 0.2254 m / s2.

Weight of the equipment on the satellite = 120 kg.

Weight = Mass × Acceleration due to gravity at that place.

Weight = 120 × 0.2254

Weight = 27.048 N.

Weight ≈ 27 N.

The body weight in the Geostationary Satellite is therefore 27 N.  

Answer 2

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The answer will be 27 N