Calculating Acceleration Due to Gravity at Altitudes Far from the Earth's Surface

Understanding the Problem

Understanding the acceleration due to gravity at various distances from the Earth's surface is essential in

astronomy, space exploration, and even in daily applications like air traffic control and GPS navigation. When considering the acceleration due to gravity at a height of 3600 km above the Earth's surface, we need to account for the Earth's mass and the distance from the center of the Earth to the point of interest.

The Universal Gravitational Constant and Its Impact

The universal gravitational constant, (G) (6.674 x 10-11 m3/kg2·s2), is a key parameter in the formula for gravitational force. It represents the strength of the gravitational force between two objects with a given mass. The gravitational force is directly proportional to the product of the masses of the two objects and inversely proportional to the square of the distance between their centers.

Formulating the Equation

The formula for the acceleration due to gravity ((g_{h})) at a height (h) above the Earth's surface is:

[g_{h} frac{G cdot M}{r^2}]

Where:

(G) is the universal gravitational constant. (M) is the mass of the Earth (6 x 1024 kg). (r) is the distance from the center of the Earth to the point where we are calculating (g).

Converting Heights to Meters

The radius of the Earth, (R), is 6400 km (6.4 x 106 m). The height above the Earth's surface is 3600 km (3.6 x 106 m).

The distance from the center of the Earth to the point at height (h) is:

[r R h 6.4 times 10^6 m 3.6 times 10^6 m 10 times 10^6 m 10^7 m]

Solving for Gravitational Acceleration

Substituting the values into the formula:

[g_{h} frac{6.674 times 10^{-11} cdot 6 times 10^{24}}{(10^7)^2} frac{4.0044 times 10^{14}}{10^{14}} 4.0044 m/s^2]

Thus, the acceleration due to gravity at a height of 3600 km above the Earth's surface is approximately

[boxed{4.00 m/s^2}]

This result highlights the significant reduction in gravitational acceleration at higher altitudes due to the inverse square relationship with distance.