A satellite, of mass m=1.32E3 kg, falls from rest directly tothe earth from a di
ID: 1744975 • Letter: A
Question
A satellite, of mass m=1.32E3 kg, falls from rest directly tothe earth from a distance of R=3.00 earth radii from the center ofthe earth. (Mass of earth: 5.976*1024 kg. Radius ofearth: 6.37*106 m. G: 6.67*10 -11N*m2/kg2.)a) What is the speed of the satellite just before it hitsearth?
b) What would the speed have been if the acceleration due togravity was a constant g=9.8 m/s2?
I will give top score for detail explained answers... Thank you,
A satellite, of mass m=1.32E3 kg, falls from rest directly tothe earth from a distance of R=3.00 earth radii from the center ofthe earth. (Mass of earth: 5.976*1024 kg. Radius ofearth: 6.37*106 m. G: 6.67*10 -11N*m2/kg2.)
a) What is the speed of the satellite just before it hitsearth?
b) What would the speed have been if the acceleration due togravity was a constant g=9.8 m/s2?
I will give top score for detail explained answers... Thank you,
Explanation / Answer
The final mechanical energy(when satelite reaches theEarth) is equal to the initial mechanical energy. Kf+ Uf = Ki + Ui The falls initially from rest, so that Ki =0 Therefore Kf = Ui - Uf 0.5mvf2 = Ui - Uf 0.5mvf2= (GMm / Re ) - (GMm /3Re ) vf2 = 2 ( (GM / Re ) -(GM / 3Re ) ) = (4/3)(GM / Re ) Where G = 6.67 x10-11m3 / kg.s2 M= 5.98 x 1024 kg Re = 6.37 x 106 m Substitute values. 0.5mvf2= (GMm / Re ) - (GMm /3Re ) vf2 = 2 ( (GM / Re ) -(GM / 3Re ) ) = (4/3)(GM / Re ) Where G = 6.67 x10-11m3 / kg.s2 M= 5.98 x 1024 kg Re = 6.37 x 106 m Substitute values.Related Questions
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