Question
Calculate a space shuttle territory 250 km above Earth's surface. Earth's mass is 6.0 * 10^24kg and its radius is 6.38 * 10^6 m (6380km)
Answers
Damon
r = 6380+250 = 6630 km = 6.63 * 10^6 m
g there = 9.81 (6.38/6.63)^2 = 9.08 m/s^2
if constant altitude g = centripetal a
v^2/r = 9.08 = v^2/(6.63*10^6)
so
v = 7.76*10^3 m/s
for orbit time
2 pi r = v t = 7.76*10^3 t
t = 5368 seconds
/3600 = 1.5 hours
g there = 9.81 (6.38/6.63)^2 = 9.08 m/s^2
if constant altitude g = centripetal a
v^2/r = 9.08 = v^2/(6.63*10^6)
so
v = 7.76*10^3 m/s
for orbit time
2 pi r = v t = 7.76*10^3 t
t = 5368 seconds
/3600 = 1.5 hours
Damon
By the way if you google "low altitude earth orbit" I think you will find lots on this.