Answer:
35.869 km.
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.
