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lubo
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Homework Statement
An Earth satellite is to be put into orbit at a height of 2 Mm above the Earth's surface. Determine the velocity it must have. Make the following assumptions: (a) the orbit is circular; (b) the Earth is a sphere of radius 6•37 Mm; (c) gravity decreases as an inverse square law.
(MY main question is to get an idea of how to find the height above the Earth's surface of the satellite, when you are only given the Velocity of a satellite in orbit).
Homework Equations
g' (acceleration due to gravity at a distance R to the centre of the earth)=g*r earth^2/R orbit^2
a=V^2/r orbit
The Attempt at a Solution
g'=?
g=9.81
r earth=6.37Mm
R orbit=6.37 + 2(height)
g' = 9.81*6.37^2/(2+6.37^2)
g'=5.682m/s^2
a=v^2/R therefore v^2=8.37*10^6 * 5.682 Therefore V=6896 m/s
However if I try to work back using V only V=6896 m/s and find the height 2Mm, I can only use g=9.81. Therefore I cannot find g' to allow me to find Ro (re + h). Then I could take Ro from re and I would have the height.
i.e. a=v^2/R = 6896^2/r Earth as I don't have R orbit = 6896^2/6.37*10^6 = 7.465m/s^2
If I try g' =mg then I do not have a mass.
If I try a=rw^2 and find w then I can only use re and a=9.81. Therefore I cannot find the g'= 5.682m/s^2
If I substitute g' with rw^2 and get Ro^3=g*r^2/w^2 I am still only using r of Earth and w from r and not a time period or R orbit
I hope all this makes sense, Thanks for any help.