- #1
gcombina
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- 3
Warning: Infraction points awarded to gcombina for not using the homework template.
A satellite in orbit around the Earth has a period of one hour. An identical satellite is placed in an orbit having a radius that is nine times larger than that of the first satellite. What is the period of the second satellite?
(a) 27 h (c) 4 h (e) 0.04 h
(b) 3 h (d) 9 h** I truly don't know where to start.
I have seen (online) people using equation for force of gravity, and centripetal acceleration but they don't explain why
**Please do not use Keplen laws cause I haven't studied that
My attempt:
The period is the distance of the orbit which is 2piR
The first satellite moves because it has centripetal force which is Fc= mv^2/R
The first satellite also has acceleration which is ac = v^2/R
(a) 27 h (c) 4 h (e) 0.04 h
(b) 3 h (d) 9 h** I truly don't know where to start.
I have seen (online) people using equation for force of gravity, and centripetal acceleration but they don't explain why
**Please do not use Keplen laws cause I haven't studied that
My attempt:
The period is the distance of the orbit which is 2piR
The first satellite moves because it has centripetal force which is Fc= mv^2/R
The first satellite also has acceleration which is ac = v^2/R