Question

An earth satellite makes a complete circuit around the earth in 90 minutes. If the orbit is circular, calculate the height of the satellite above the earth.​

Answer 1

Answer:
`6.654*10^6 m`

Step-by-step explanation:

Equation Used:

`(r^3)/(T^2)=(GM)/(4pi^2)`

• r = The Orbital Radius
• T = The Period It Takes the Satellite to Orbit Earth
• G = Universal Gravitational Constant
• M = Mass of the Earth

Given:

• G = 6.67*10^-11
  `(Nm^2)/(kg^2)`
• M = 5.98*10^24 kg
• T = 90 Minutes

Steps:

1. Convert 90 minutes into seconds:

`s = (90 min)((60 s)/(1 min) ) = 5400 s`

2. Isolate
`r^3` in the
`(r^3)/(T^2) = (GM)/(4pi^2)` equation:

`r^3 = (GMT^2)/(4pi^2)`

3. Plug in all values from "Given" and solve
`r^3 = ((6.67*10^-^1^1(Nm^2)/(kg^2))(5.98*10^2^4kg)(5400s)^2 )/(4pi^2) \\= ((6.67*10^-^1^1(Nm^2)/(kg^2))(5.98*10^2^4kg)(2.916*10^7s^2) )/(4(9.870))\\= ((6.67*10^-^1^1(Nm^2)/(kg^2))(5.98*10^2^4kg)(2.916*10^7s^2) )/(39.478)\\\\= (1.163*10^2^2m^3 )/(39.478)\\\\= 2.946*10^2^0m^3\\\\r = \sqrt[3]{ 2.946*10^2^0m^3} \\= 6.654*10^6 m`:

Answer 2

17,500~ Mph

Step-by-step explanation:

for a satalite to orbit without assist(RCS Systems or light rocket engines), you need to aim for an altitude of 99~ to 1250~ Miles(160~ to 200~ km) above sea level, and the escape velocity for earth is 25,000 Mph(40233.6 Km/h) and to get to a hieght with the average rocket, you need to go around 28,000 Mph(45061.632 Km/h) to reach orbital velocity. then slow down at around 91 Miles from 19,000 Mph to 17,500 mph to reach stable orbit