Question

What is the ratio of Earth's gravitational force on a satellite when it is on the ground to the gravitational force exerted when the

satellite is orbiting at an altitude of 484 km?

Answer 1

The ratio g :
`g_(h)` = 3:1

Step-by-step 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,
`g_(h)` = GM/(R+h)²

Therefore,

g :
`g_(h)` = (R+h)² : R²

where R is the radius of the Earth, R = 6400000 m

Substituting in the above equation

g :
`g_(h)` = (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 :
`g_(h)` = 3:1