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At a certain instant, the earth, the moon, and a stationary 1450kg spacecraft lie at the...

At a certain instant, the earth, the moon, and a stationary 1450kg spacecraft lie at the vertices of an equilateral triangle whose sides are 3.84

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Answer #1

a) gravitational force =Gm1m2/r^2

mass of earth=5.97*10^24 kg

F = 6.673*10^-11*5.97*10^24*1450 / (3.84*10^5)^2

F = 3.92*10^6 N

b) gravitational force between the moon and the spacecraft =   6.673*10^-11*7.34*10^22*1450 / (3.84*10^5)^2

            = 48164N

angle between the force exerted by the moon and the force exterted by the earth is 60

So Fnet = sqrt(F1^2+F2^2 + 2F1F2cos a)

Fnet = sqrt ( (3.92*10^6)^2 + 48164^2 + 2*3.92*10^6*48164*cos 60 )

Fnet = 3.94*10^6N

c) minimum work needed = total energy of the two positions

   Work done = E(final)-E(initial)

W= GM/r (Mearth- Mmoon)

W= 6.67*10^-11*1450 / 3.84*10^5    * ( 5.97*10^24 - 7.34*10^22)

W=1.48*10^12J

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