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[asterroc]

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[best-ken-ever.livejournal.com]

I had the impression that moving a certain distance from Earth reduces gravitational forces, but I don't know how by how much.

Examples that might help me understand this better would probably be the answers to the following problems (not that I know how to solve them)

1a) Assume a spaceship leaves Earth and travels 100,000 miles away, then fires retrorockets in order to come to a complete stop position relative to Earth. What sort of pull effect will the Earth have on it at this time (ignoring the effects of the Moon)? What sort of gravity will any astronauts feel?

1b) Assume the above, except the spaceship travels 1,000,000 miles away before stopping. What sort of pull effect will the Earth have? What sort of gravity will any astronauts feel?

2) Re-assume the existence of Moon, and the spaceship flies n miles away from Earth, between Earth and Moon. At about what n does the spaceship get pulled equally by Earth and Moon such that it stays more or less in that spot? I think, to make this make sense to me, we must assume the spaceship also keeps itself in line with the fraction of Earth-orbiting speed it needs to maintain a line between the Moon, itself, and Earth. I feel the necessary velocity is more or less the fractional speed of Moon rotation based on how big of an orbit the spaceship is tracing. Will astronauts notice any gravitational effects? If so, how much?

I also am not sure how difficult these problems are to solve, but if anyone feels like solving them and improving my understanding of mechanics, that would be awesome (:

[zandperl.livejournal.com]

To start off with, the formula to determine the force of gravity is F=GMm/r^2. F is Force, G is the gravitational constant (it's just a number, don't worry about it), M is the mass of one object (here the Earth), m is the mass of the other object (if you're looking at the acceleration due to gravity this term gets dropped), and r is the distance between the two objects' centers (or the radius of the orbit). The answers to all of your questions are well known, and should be calculable by anyone taking a calc-based Physics I course, or by most algebra-based Physics I students.

1a) [FYI: The speed of the spaceship when it crosses the 100,000 mi line is not relevant.] The radius of the Earth is around 4,000 mi, so this distance is 25 times further, so the strength of gravity will be 1/25^2 what it would be on the surface of the Earth. It will not be directly noticeable however unless there is a solid surface under the astronauts' feet, for example if the ship's thrusters were turned on just strongly enough to keep it at the same constant 100,000mi from Earth.

Note that the ISS is located only around 20 miles above the Earth's surface, so that its orbital radius is around 4,020mi, so the strength of gravity on the ISS is only a tiny bit weaker than it would be on the surface of the Earth [(4000/4020)^2 times weaker].

1b) Same as 1a, but now gravity is 1/250^2 weaker.

2) There are two different points in space that you are conflating. The first is the L1 Lagrange point, which is the spot in space between two objects where another thing (a spaceship) can orbit the bigger thing (the Earth) at the same rate as the smaller thing (the Moon). Some numbers related to this: Earth's radius ~6,400km, Moon's radius ~1,700km, Earth-Moon distance ~380,000km, L1 point ~60,000km from Moon (~320,000km from Earth).

The second is sometimes called the turnover point or the neutral gravity point, which is the spot in space at which the net gravitational force from the Earth and Moon are balanced. It is called the turnover because when traveling between the two objects as the Apollo ships did, when you pass through that point you transition from working against Earth's gravity, to instead working against the Moon's gravity. This is located approximately ~340,000km from Earth.

[best-ken-ever.livejournal.com]

Thanks! This is really helpful (: