Question

At what distance from Earth does the force of gravity exerted by Earth on the coasting spacecraft cancel the force of gravity exerted by the Moon on the spacecraft? Use the Moon's orbit semimajor axis as a distance between planets.

Answer 1

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

Answer 2

The distance from Earth where the force of gravity exerted by Earth on the coasting spacecraft cancel the force of gravity exerted by the Moon on the spacecraft will be 3.47 * 10⁸ m

The distance from Earth where the force of gravity exerted by Earth on the coasting spacecraft cancel the force of gravity exerted by the Moon on the spacecraft will be at point between the distance of the Earth and the Moon.

The distance between the earth and the moon, d = 3.85 x 10⁸ m.

Let the distance from Earth to the coasting spacecraft be r

distance between the moon and spacecraft = d - r

The mass of the earth is 81.3 times greater than that of the moon.

Let mass of the moon be Mm

mass of the Earth, Me = 81.3 * Mm

mass of spacecraft = M

From Newton's law of gravitation, F = GM₁M₂/d²

Force of attraction between earth and spacecraft Fe = GMeM/r²

Force of attraction between moon and spacecraft Fm = GMmM/(d-r)²

Since the forces cancel each other, they must be equal

Equating the two forces and solving for r:

Fe = Fm

GMeM/r²  = GMmM/(d-r)²

Me/r² = Mm/(d-r)²

r² = (d-r)²* Me/Mm

take square root of both sides

r = (d - r)√Me/Mm

r + r(√Me/Mm) = d(√Me/Mm)

r (1 + √Me/Mm) = d(√Me/Mm)

r = d(√Me/Mm) / (1 + √Me/Mm)

substituting the values

r = 3.85 * 10⁸ (√81.3 * Mm/Mm) / (1 + √81.3 * Mm/Mm)

r = 3.47 * 10⁸ m

Therefore, the distance from Earth where the force of gravity exerted by Earth on the coasting spacecraft cancel the force of gravity exerted by the Moon on the spacecraft will be 3.47 * 10⁸ m

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