Question

A geosynchronous satellite is a satellite whose orbital period matches the rotation of the Earth. Calculate the height above the surface of the Earth which a satellite must have in order to be in a geosynchronous orbit. (Please enter your answer in units of kilometers)

Answer 1

35.869 km.

Step-by-step explanation:

The orbital period (T) is the time a body takes to make a complete lapse. Earth takes 24h, since the satellite must have the same orbital period, it's as well 24h.

T =24hrs = 86400 s

We will derive the equation to get the height using Newton laws.

Let's define the varaibles: say h is height of the satellite from the surface of the Earth, R is Earth radius, r is h+r or total distance from centre of Earth to satellite, G is gravitational constant, a is acceleration, M is Earth mass and m is satellite mass.

First Newton's Law says the force the Earth exerts in the satellite is

F = G* (M*m)/r^2

Second Newton's Law gives the force must be

F=m*a

Combining both we get

G*(M*m)/r^2 =m*a

We substitute a= v^2 /r

G*(M*m)/r^2=m*(v^2 /r)

The m and one r cancells. We move the r left multiplying the other side of the equation. We get

G*M = r*v^2

Here we know the orbital speed is v=2πr/T . So we substitute

G*M = r*(2πr/T)^2

And rearrange

r^3 = (G*M*T^2)/ 4π^2

Solve for r

r= [(G*M*T^2)/ 4π^2]^(1/3)

Note the 1/3 exponent is in fact a cube root.

We know that r= h + R

So we solve for h

h= r - R

And substitute the derived equation for r

h = [(G*M*T^2)/ 4π^2]^(1/3) - R

Now we plug in the values, because we know them.

G = 6.674×10^(-11) (m^3 /kg s^2)

M = 5,972 × 10^24 (kg)

T = 86.400 (s)

R = 6.371.000 (m)

Finally, our result should be

35.869.065 m

Which is 35.869 km.