Question

At a certain instant, the earth, the moon, and a stationary 1250 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.

Answer 1

Answer:

4.825 N

Explanation:

mass of earth. Me  = 6 x 10^24 kg

mass of moon, Mm = 7.34 x 10^22 kg

Mass of satellite, Ms = 1250 kg

length of side of triangle, a = 3.84 x 10^5 km = 3.84 x 10^8 m

A.

force on space craft by the earth

$F_{1}=\frac{GM_{e}M_{s}}{a^{2}}$

By substituting the values

$F_{1}=\frac{6.67\times 10^{-11}\times 6\times 10^{24}\times 1250}{\left (3.84\times 10^_{8}  \right )^{2}}$

F1 = 3.39 N

force on space craft by the moon

$F_{1}=\frac{GM_{m}M_{s}}{a^{2}}$

By substituting the values

$F_{1}=\frac{6.67\times 10^{-11}\times 7.34\times 10^{22}\times 1250}{\left (3.84\times 10^_{8}  \right )^{2}}$

F2 = 0.042 N

These two forces act at 60° with each other

$F = \sqrt{F_{1}^{2}+F_{2}^{2}+2F_{1}F_{2}Cos60}$

$F = \sqrt{3.39^{2}+0.042^{2}+2\times 3.39\times 0.042\times Cos60}$

F = 4.825 N

Answer 2

The magnitude of the net gravitational force exerted on the spacecraft by the earth and moon is F= 4.825 N

Calculations and Parameters:

A. Force on space craft by the earth

Using Newton's gravitational law,

F= mmEG/a^2 + mmMG/a^2

Where

• mass of earth. Me  = 6 x 10^24 kg
• mass of moon, Mm = 7.34 x 10^22 kg
• Mass of satellite, Ms = 1250 kg
• length of side of triangle, a = 3.84 x 10^5 km = 3.84 x 10^8

We would input the values:

6.67 x 10^-11 x 6 x 10^24 x 1250/(3.84 x 10^8)

=> 3.39 N

B. Force on space craft by the moon

F1= 6.67 x 10^-11 x 6 x 10^22 x 1250/(3.84 x 10^8)^2

=> F2 = 0.042 N

Hence, the forces at 60° with each other is

=> F= 4.825 N

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