According to the Newton's law of universal gravitation:
Any two bodies in the universe attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
In view of this, astronauts in a space station should be attracted by nearby objects. So they must be attracted towards heavy objects like machines etc.
How are they floating without the attraction of gravitational force of nearby objects?
Objects in orbit are attracted to each other, it's just their mass is small enough that the force of gravity between them is infinitesimal. Gravitational acceleration is dependent on mass and distance. In a scenario where a 150 kg astronaut is 10 m from a 80,000 kg Space Shuttle, the astronaut would be pulled toward the Shuttle at 5.336e-8 m/second squared. That's 0.00000005336 as opposed to the Earth's surface gravity of 9.8 m/second squared, and ~ 8.7 m/second squared at orbital altitude of 375 km. In other words, the Earth's force on the astronaut would be over 160 million times stronger, if it wasn't balanced by the exactly opposing centrifugal force due to their orbital speed of 7,682.5 m/s at that same orbital altitude.
Jupiter's influence on the astronaut (and the Shuttle and the Earth) when the Earth and Jupiter are close together is 3.2e-7 m/second squared, six times more than gravitational attraction of the Shuttle on the astronaut, but still very little.
Consider this equation for gravitational attraction between two bodies:
$$F = G \frac{m_1 m_2}{r^2}$$
where:
- $F$ is the force between the masses;
- $G$ is the gravitational constant (6.674×10−11 N · (m/kg)2);
- $m1$ is the first mass;
- $m2$ is the second mass;
- $r$ is the distance between the centers of the masses.
So if we say that an astronaut in an EVA suit has a mass of 150 kg, the International Space Station (ISS) itself has a mass of 390,000 kg, and the distance between their centers of mass is 5 meters, then their gravitational attraction between them is $1.042 × 10^{-6} \text{ m/s}^2$, or roughly one ten millionth (1/10,000,000) that of standard gravity at the surface of the Earth ($9.80665 \text{ m/s}^2$).
Now, neither the astronaut nor the station are point masses, so at such close proximity to each other, their mass distribution will play a major role and we have to account for it. Problem is, we don't really have exact mass distribution of the station, even if we neglected astronauts own non-uniformity as largely irrelevant due to small relative mass to the station. But, since we also don't have astronaut's angle to it, I'll just assume uniform mass for the 100 m long and 5 m in girth (r=2.5 m) station, and astronaut's position tangential to the station and orthogonal to its center of mass. I.e. the astronaut will be outside the station, somewhere near Node 1's outer truss;
In our case (with a few assumptions making this easier to calculate), combined gravity vector amplitude will change with cosine to the mean angle to the station's mass. That is, angle to the centroid of each 50 m sides. In our case, using the SOHCAHTOA mnemonic, that would be, 78.69°. So our acceleration to our uniform mass center of given dimensions and distribution would be $\text{cos}(78.69°) \cdot 1.042 × 10^{-6} \text{ m/s}^2$ or $2.04264874 × 10^{-7} \text{ m/s}^2$.
That is only $2.36349805 × 10^{-8}$ times (24 one billionths) the Earth's gravity at ISS mean orbital altitude (semi-major axis) of $\approx 8.64 \text{ m/s}^2$.
So not exactly nothing, but you'll grow a really long beard waiting for any noticeable change in your position relative to the station to happen due to your mutual gravitational attraction alone. Slight difference in your orbit relative to the station, combined with gravity gradient with respect to the station's center of mass will move you relative to it much faster than that, and we still call it a microgravity environment (i.e. difference measurable on a micro-g scale for its entire volume).
Also, for what is worth, distant gravitational perturbers like other celestial bodies will act on the station in exactly the same way as on our EVA astronaut, so their position relative to each other doesn't change because of that. There might be other, non-gravitational perturbing effects like solar wind, radiation pressure and exospheric drag, affecting station's movement slightly differently than our astronaut's, but that's not what the question asks.
Gravity is essentially a really weak force and it takes a whole lot of mass for its effects to be appreciable without doing extremely precise measurements (and growing a long beard). If you don't believe me, consider this: You can pick up relatively heavy objects off the Earth's surface while the whole planet is gravitationally pulling onto them. Conversely, take two small rare earth magnets, join them together by opposite magnetic poles, and you'll have really big problems separating them again.
Don't worry, they are attracted by other objects.
Only problem is, that attraction is ridiculously weak, so you won't notice it.
Most high school students, when they learn about the four forces, are a bit surprised that gravity is the weakest one – it seems so strong?
Well, consider this: it takes a whole planet's gravity for you to even notice it. A planet is a very big and heavy thing...not so much a space station or a piece of machinery.
Also, as someone mentioned, the magnetism of a tiny magnet can easily overcome the gravitational pull of the whole planet.
However, gravity is there, and it does exist even between small objects. And it does affect us in interesting if unexpected ways.
Take pool, for example. If you try to do a trick shot, gravity is there to mess it up for you. Well, you wouldn't notice normally, but with 4-5 or more collisions, you are hosed – there is no way for you to predict where a ball will go after that. Not even a pro can do that.
The reason is that gravity affects the ball's trajectory. Yes, even the tiny gravity of the people around the table is big enough to make it impossible to know where things will end up after more than five or so bounces :)
(Berry, M V, 1988, ‘The electron at the end of the universe’, p. 44)
So, don't think gravity has no effect between small objects!
Douglas Adams once said:
Space is big. Really big. You just won't believe how vastly, hugely, mind-bogglingly big it is. I mean, you may think it's a long way down the road to the chemist, but that's just peanuts to space.
Gravity works well with very massy things. With low mass things it has very very minor effects.
A US aircraft carrier, floating next to a battleship, experiences a pull of gravity between the two masses but even then it is so low as to be immaterial. (About the biggest mobile structures on earth by mass, maybe an oil tanker masses more). Scale that down to a person in a fairly low mass ISS and you will see that yes, they do experience a pull, but it is incredibly low.
I just realized the issue INSIDE the space station is terminal velocity.
On earth with a gravitational pull of ~9.8 m/s2 the terminal velocity of the human body is about 53 m/s or 195 kph.
The gravitational pull between objects on the station is at least a billion times less, so too is the terminal velocity (see the wikipedia article on "free fall"). The terminal velocity on the station is less than 53 nanometers/s or 18 mm/hour.
You might say they are gliding, or even flying if they flap something effectively.
Outside the station, no air, terminal velocity not an issue.
Gravity has a tendency to sneak up on you. Consider the 150 Kg EVA Suited astronaut hanging about the 390 ton space station about 50 meters from the station and lulled by the beautiful starscape falls asleep. So what happens ?
Relevant Equations .
Acceleration due to Gravity
ISS mass (350,000) * Gravitation constant (6.674 ) * 1/(100,000,000,000) / (distance * distance )
Distance Moved at Constant Acceleration
Acceleration due to Gravity * (time * time)/2 + starting velocity * time
slot time distance acceleration velocity
1 0 50 .000 000 0093 0
2 12 hours 42 .000 000 014 0.0004
3 18 hours 29 .000 000 027 0.0007
4 24 hours 10 .000 000 29 0.0013
So in less than 24 hours, the sleeper is dragged to within 10 meters of station and drifting into the station at 1.3 mm per second. Just by "weak" gravity. Less than another 1.5 hours to reach the station.
I apologize, it is late here, and I only calculated the acceleration at each slot, instead of continuous change. The actual results would significantly reduce the time to cross the distance by at least 25%.