Respuesta :
Answer:
The ratio g : [tex]g_{h}[/tex] = 3:1
Explanation:
Given that,
Height of the orbiting satellite from the ground, h = 484 km
h = 484000 m
The gravitational force of Earth is proportional to acceleration at the location.
The acceleration at the surface is given by relation
g = GM/R²
Similarly, acceleration at height, [tex]g_{h}[/tex] = GM/(R+h)²
Therefore,
g : [tex]g_{h}[/tex] = (R+h)² : R²
where R is the radius of the Earth, R = 6400000 m
Substituting in the above equation
g : [tex]g_{h}[/tex] = (6400000 +484000)² : 6400000²
= 12.63 : 4.096
Approximating to the nearest real values
= 3 : 1
Thus, the ratio of the Earth's gravitational force on the satellite when it is on ground to the gravitational force at an altitude 484 is g : [tex]g_{h}[/tex] = 3:1
