orbits (1 Viewer)

[shkspeare]

"Draw a diagram of a satellite orbiting the earth, showing all forces acting on the satellite"

mm wot do i draw

just a satellite in a path of a circle about the earth, with an arrow pointing from the satellite towards the earth?

and this arrow represents the earth's gravitational pull or something??

 

[...]

all forces...

in this case, the only force acting on the satellite is gravitational pull..so u draw a line from the satellite to the centre of the earth

 

[shkspeare]

woo ok i was right.. thx

edit : mm now heres another q >.<

ok say you're given the radius of the earth's orbit is 150 x 10⁶km

how would u find the time for mercury in earth years to orbit the sun if the orbit radius 58.5 x 10⁶km

 

[...]

does using Kepler's third law work?

providing u know the mass of the planets..err..yea

 

[Xayma]

Originally posted by ...
does using Kepler's third law work?

providing u know the mass of the planets..err..yea

Yeah you use Kepler's third law, T^2/R^3=k
So for the solar system k=1/((150*10^6)^3) yrs^2km^-3
Rearranging T=Sqrt(kR^3)
T=Sqrt([1/{(150*10^6)^3}][{58.5*10^6}^3]
T=.244 Earth Years.

 

[shkspeare]

woo thxthx

 

[metalicarulez]

hmmm on the topic of orbits

can someone explain to me why geostationary orbits have a period of 24hrs MINUS 3minutes 56 seconds ?!?!

i know it may sound dumb but i dont get it !

 

[Constip8edSkunk]

each earth day is actually 23 hours, 56 minutes and 4 seconds ...but is generally rounded off to 24 hrs.

 

[Dash]

Originally posted by metalicarulez
can someone explain to me why geostationary orbits have a period of 24hrs MINUS 3minutes 56 seconds ?!?!

In short: They are put at a certain altitude I believe that allows them to have a period of 23 hours 56 minutes and 4 seconds.
Scientists use kepler's law of periods to determine it