Why do astronauts in space experience less gravitational force than they do on earth?

2 Answers


As #r#, the distance between the astronaut and the Earth increases in #F_g=G(m_1m_2)/r^2#, the gravitational force between the two masses. The weightlessness of an astronaut is actually a continual falling towards but never landing on Earth.


There are a couple of ways of answering this question: one is to answer your specific question about the force of gravity experienced by an astronaut, and the other is to talk about the weightlessness experienced in space.

Let's first talk gravity:

Gravity is a force that one mass exerts on another mass. The equation for that force is:


where #G# is the gravitational constant, the #2color(white)(0)m# terms are the #2# masses exerting the force on each other, and #r# is the distance between them.

The question being asked is about the experience of gravity by an astronaut in space. So let's work this out: as a person rises up into the sky, what in the equation is changing? #G# is the same, as are the mass of the person and the mass of the Earth. It's #r# that is increasing and it increases at an exponential rate.

So as the astronaut rises up above the Earth, the denominator in the equation gets bigger, resulting in less gravitational force the astronaut experiences from the Earth (and coincidentally the Earth experiences less gravity from the astronaut - but the Earth is so massive it hardly notices!)

Ok - so now let's talk about weightlessness:


We've seen on TV and youtube and movies how an astronaut in space, say for instance one living aboard the ISS (International Space Station), seemingly floats in midair. They can take a water droplet and play with the little spherical drop and can add more and more water to make the sphere bigger. So - how is this possible? Is it a result of experiencing a lesser amount of gravity? The answer is - partly.

Remember the story about Sir Issac Newton and the apple - how the apple hit him on the head and suddenly he had the framework for the theory of gravity? What's far more likely in that story is that after the apple hit him on the head, he wondered why the Moon didn't do the same thing - why it keeps moving around the Earth and not plummeting onto it.

The answer here is what @SCooke is getting to in his answer: the movement of someone (or something) in orbit is one part falling towards Earth and one part moving tangentially away from Earth, with the two vectors adding up to being a sort of continuous falling - you fall but never land. Everything in orbit, whether the astronaut or the water drop or the Moon, they are all falling towards the Earth but will never hit it. And it's that continual falling that is the experience of weightlessness.

May 21, 2016


“Weightlessness” is really the feeling caused by falling at constant speed around the earth.


While the answer provided by Parzival S. is also correct, physically and mathematically, you can see from the calculation that in practice the change in gravitational attraction due to mass and difference is virtually unnoticeable to a person, and it is not the explanation for “weight” (mass x gravitational acceleration) compared to “weightlessness” experienced in orbit.

However, in orbit around the earth an astronaut is actually falling towards the earth, but with a sufficient velocity to remain in orbit above it. The NET forces acting on the astronaut result in microgravity, or weightlessness.