At what distance from the Earth will a spacecraft traveling directly from the Earth to the Moon experience zero net force because the Earth and Moon pull with equal and opposite forces?

Answer 1

First of all, once you are in a spaceship, you won't feel gravitation, because you are in "free fall".

Anyway, to calculate this, consider that the Sun has 330,000 times the mass of Earth. Also consider the inverse-square relation. Therefore, for the two forces to cancel, the object would have to be square-root(330,000) or about 574 times farther from the Sun, than from the Earth - a distance of about 260,000 kilometers from Earth.

First of all, once you are in a spaceship, you won't feel gravitation, because you are in "free fall".

Anyway, to calculate this, consider that the Sun has 330,000 times the mass of Earth. Also consider the inverse-square relation. Therefore, for the two forces to cancel, the object would have to be square-root(330,000) or about 574 times farther from the Sun, than from the Earth - a distance of about 260,000 kilometers from Earth.

First of all, once you are in a spaceship, you won't feel gravitation, because you are in "free fall".

Anyway, to calculate this, consider that the Sun has 330,000 times the mass of Earth. Also consider the inverse-square relation. Therefore, for the two forces to cancel, the object would have to be square-root(330,000) or about 574 times farther from the Sun, than from the Earth - a distance of about 260,000 kilometers from Earth.

First of all, once you are in a spaceship, you won't feel gravitation, because you are in "free fall".

Anyway, to calculate this, consider that the Sun has 330,000 times the mass of Earth. Also consider the inverse-square relation. Therefore, for the two forces to cancel, the object would have to be square-root(330,000) or about 574 times farther from the Sun, than from the Earth - a distance of about 260,000 kilometers from Earth.

Answer 2

When the Earth/Moon distance is 238,000 miles, it's in the neighborhood of

214,100 miles from Earth, 23,900 miles from the moon.

The answer can't be rock-solid, because the Earth/Moon distance is so variable.

Answer 3

Here's the answer we worked out the last time the same question came up ...

22 hours before you posted it. If our answer was correct the first time, then

it's still good, as neither the solar mass nor the earth's mass has changed much

since then.

======================================

Mass of the sun = 1.98 x 1030 kg

Mass of the earth = 5.97 x 1024 kg

Ratio of sun's mass to earth's mass = 3.32 x 105

Ratio of the respective distances probably ought to be the square-root of the mass ratio = 575.9

Average distance between earth and sun = 1.496 x 1011 meters.

So there are the two simultaneous equations we have to crunch: There are

two numbers. Their ratio is 575.9 and their sum is 1.496 x 1011, and we want

to know what the smaller one is.

Hey, that's not difficult. If their ratio is 575.9, then the smaller one is (1/576.9)

of the total .

(1.496 x 1011) / (576.9) = 259.3 thousand kilometers.

Answer 4

First of all, once you are in a spaceship, you won't feel gravitation, because you are in "free fall".

Anyway, to calculate this, consider that the Sun has 330,000 times the mass of Earth. Also consider the inverse-square relation. Therefore, for the two forces to cancel, the object would have to be square-root(330,000) or about 574 times farther from the Sun, than from the Earth - a distance of about 260,000 kilometers from Earth.

Answer 5

This point is known as the L1 Lagrange Point, located about 56,000 km from the center of the Earth-Moon system in the direction opposite to the Moon. At this point, the gravitational forces from both the Earth and the Moon balance out, creating a stable equilibrium.