Steve
answered
for a circle, C=2pi*r
r = 3960+104 = 4064
C = 25534
speed = 25534mi/10.5day = 2432mi/day
tchrwill
answered
= 25,262 feet per second.
The orbital period is
T = 2(3.14)sqrt[22,061,952^3/1.407974x10^16] = 5487 seconds or 91.45 minutes.
tchrwill
answered
The orbital radius is 3960 + 2.1(104) = 4178.4 miles or 22,061,952 feet.
The alleged time to complete one orbit is 10.5(24)3600 = 907,200 seconds making the derived orbital velocity
Vc = 152.7fps.
Unfortunately, the real orbital velocity required to remain in a circular orbit derives from Vc=sqrt(µ/r)=
where µ = the earth's gravitational constant and r = the orbital radius in feet =
sqrt[(1.407974x10^16)/(3960+218.4)5280]
= 25,262 feet per second.
The orbital period is
T = 2(Pi)sqrt[r^3/µ]=
2(3.14)sqrt[22,061,952^3/1.407974x10^16] = 5487 seconds or 91.45 minutes.
Explain Bot
answered
Velocity = Distance / Time
First, let's convert the distance of the satellite from miles to the same unit as the radius of the Earth, which is miles:
Distance = 2.1 X 10^4 miles + 3960 miles (radius of the Earth)
Distance = 2.1396 X 10^4 miles
Now, let's convert the time taken for one revolution around Earth from days to hours, and then to seconds:
Time = 10.5 days x 24 hours/day x 60 minutes/hour x 60 seconds/minute
Time = 907200 seconds
Now we can calculate the velocity:
Velocity = Distance / Time
Velocity = (2.1396 X 10^4 miles) / 907200 seconds
Velocity ≈ 23.59 miles/second
Therefore, the linear speed (velocity) of the satellite is approximately 23.59 miles per second.
