At a certain instant, the earth, the moon, and a stationary 1070 kg spacecraft l

Question

At a certain instant, the earth, the moon, and a stationary 1070 kg spacecraft lie at the vertices of an equilateral triangle whose sides are 3.84×105km in length.

Find the magnitude of the net gravitational force exerted on the spacecraft by the earth and moon.

Find the direction of the net gravitational force exerted on the spacecraft by the earth and moon. State the direction as an angle measured from a line connecting the earth and the spacecraft.

What is the minimum amount of work that you would have to do to move the spacecraft to a point far from the earth and moon? You can ignore any gravitational effects due to the other planets or the sun

Explanation / Answer

a)

me=5.98*10^24 kg, mm= 3.35*10^22kg, ms =1070kg,r=3.84*10^8 m

Force on spacecraft by earth Fse= (G*mm*mm)/r^2 = (6.67*10^-11*5.98*10^24*1070)/(3.84*10^8)^2 = 2.89N   directed along x axis

Fsex =2.89 N

Fsey =0.0 N

Force on spacecraft by earth Fsm= (G*mm*mm)/r^2 = (6.67*10^-11*3.35*10^22*1070)/(3.84*10^8)^2 = 0.016N

Force on spacecraft by earth Fsm along x axis = Fsmx = Fsm*cos60 = 0.016cos60 = 0.008 N

Force on spacecraft by earth Fsm along x axis = Fsmx = Fsm*cos60 = 0.016sin60 = 0.0138 N

Fsnetx = Fsex + Fsmx = 2.89+0.08 = 2.97 N

Fsnety = Fsey + Fsmy = 0.00+0.08 = 0.0138 N

Fsnet = sqrt(Fsnetx^2 + Fsnety^2) = sqrt(2.97^2+0.0138^2) = 2.97 N

b)

direction= =tan^-1(Fsnety/Fsnetx) = tan^-1(0.0138/2.97)= 0.27 deg

c)

W = MEf – MEi

W = [0] – [-Gmeme/r-Gmmme/r]

W = (Gme/r)*(me+ms)

W= (6.67*10^-11*1070)/(3.84*10^8)*(5.98*10^24+3.35*10^22)

W=1.12*10^9 J

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