Respuesta :
Explanation:
It is known that the value of free fall acceleration is g = 6.44 [tex]m/s^{2}[/tex]. And, the value of radius of Earth is [tex]6.4 \times 10^{6} m[/tex].
Let us assume that height of the satellite is h.
It is known that,
g = [tex]\frac{GM}{r^{2}}[/tex]
where, r = (R + h)
Hence,
r = [tex]\sqrt{\frac{GM}{g}}[/tex]
r = [tex]\sqrt{\frac{6.67 \times 10^{-11} \times 5.98 \times 10^{24}}{6.44 m/s}}[/tex]
= [tex]7.9 \times 10^{6} m[/tex]
Now, formula for height of satellite above the Earth's surface is as follows.
h = r - R
= [tex]7.9 \times 10^{6} - 6.4 \times 10^{6}[/tex]
= [tex]1.5 \times 10^{6} m[/tex]
Thus, we can conclude that the satellite is [tex]1.5 \times 10^{6} m[/tex] high above Earth's surface.
The height of the satellite moving in a circular orbit above the Earth's surface is the satellite is; 1.5 × 10⁶ m
What is the height of gravitational fall?
The formula to find the height of the satellite above the earth's surface is; h = [√(GM/g)] - R
where;
G is gravitational constant = 6.67 × 10⁻¹¹ N.m²/kg²
M is mass of earth = 5.98 × 10²⁴ kg
g is free fall acceleration = 6.44 m/s²
R is radius of earth 6400000 m
Thus;
h = [√(6.67 × 10⁻¹¹ × 5.98 × 10²⁴/6.44)] - 6400000
h = 1.5 × 10⁶ m
Read more about height of gravitational fall at; https://brainly.com/question/14460830
