Physics problem on variable gravitation field !!!

okay so it was like that I was learning the concept of centre of gravitation n I got to know that after it hokds good only if there is variable gravitational field unless its just like centre of mass n my teacher gave me example that even on earth after a certain height the gravitation starts varying... can somebody calculate me the exact distance from where it comes in real effect for more practical clearification about how different machines like sattelites,rockets are designed to make ........

The Universal Gravitation Force = (G * M1 * M2) ÷ r^2

G = 6.67 * 10^-11
M1 = mass of object #1
M2 = mass of object #2
r = distance from center of mass of Object #1 to center of mass of Object #2

The Universal Gravitation Force = Weight of M2
Weight of M2 = M2 * g

On the Earth:
M1 = mass of Earth = 5.98 * 10^24 kg
M2 = Your mass
r = 6.38 * 10^6 meters


The Universal Gravitation Force = 6.67 * 10^-11 * 5.98 * 10^24 * (Your mass) ÷ (6.38 * 10^6) ^2

The Universal Gravitation Force = (Your mass) * 9.8 m/s^2

Do you recognize the expression, (Your mass) * 9.8 m/s^2??

9.8 m/s^2 = acceleration due to gravity


acceleration = 6.67 * 10^-11 * 5.98 * 10^24 ÷ r ^2 8


If the acceleration due to gravity to decrease by 1%:
g = 0.99 * 9.8 = 9.702

9.702 = 6.67 * 10^-11 * 5.98 * 10^24 ÷ r^2


r^2 = 6.67 * 10^-11 * 5.98 * 10^24 ÷ 9.702
r = (6.67 * 10^-11 * 5.98 * 10^24 ÷ 9.702)^0.5
r = 6.41* 10^6 meters

Height above surface = 6.41 * 10^6 – 6.38 * 10^6 = 3 * 10^4 meters = 30 km
At a height of 38 km above the surface of the earth, the acceleration due to gravity is 1% less than at the surface.

At a height of 38 km above the surface of the earth, the force due to gravity is 1% less than at the surface.

At a height of 38 km above the surface of the earth, the weight of an object is 1% less than at the surface.