Answer:
a) The altitude of the satellite is approximately 2129 kilometers.
b) The gravitational acceleration at the location of the satellite is 5.517 meters per square second.
Step-by-step explanation:
a) At first we assume that Earth is a sphere with a uniform distributed mass and the satellite rotates on a circular orbit at constant speed. From Newton's Law of Gravitation and definition of uniform circular motion, we get the following identity:
(Eq. 1)
Where:
- Gravitational constant, measured in newton-square meters per square kilograms.
- Mass of the satellite, measured in kilograms.
- Mass of the Earth, measured in kilograms.
- Angular speed of the satellite, measured in radians per second.
- Radius of the Earth, measured in meters.
- Height of the satellite above surface, measured in meters.
Then, we simplify the formula and clear the height above the surface:


![R_(E)+h = \sqrt[3]{(G\cdot M)/(\omega^(2)) }](https://img.qammunity.org/2021/formulas/physics/college/ayew4u9lv2ixd9izhkgpg1y48r5te2ks5i.png)
![h = \sqrt[3]{(G\cdot M)/(\omega^(2)) }-R_(E)](https://img.qammunity.org/2021/formulas/physics/college/fpod94h8fwj0lbgcqjpzalh3sbqi2jyegj.png)
From Rotation physics, we know that angular speed is equal to:
(Eq. 2)
And (Eq. 1) is now expanded:
(Eq. 3)
If we know that
,
,
and
, then the altitude of the satellite is:
![h = \sqrt[3]{(\left(6.674* 10^(-11)\,(N\cdot m^(2))/(kg^(2)) \right)\cdot (5.972* 10^(24)\,kg)\cdot (7800\,s)^(2))/(4\pi^(2)) }-6.371* 10^(6)\,m](https://img.qammunity.org/2021/formulas/physics/college/txazscjrrycnawsjsnoeenr9sawaxmtvyh.png)


The altitude of the satellite is approximately 2129 kilometers.
b) The value for the gravitational acceleration of the satellite (
), measured in meters per square second, is derived from the Newton's Law of Gravitation, that is:
(Eq. 4)
If we know that
,
,
and
, then the value of the gravitational acceleration at the location of the satellite is:


The gravitational acceleration at the location of the satellite is 5.517 meters per square second.