Gravitational pull of Earth and Moon on spaceship

[mckayl2]

Homework Statement

A spaceship is launched and starts moving directly towards the Moon. At what distance from the Earth will the pull of the Moon on the spaceship exceed the pull of the Earth? Ignore the effect of the Sun in this calculation.

Homework Equations

F = Gm¹m²/r²
F = ma = mg
g=9.80m/s²
Mass of Earth = 5.86 x 10²⁴
Mass of moon = 7.35 x 10²²

The Attempt at a Solution

For the earth:
Using F=mg=Gm¹m²/r², radius of Earth to Moon is calculated to be 2.24 x 10⁶km

For the moon:
Using mg = Gm¹m²/r², the gravitational pull of the moon is calculated to be 1.618m/s²

I'm not sure where to go from here, any help is much appreciated :)

 

[Doc Al]

Hint: If the spaceship is a distance X from the center of the earth, how far is it from the center of the moon? If you call the mass of the spaceship m, what's the Earth's gravitational force on the ship in terms of X? What's the moon's gravitational force? Set up an equation and solve for X.

 

[tomcue]

To determine at what distance from the Earth the gravitational pull of the Moon on the spaceship will exceed the pull of the Earth, we can use the concept of gravitational force. The gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

In this case, the mass of the spaceship is negligible compared to the masses of the Earth and Moon, so we can assume that the spaceship will be affected by the gravitational pull of both objects. As the spaceship moves closer to the Moon, the distance between the Earth and the spaceship will increase, causing the gravitational force of the Earth on the spaceship to decrease.

To find the distance at which the gravitational pull of the Moon exceeds that of the Earth, we can set the two forces equal to each other and solve for the distance.

F(earth) = F(moon)

Gm(earth)m(spaceship)/r(earth)^2 = Gm(moon)m(spaceship)/r(moon)^2

Solving for r(moon), we get:
r(moon) = √(r(earth)^2m(earth)/m(moon))

Plugging in the values given in the problem, we get:
r(moon) = √(2.24 x 10^6 km)^2(5.86 x 10^24 kg)/(7.35 x 10^22 kg) = 2.5 x 10^5 km

Therefore, at a distance of 2.5 x 10^5 km from the Earth, the gravitational pull of the Moon on the spaceship will exceed that of the Earth. This is approximately 4 times the distance between the Earth and the Moon, which is 3.84 x 10^5 km.

It's important to note that this calculation ignores the effect of the Sun on the spaceship's trajectory. In reality, the gravitational pull of the Sun would also affect the spaceship's path towards the Moon. However, for the purposes of this calculation, we can assume that the Sun's effect is negligible.