# How can astronauts float in space without being affected by the gravitational force of nearby objects?

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?

• Gravity is very, very, very weak. Commented Oct 28, 2015 at 14:36
• @gerrit Until you want to get something heavy off the ground using a rocket, or blow stuff up (warning: Worldbuilding SE link). Then gravity is a real pain.
– user
Commented Oct 28, 2015 at 16:03
• @MichaelKjörling But with a $10^{-1}$ kg magnet I can overcome (for an object small and near) of the $6 \cdot 10^{24}$ kg Earth... Commented Oct 28, 2015 at 16:07
• In fact they do attract each other. There is more gravity between two freight ships in harbour than between an astronaut and a shuttle in space. Mostly because the masses involved are much greater. Commented Oct 28, 2015 at 18:26
• If you think about it, nothing at all in your question depends on the astronaut and the heavy object being in space. You could equally well ask why an astronaut who is walking down the street isn't attracted to nearby heavy objects like buildings. The answer in space is exactly the same as the answer here on earth. Commented Oct 29, 2015 at 19:00

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.

• In space the acceleration due to Earth wouldn't be -9.8m/s^2. It would probably be close, though. Additionally, it is important to note centripetal force -- as the shuttle and astronaut orbit Earth, they are pulled inward by the gravitational force; however, their semi-perpendicular velocity allows them to continuously orbit. Lastly, gravity is a really weak force, so -- for the questioner -- objects that are smaller (such as humans and shuttles) don't posses much gravitational pull. Planets, on the other hand, posses quite a bit and suns even more. Commented Oct 28, 2015 at 15:03
• @dylnmc I take it we're assuming that the space station is in orbit around the Earth. The International Space Station is orbiting about 250 miles above the surface. The difference between the force of gravity on the ground and the force of gravity at 250 miles up is 1/3960^2 versus 1/(3960+250)^2. I calculate that would come to about 8.7 m/s^2. Actually more difference than I expected before I did the calculation.
– Jay
Commented Oct 28, 2015 at 16:25
• Jay and dylnmc, you both have a point, at 250 miles up acceleration due to gravity is less.
– GdD
Commented Oct 28, 2015 at 20:10
• I think it is a bit more than infinitesimal. Commented Oct 29, 2015 at 18:54

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.

• I love your comparison between lifting heavy objects on Earth and separating magnets! Commented Mar 24, 2020 at 0:40

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 :)

So, don't think gravity has no effect between small objects!

• "The reason is that gravity affects the ball's trajectory" -- well, that's one of the reasons. There's a boatload of other tiny influences with similar effects. Commented Oct 30, 2015 at 12:08
• @SteveJessop Of course, but we are talking about the effects of gravity here, and they are sufficient to mess up the trajectories. Commented Oct 30, 2015 at 12:22

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.

• This doesn't really answer the question, you are not making the relationship between mass and distance clear.
– GdD
Commented Oct 28, 2015 at 13:18
• @GdD Agreed inverse square of distance, but disregard that at close distances since the mass is so low regardless. Commented Oct 28, 2015 at 13:59
• Douglas Adams also once said (paraphrased, I think - and in the original BBC radio series): "The idea is that, if every atom of the universe is affected by every other atom of the universe, then it is theoretically possible to extrapolate a model of the entire universe using any single piece of matter as a starting point. The Vortex does this using a piece of fairy cake as its extrapolatory base." It is probably mostly gravity that is referred to here (given the distance). Commented Oct 29, 2015 at 18:40
• OK, a more direct quote (#18): "The Total Perspective Vortex derives its picture of the whole Universe on the principle of extrapolated matter analyses.To explain — since every piece of matter in the Universe is in some way affected by every other piece of matter in the Universe, it is in theory possible to extrapolate the whole of creation — every sun, every planet, their orbits, their composition and their economic and social history from, say, one small piece of fairy cake." Commented Oct 29, 2015 at 18:52

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%.

• It's not that easy. What you also didn't consider is that while you're attracted towards your common barycenter, the station's dimensions are much larger than those 50 m distance you started with. So as you approach much closer, while the vector direction doesn't change, its amplitude actually decreases, not increases. If that wasn't the case, then all the astronauts would constantly float towards the station's center of mass and wouldn't be able to escape it as the gravity acceleration approached infinity with ever decreasing distance to the center of mass. I.e gravity at center is 0, not ∞. Commented Oct 30, 2015 at 1:30
• @Lois you are calculating the gravity only for space station, but there will be some attraction with other planet (space.stackexchange.com/questions/12467/…) and sun's mass also make some attraction on the astronaut. Commented Oct 30, 2015 at 6:46
• @Ayyappan That's largely irrelevant when considering relative position and gravitational attraction of nearby objects, as per the question at the top of the page. An astronaut next to the station will be gravitationally attracted towards other distant celestial bodies at exactly the same amplitude and direction than the station. That doesn't change anything with respect to their relative movement between themselves. Neither does it remove their mutual gravitational attraction. There are perturbing effects that might act differently on the station and the astronaut, but not distant mass bodies. Commented Oct 30, 2015 at 13:08
• BTW there's no way to calculate this at certain intervals and average (smooth) it out. You'll have to integrate it (cumulative change is in relation with the surface area prescribed by the function), and also consider mass distribution. See update to my answer for an example of the latter. Commented Oct 30, 2015 at 14:29