Question
What linear speed must an Earth satellite have to be in a circular orbit at an altitude of 208 km above Earth's surface? (b) What is the period of revolution?
I got a to be 7786 m/s^2 but im not sure how to get b.
I got a to be 7786 m/s^2 but im not sure how to get b.
Answers
m/s is speed
m/s^2 is acceleration
G m Me/r^2 = m v^2/r
G Me/r = v^2
r = .208*10^6 + 6.38*10^6 = 6.59 * 10^6 meters
v^2 = 6.67*10^-11*5.98*10^24/6.59*10^6
v^2 = 6.05*10^7 = 60.5*10^6
v = 7780 m/s agree
now
circumference = 2 pi r
time = circumference/speed
m/s^2 is acceleration
G m Me/r^2 = m v^2/r
G Me/r = v^2
r = .208*10^6 + 6.38*10^6 = 6.59 * 10^6 meters
v^2 = 6.67*10^-11*5.98*10^24/6.59*10^6
v^2 = 6.05*10^7 = 60.5*10^6
v = 7780 m/s agree
now
circumference = 2 pi r
time = circumference/speed
is the time is seconds or minutes?