Equate the gravitational force,
GmM/R^2, to the centripetal force, m V^2/R
M is the Earth's mass and G is the universal constant, which you need to look up. Satellite mass m cancels out.
Substitute 2 pi R/(139*60) for the velocity, V (in m/s).
Solve for the remaining variable R.
The satellite altitude is H = R - Rearth
An artificial satellite circling the Earth completes each orbit in 139 minutes. (The radius of the Earth is 6.38 106 m. The mass of the Earth is 5.98 1024 kg.)
(a) Find the altitude of the satellite.
2 answers
The orbital period of a satellite derives from
T = 2(Pi)sqrt(r^3/µ) where T = the period in seconds
Pi = 3.14
r = the orbital radius in feet and
µ = the earth's gravitational constant = 1.407974x10^16 ft^3/sec^2.
139(60)= 2(3.14)sqrt(r^3/1.407974x10^16)
Solving for r = 5523.6 miles making the altitude 1560.6 miles = 2511km.
T = 2(Pi)sqrt(r^3/µ) where T = the period in seconds
Pi = 3.14
r = the orbital radius in feet and
µ = the earth's gravitational constant = 1.407974x10^16 ft^3/sec^2.
139(60)= 2(3.14)sqrt(r^3/1.407974x10^16)
Solving for r = 5523.6 miles making the altitude 1560.6 miles = 2511km.
0 answers