At a certain instant, the earth, the moon, and a stationary 1060 kg spacecraft l

Question

At a certain instant, the earth, the moon, and a stationary 1060 kg spacecraft lie at the vertices of an equilateral triangle whose sides are 3.84×10^5 km in length.

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

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.

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.

Explanation / Answer

Gravitational attraction in newtons
F = G mm/r²
G = 6.67e-11 m³/kgs²
m and m are the masses of the two objects in kg
r is the distance in meters between their centres
(center of mass)
earth mass = Me = 5.97e24 kg
moon mass = Mm = 7.35e22 kg

m = 1060 kg

D = 3.84*10^8 m

Calculate the two forces separately.
To the moon,
Fm = (6.67e-11)(1060)(7.35e22) / (3.84e8)² = 0.03524 N
to the earth
Fe = (6.67e-11)(1060)(5.97e24) / (3.84e8)² = 2.86 N

now use vector arithmetic to add the two vectors. If you take the baseline (x axis) the line connecting spaceship to earth, then the sum is
Fx = Fe + Fm(cos60) = 2.87762 N
Fy = Fm(sin60) = 0.0305 N
total force is F = (Fx² + Fy²) = 2.87778 N
angle is arctan (Fy/Fx) = 0.607 degree

c) The minimum work is equal to the gravitational potential energy at infinity minus that of the spaceship at its location. The gravitational PE at infinity is zero.

PEe = -G*Me*m/D
PEm = -G*Mm*m/D
Total PE = [-G*m/D]*(Me + Mm)

Work = [G*m/D]*(Me + Mm)

Work = 1.11x10^9 J

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