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Marrrta [24]
3 years ago
15

At what distance from Earth does the force of gravity exerted by Earth on the coasting spacecraft cancel the force of gravity ex

erted by the Moon on the spacecraft? Use the Moon's orbit semimajor axis as a distance between planets.
Physics
1 answer:
telo118 [61]3 years ago
4 0

Answer:

346 * 10⁶ m

Explanation:

The force of gravity of the earth that will cancel the the force of gravity exerted by the moon will be equal to each other

Let F_{e} be the force of gravity exerted by the earth

and let F_{m} be the force of gravity exerted by the moon

According to Newton's law of universal gravitation, the force of attraction between two different masses, m₁ and m₂ separated by a distance, d,  is given by:

F = \frac{Gm_{1} m_{2} }{d^{2} }

Mass of the earth, m_{e} = 5.97 * 10^{24} kg

Mass of the moon, m_{m} = 7.348 * 10^{22} kg

Mass of the satellite, m_{s} = ?

F_{e}  = \frac{G*5.97 * 10^{24} M }{d^{2} }...............................(1)

The earth and the moon are separated by a distance, 3.844 * 10⁸ m

F_{m}  = \frac{G*7.348 * 10^{22} M }{(3.844 * 10^{8} - d) ^{2} }............................(2)

Equating equations (1) and (2)

\frac{5.97 * 10^{24} }{d^{2} } = \frac{7.348 * 10^{24} }{(3.844* 10^{8} -d)^{2} }

(5.97 * 10^{24})(14.78 * 10^{16}  -7.688*10^{8}d + d^{2}) = 7.348 * 10^{24} d^{2} \\88.24*10^{40} - 45.9 * 10^{32}d +  5.97 * 10^{24}d^{2} =  7.348 * 10^{24} d^{2}\\ 1.378 * 10^{24}d^{2} + 45.9 * 10^{32}d + 88.24*10^{40} = 0\\

Factorising out 10^{24}

1.378d^{2} + 45.9 * 10^{8}d + 88.24*10^{16} = 0

Solving for d in the quadratic equation  above:

d = 346 * 10⁶ m

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