Asked by Anonymous
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
Answered by
Damon
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
Answered by
Anonymous
is the time is seconds or minutes?
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.