At a certain instant, the earth, the moon, and a stationary 1010 kg spacecraft l
ID: 1494416 • Letter: A
Question
At a certain instant, the earth, the moon, and a stationary 1010 kg spacecraft lie at the vertices of an equilateral triangle whose sides are 3.84×105km in length.Part A Find the magnitude of the net gravitational force exerted on the spacecraft by the earth and moon. Express your answer to three significant figures.Part B 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. Express your answer to three significant figures. Part C 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. Express your answer to three significant figures.
Explanation / Answer
given that
mass of spacecraft m = 1010 kg
d = 3.84*10^5*10^3 m
mass of moon M1 = 7.35*10^22 kg
mass of earth M2 = 5.97*10^24 kg
we know that
gravitational force = G*M1*M2 / d^2
The force due to the moon is
Fm = G*M1*m/d^2
Fm = 6.674*10^(-11)* 7.35*10^22*1010 / (3.84*10^8)^2
Fm = 3359.94*10^(-5)
Fm = 0.0335 N
force due to earth
Fe = 6.674*10^(-11)* 5.97*10^24*1010 / (3.84*10^8)^2
Fe = 2729.10 *10^(-3)
Fe = 2.729 N
for the magnitude of net force decomposing each force into orthogonal parts, summing parallel vectors, then combining by Pythagorean theorem.
Let earth-to-spacecraft line as x-axis.so
F = sqrt[ (Fe+Fm*cos60)^2 + (Fm *sin60)^2 ]
F = sqrt [ (2.729 + 0.0335*0.5)^2 + (0.0335*0.86)^2 ]
F = sqrt [ 7.53 + 0.000830 ]
F = 2.744 N
the direction of net force
tan(theta) = ( (Fm*sin60) / (Fe + Fm*cos60))
tan(theta) = 0.028 / 2.74
theta= tan^-1(0.010)
theta= 0.57 degree
W = F*d
W = 2.744*3.84*10^8
W = 10.53*10^8 J
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